\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 162, pp. 1--50.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/162\hfil Family of quadratic differential systems]
{Family of quadratic differential systems with invariant hyperbolas:
a complete classification in the space $\mathbb{R}^{12}$}

\author[R. D. S. Oliveira, A. C. Rezende, N. Vulpe \hfil EJDE-2016/162\hfilneg]
{Regilene D. S. Oliveira, Alex C. Rezende, Nicolae Vulpe}

\address{Regilene D. S. Oliveira \newline
 Instituto de Ci\^encias Matem\'aticas e de Computa\c{c}\~{a}o,
 Universidade de S\~{a}o Paulo, Brazil}
\email{regilene@icmc.usp.br}

\address{Alex C. Rezende \newline
Instituto de Ci\^encias Matem\'aticas e de Computa\c{c}\~{a}o,
Universidade de S\~{a}o Paulo, Brazil}
\email{alexcrezende@gmail.com}

\address{Nicolae Vulpe \newline
 Institute of Mathematics and Computer Science,
 Academy of Sciences of Moldova, Moldova}
\email{nvulpe@gmail.com}

\thanks{Submitted February 5, 2015. Published June 27, 2016.}
\subjclass[2010]{34C23, 34A34}
\keywords{Quadratic differential systems; invariant hyperbola;
\hfill\break\indent affine invariant polynomials; group action}

\begin{abstract}
 In this article we consider the class $QS$ of all
 non-degenerate quadratic systems. A quadratic polynomial
 differential system can be identified with a single point of
 ${\mathbb{R}}^{12}$ through its coefficients. In this paper using
 the algebraic invariant theory we provided necessary and
 sufficient conditions for a system in $QS$ to have at least
 one invariant hyperbola in terms of its coefficients. We also
 considered the number and multiplicity of such hyperbolas. We
 give here the global bifurcation diagram of the class $QS$
 of systems with invariant hyperbolas. The bifurcation diagram is
 done in the 12-dimensional space of parameters and it is
 expressed in terms of polynomial invariants. The results can
 therefore be applied for any family of quadratic systems in this
 class, given in any normal form.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction and statement of main results}

In this article, we consider differential systems of the form
\begin{equation} \label{sys:PolPQ}
 \frac {dx}{dt}= P(x,y),\quad
 \frac {dy}{dt}= Q(x,y),
\end{equation}
where $P,Q\in \mathbb{R}[x,y]$, i.e. $P, Q$ are polynomials in $x, y$
over $\mathbb{R}$ and their associated vector fields
\begin{equation} \label{vectFild:X}
X=P(x,y)\frac{\partial}{\partial x} +
Q(x,y)\frac{\partial}{\partial y}.
\end{equation}
We call \emph{degree} of a system
\eqref{sys:PolPQ} the integer $m=\max(\deg P, \deg Q)$. In
particular we call \emph{quadratic} a differential system
\eqref{sys:PolPQ} with $m=2$. We denote here by $QS$ the whole
class of real \emph{non-degenerate quadratic systems}, i.e. we
assume that the polynomials $P$ and $Q$ are coprime.

Quadratic systems appear in the modeling of many natural
phenomena described in different branches of science, in
biological and physical applications and applications of these
systems became a subject of interest for the mathematicians. Many
papers have been published about quadratic systems, see for
example \cite{Sch2014} for a bibliographical survey.


Let $V$ be an open and dense subset of $\mathbb{R}^2$, we say that
a nonconstant differentiable function $H : V \to \mathbb{R}$ is a
first integral of a system \eqref{sys:PolPQ} on $V$ if $H(x(t), y(t))$ is
constant for all of the values of $t$ for which $(x(t), y(t))$ is
a solution of this system contained in $V$. Obviously $H$ is a
first integral of systems \eqref{sys:PolPQ} if and only if
\begin{equation}\label{il2}
X(H)=P \frac{\partial H}{\partial x} + Q \frac{\partial
H}{\partial y} = 0,
\end{equation}
for all $(x,y)\in V$. When a system \eqref{sys:PolPQ} has a first
integral we say that this system is integrable.

The knowledge of the first integrals is of particular interest in
planar differential systems because they allow us to draw their phase
portraits.

On the other hand given $f \in \mathbb{C}[x, y]$ we say that the
curve $f(x,y)=0$ is an \emph{invariant algebraic curve} of systems
\eqref{sys:PolPQ} if there exists $K \in \mathbb{C}[x, y]$ such
that \label{ref:page}
\begin{equation}\label{algebraicinv}
P \frac{\partial f}{\partial x} + Q \frac{\partial f}{\partial y}
= K f.
\end{equation}
The polynomial $K$ is called the \emph{cofactor} of the invariant
algebraic curve $f=0$. When $K=0$, $f$ is a polynomial first
integral.

Quadratic systems with an invariant algebraic curve have been
studied by many authors, for example Schlomiuk and Vulpe 
\cite{SchVul04-QTDS, SchVul08-RMJM} have studied quadratic systems
with invariant straight lines, Qin Yuan-xum \cite{Qin-Yuan} has
investigated the quadratic systems having an ellipse as limit
cycle, Druzhkova \cite{druz} has presented 
necessary and sufficient conditions for existence and uniqueness
of an invariant algebraic curve of second degree in terms of the
coefficients of quadratic systems, and Cairo and Llibre 
\cite{Cairo-Llibre} have studied the quadratic systems
having invariant algebraic conics in order to investigate the
Darboux integrability of such systems.

The motivation for studying the systems in the quadratic class is
not only because of their usefulness in many applications but also
for theoretical reasons, as discussed by Schlomiuk and Vulpe in
the introduction of \cite{SchVul04-QTDS}. The study of
non--degenerate quadratic systems could be done using normal forms
and applying the invariant theory.

The main goal of this paper is to investigate non--degenerate
quadratic systems having invariant hyperbolas and this study is
done applying the invariant theory. More precisely in this paper
we give necessary and sufficient conditions for a quadratic system
in $QS$ to have invariant hyperbolas. We also determine
the invariant criteria which provide the number and multiplicity
of such hyperbolas.

\begin{definition} \label{def:multipl} \rm
 We say that an invariant conic
$\Phi(x,y)=p+qx+ry+sx^2+2txy+uy^2=0$, $(s,t,u)\ne(0,0,0)$,
$(p,q,r,s,t,u)\in \mathbb{C}^6$ for a quadratic vector field $X$ has
\emph{multiplicity $m$} if there exists a sequence of real
quadratic vector fields $X_k$ converging to $X$, such that each
$X_k$ has $m$ distinct (complex) invariant conics
$\Phi^1_k=0,\ldots, \Phi^m_k=0$, converging to $\Phi=0$ as
$k\to\infty$ (with the topology of their coefficients), and this
does not occur for $m+1$. In the case when an invariant conic
$\Phi(x,y)=0$ has multiplicity one we call it \emph{simple}.
\end{definition}

Our main results are stated in the following theorem.


\begin{theorem} \label{mainthm} \quad

\noindent\textrm{(A)}
The conditions $\gamma_1=\gamma_2=0$
and either $\eta\ge0$, $M\ne0$ or $C_2=0$ are necessary for a
quadratic system in the class $QS$ to possess at least one
invariant hyperbola.

\noindent\textrm{(B)}
Assume that for a system in the class $QS$ the condition $\gamma_1=\gamma_2=0$
is satisfied.
\begin{enumerate}
\item  If $\eta>0$ then the necessary and sufficient
conditions for this system to possess at least one invariant
hyperbola are given in Figure \ref{diagr:eta-poz},
where we can also find the number and multiplicity of such
hyperbolas.

 \item In the case $\eta=0$ and either $M\ne0$ or $C_2=0$
the corresponding necessary and sufficient conditions for this
system to possess at least one invariant hyperbola are given
in Figure \ref{diagr:eta=0}, where we can also find the
number and multiplicity of such hyperbolas.
\end{enumerate}

\noindent\textrm{(C)}
Figures \ref{diagr:eta-poz} and
\ref{diagr:eta=0} actually contain the global bifurcation diagram
in the 12-dimensional space of parameters of the systems
belonging to family $QS$, which possess at least one
 invariant hyperbola. The corresponding conditions
are given in terms of invariant polynomials with respect to the
group of affine transformations and time rescaling.
\end{theorem}



\begin{remark} \label{rem:Hp} \rm
An invariant hyperbola is denoted by $\mathcal{H}$ if it is real
and by $\overset{c}{\mathcal{H}}$ if it is complex. In
the case we have two such hyperbolas then it is necessary to
distinguish whether they have parallel or non-parallel asymptotes
in which case we denote them by $\mathcal{H}^p$
($\overset{c}{\mathcal{H}^p}$) if their asymptotes are
parallel and by $\mathcal{H}$ if there exists at least one pair of
non-parallel asymptotes. We denote by $\mathcal{H}_k$ ($k=2,3$) a
hyperbola with multiplicity $k$; by $\mathcal{H}_2^p$ a double
hyperbola, which after perturbation splits into two
$\mathcal{H}^p$; and by $\mathcal{H}_3^p$ a triple hyperbola
which splits into two $\mathcal{H}^p$ and one $\mathcal{H}$.
\end{remark}

The term ``complex invariant hyperbolas'' of a real system requires
some explanation. Indeed the term hyperbola is reserved for a real
irreducible affine conic which has two real points at infinity.
This distinguishes it from the other two irreducible real conics:
the parabola with just one real point at infinity and the ellipse
which has two complex points at infinity. We call ``complex
hyperbola'' an irreducible affine conic $\Phi(x,y)=0$, with $
\Phi(x,y)=p+qx+ry+sx^2+2txy+uy^2=0$ over $\mathbb{C}$, such that there
does not exist a non-zero complex number $\lambda$ with
$\lambda(p,q,r,s,t,u)\in \mathbb{R}^6$ and in addition this conic has
two real points at infinity.

The invariants and comitants of differential equations (see
Subsection \ref{subs: main pol-inv}) used for proving our main
result are obtained following the theory of algebraic invariants
of polynomial differential systems, developed by Sibirsky and his
disciples (see for instance
\cite{Sib1,Vul86-Book,Popa,Baltag,Calin}).

\begin{figure}[ht]
\begin{center}
\includegraphics[width=1\textwidth]{fig1} % Diagram-Hyp-eta-poz-New.eps}}
\end{center}
\caption{Existence of invariant hyperbolas: the case
$\eta>0$} \label{diagr:eta-poz}
\end{figure}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=1\textwidth]{fig2}% Diagram-Hyp-eta-zero-New.eps}}
\end{center}
 \caption{Existence of invariant hyperbolas:
the case $\eta=0$} \label{diagr:eta=0}
\end{figure}

\section{Preliminaries} \label{sec:Invariant polyn.}

Consider real quadratic systems of the form:
\begin{equation} \label{sys:QSgen}
\begin{gathered}
 \frac {dx}{dt} =p_0+ p_1(x,y)+\,p_2(x,y)\equiv P(x,y), \\
 \frac {dy}{dt} =q_0+ q_1(x,y)+\,q_2(x,y)\equiv Q(x,y)
\end{gathered}
\end{equation}
with homogeneous polynomials $p_i$ and $q_i$
$(i=0,1,2)$ of degree $i$ in $x,y$:
\begin{gather*}
p_0=a_{00},\quad p_1(x,y)= a_{10}x+ a_{01}y,\quad
p_2(x,y)= a_{20}x^2 +2a_{11}xy + a_{02}y^2,\\
q_0=b_{00},\quad  q_1(x,y)= b_{10}x+ b_{01}y,\quad
q_2(x,y)=b_{20}x^2 +2 b_{11}xy + b_{02}y^2.
\end{gather*}
Such a system \eqref{sys:QSgen} can be identified with a point in
$\mathbb{R}^{12}$. Let
\[
 \tilde a=\big(a_{00},a_{10},a_{01},a_{20},a_{11},a_{02},b_{00},b_{10},b_{01},b_{20},
b_{11},b_{02}\big)
\]
 and consider the ring
$\mathbb{R}[a_{00},a_{10},\ldots,a_{02},b_{00},b_{10},\ldots,b_{02}, x,y]$
which we shall denote $\mathbb{R}[\tilde a,x,y]$.

\subsection{Group actions on quadratic systems \eqref{sys:QSgen}
and invariant polynomials with respect to these actions}

On the set QS of all quadratic differential systems
\eqref{sys:QSgen} acts the group $\operatorname{Aff}(2,\mathbb{R})$ of affine
transformations on the plane. Indeed for every
$ g\in \operatorname{Aff}(2,\mathbb{R})$,
$g: \mathbb{R}^{2}\to \mathbb{R}^{2}$ we have:
$$
 g: \begin{pmatrix}\tilde{x} \\ \tilde{y}\end{pmatrix}
 =  M\begin{pmatrix} x \\ y\end{pmatrix} +B;\quad
 g^{-1}:\begin{pmatrix} x\\  y\end{pmatrix} =
 M^{-1} \begin{pmatrix} \tilde{x}\\ \tilde{y} \end{pmatrix} -M^{-1}B.
$$
where $M=\| M_{ij} \| $ is a $2\times 2$ nonsingular matrix and
$B$ is a $2\times 1$ matrix over $\mathbb{R}$. For every
$ S\in QS$ we can form its transformed system $\tilde S=g S$:
\begin{equation}
 \frac{d\tilde{x}}{dt} =\tilde P(\tilde{x},\tilde{y}),\quad
 \frac{d\tilde{y}}{dt} =\tilde Q(\tilde{x},\tilde{y}),\label{etildeS}
\end{equation}
where
$$
 \begin{pmatrix} \tilde P(\tilde{x},\tilde{y})\\
\tilde Q(\tilde{x},\tilde{y})\end{pmatrix}
 = M\begin{pmatrix} (P\circ  {g^{-1}})(\tilde{x},\tilde{y})\\
 (Q\circ  {g^{-1}})(\tilde{x},\tilde{y}) \end{pmatrix}.
$$
The map $\operatorname{Aff}(2,\mathbb{R})\times QS \to QS$ defined by
\[
(g, S)\to \tilde S=g S
\]
satisfies the axioms for a left group action. For every subgroup
$G\subseteq \operatorname{Aff}(2,\mathbb{R})$ we have an induced action of $G$
on QS. We can identify the set QS of systems \eqref{sys:QSgen}
with a subset of $\mathbb{R}^{12}$ via the embedding $QS
\hookrightarrow\mathbb{R}^{12}$ which associates to each system
\eqref{sys:QSgen} the 12-tuple $(a_{00},\ldots,b_{02})$ of its
coefficients.

 On systems $(S)$ such that
$\max(\deg(p),\deg(q))\le2$ we consider the action of the group
$\operatorname{Aff}(2,\mathbb{R})$ which yields an action of this group on $\mathbb{R}^{12}$.
 For every $ g\in \operatorname{Aff}(2,\mathbb{R})$ let $r_g:\mathbb{R}^{12}\to \mathbb{R}^{12}$ be the map which
corresponds to $g$ via this action. We know (cf. \cite {Sib1}) that $r_g$
is linear and that the map $r:\operatorname{Aff}(2,\mathbb{R})\to GL(12,\mathbb{R})$ thus obtained is
a group homomorphism. For every
subgroup $G$ of $\operatorname{Aff}(2,\mathbb{R})$, $r$ induces a representation of $G$
onto a subgroup $\mathcal G$ of $GL(12,\mathbb{R})$.

We shall denote a polynomial $U$ in the ring $\mathbb{R}[\tilde a,x,y]$ by
$U(\tilde a,x,y)$.

\begin{definition} \label{def:comit} \rm
A polynomial $U(\tilde a,x,y)\in
\mathbb{R}[\tilde a,x,y]$ is a \emph{comitant} for systems
\eqref{sys:QSgen} with respect to a subgroup $G$ of $\operatorname{Aff}(2,\mathbb{R})$,
if there exists $\chi\in \mathbb{Z}$ such that for every $(g,\, \tilde
a)\in G\times\mathbb{R}^{12}\ $ and for every $(x,y)\in \mathbb{R}^2$ the
following relation holds:
$$
U(r_g(\tilde a) , g(x,y)\,)\equiv (\det g)^{-\chi} U(\tilde a,x,y).
$$
If the polynomial $U$ does not explicitly depend on $x$ and $y$
then it is an \emph{invariant}. The number $ \chi\in \mathbb{Z} $ is the
\emph{weight} of the comitant $ U(\tilde a,x,y)$.
If $G=GL(2,\mathbb{R})$ (or $G=\operatorname{Aff}(2,\mathbb{R})$) then the comitant
$U(\tilde a,x,y)$ of systems
\eqref{sys:QSgen} is called \emph{$GL$-comitant} (respectively, \emph{affine
comitant}).
\end{definition}


\begin{definition} \label{def_G} \rm
 A subset $X\subset \mathbb{R}^{12}$ will be called
\emph{$G$-invariant}, if  for every $ g\in G$\ we have
$r_g(X)\subseteq X$.
\end{definition}

Let $T(2,\mathbb{R})$ be the subgroup of $\operatorname{Aff}(2,\mathbb{R})$ formed by
translations. Consider the linear representation of $T(2,\mathbb{R})$ into
its corresponding subgroup ${\mathcal T}\subset GL(12,\mathbb{R})$, i.e.
for every $ \tau\in T(2,\mathbb{R})$, $\tau:\ x=\tilde{x}+\alpha,\
y=\tilde{y}+\beta$ we consider as above $r_\tau:
\mathbb{R}^{12}\to \mathbb{R}^{12}$.


\begin{definition} \label{def:T-com} \rm
A $GL$-comitant $U(\tilde a,x,y)$ of systems \eqref{sys:QSgen} is a
\emph{$T$-comitant} if for every
$(\tau, \tilde a )\in T(2,\mathbb{R})\times \mathbb{R}^{12}$ the relation
$U(r_\tau(\tilde a), \tilde{x},\tilde{y}) = U( \tilde a , \tilde{x},\tilde{y}) $
holds in $\mathbb{R}[\tilde{x},\tilde{y}]$.
\end{definition}

Consider $s$ homogeneous polynomials $U_i(\tilde a,x,y)\in
\mathbb{R}[\tilde a,x,y]$, $i=1,\ldots,s$:
$$
U_i(\tilde a,x,y)=\sum_{j=0}^{d_i} U_{ij}(\tilde a)x^{d_i-j}y^j,\quad i=1,\ldots,s,
$$
and assume that the polynomials $U_i$ are $GL$-comitants of a
system \eqref{sys:QSgen} where $d_i$ denotes the degree of the
binary form $U_i(\tilde a,x,y)$ in $x$ and $y$ with coefficients
in $\mathbb{R}[\tilde a]$. We denote by
$$
\mathcal{U}=\{U_{ij}(\tilde a)\in \mathbb{R}[\tilde a]:
i=1,\ldots,s,\, j=0,1,\ldots,d_i\},
$$
the set of the coefficients in $\mathbb{R}[\tilde a]$ of the
$GL$-comitants $U_i(\tilde a,x,y)$, $i=1,\ldots,s$, and by
$V(\mathcal U)$ its zero set:
$$
 V({\mathcal U})=\{\,\tilde a\in \mathbb{R}^{12}:
 U_{ij}(\tilde a)=0, \, \forall U_{ij}(\tilde a)\in \mathcal U\}.
$$

\begin{definition} \label{def:CT-com} \rm
 Let $U_1,\ldots, U_s$ be $GL$-comitants of a
system \eqref{sys:QSgen} .
 A $GL$-comitant $U(\tilde a,x,y)$ of this system
 is called a \emph{conditional $T$-comitant} (or
\emph{$CT$-comitant}) modulo the ideal generated by $U_{ij}(\tilde
a)$ $(i=1,\ldots,s; j=0,1,\ldots, d_i)$ in the ring $\mathbb{R}[\tilde a]$
if the following two conditions are satisfied:
\begin{itemize}
\item[(i)] the algebraic subset $V({\mathcal U})\subset \mathbb{R}^{12}$ is
affinely invariant (see Definition \ref{def_G});

\item[(ii)] for every ($\tau,\tilde a )\in T(2,\mathbb{R})\times V(\mathcal U)$
 we have
$ U(r_\tau(\tilde a), \tilde{x}, \tilde{y})=
 U(\tilde a, \tilde{x}, \tilde{y})$
 in  $\mathbb{R}[\tilde{x}, \tilde{y}]$.
\end{itemize}
\end{definition}

 In other words a $CT$-comitant $U(\tilde a ,x,y)$ is a $T$-comitant
on the algebraic subset \text{$V({\mathcal U})\subset\mathbb{R}^{12}$}.

\begin{definition} \label{def6} \rm
A homogeneous polynomial $U(\tilde a,x,y)\in \mathbb{R}[\tilde a ,x,y]$
of even degree in $x$, $y$ has \emph{well determined sign} on
$V\subset\mathbb{R}^{12}$ with respect to $x,y$ if for every
$\tilde a\in V$, the binary form $u(x,y)=U(\tilde a,x,y)$ yields a
function of constant sign on $\mathbb{R}^2$ except on a set of zero
measure where it vanishes.
\end{definition}

\begin{remark} \rm
We draw attention to the fact that if a
$CT$-comitant $U(\tilde a,x,y)$ of even weight is a binary form
of even degree in $x$ and $y$, of even degree in $\tilde a$ and
has well determined sign on some affine invariant algebraic subset
$V$, then its sign is conserved after an affine transformation and
time rescaling.
\end{remark}

\subsection{Main invariant polynomials associated with
invariant hyperbolas}\label{subs: main pol-inv}

We single out the following five polynomials, basic ingredients
in constructing invariant polynomials for systems
\eqref{sys:QSgen}:
\begin{equation} \label{expr:Ci,Di}
\begin{gathered}
 C_i(\tilde a,x,y)=yp_i(x,y)-xq_i(x,y),\ (i=0,1,2)\\
 D_i(\tilde a,x,y)=\frac{\partial p_i}{\partial x}+
 \frac{\partial q_i}{\partial y},\ (i=1,2).
\end{gathered}
\end{equation}
As it was shown in \cite{Sib1} these polynomials of degree one in
the coefficients of systems \eqref{sys:QSgen} are $GL$-comitants
of these systems.
 Let $f$, $g\in$ $\mathbb{R}[\tilde a,x,y]$ and
\[ %t \label{def:transv}
 (f,g)^{(k)}=
 \sum_{h=0}^k (-1)^h \begin{pmatrix}k\\  h \end{pmatrix}
 \frac{\partial^k f}{ \partial x^{k-h}\partial y^h}
 \frac{\partial^k g}{ \partial x^h\partial y^{k-h}}.
\]
The polynomial $(f,g)^{(k)}\in \mathbb{R}[\tilde a,x,y] $ is called
\emph{the transvectant of index $k$ of $(f,g)$} (cf.
\cite{GraYou41,Olver}).

\begin{theorem}[\cite{Vul86-Book}] \label{th:Vlp}
Any $GL$-comitant of systems \eqref{sys:QSgen} can be constructed
from the elements \eqref{expr:Ci,Di} by using the operations:
$+,\, -,\,\times$, and by applying the differential operation
$(*,* )^{(k)}$.
\end{theorem}

\begin{remark} \label{rmk3} \rm
We point out that the elements \eqref{expr:Ci,Di}
generate the whole set of $GL$-comitants and hence also the set
of affine comitants as well as the set of $T$-comitants.
\end{remark}

We construct the following $GL$-comitants of the second degree
with respect to the coefficients of the initial systems
\begin{equation} \label{expr:Ti}
\begin{gathered}
 T_1=(C_0,C_1)^{(1)},\quad
 T_2=(C_0,C_2)^{(1)},\quad
 T_3=(C_0,D_2)^{(1)},\\
 T_4=(C_1,C_1)^{(2)},\quad
 T_5=(C_1,C_2)^{(1)},\quad
 T_6=(C_1,C_2)^{(2)},\\
 T_7=(C_1,D_2)^{(1)},\quad
 T_8=(C_2,C_2)^{(2)},\quad
 T_9=(C_2,D_2)^{(1)}.
\end{gathered}
\end{equation}

Using these $GL$-comitants as well as the polynomials
\eqref{expr:Ci,Di} we construct the additional invariant
polynomials. In order to be able to calculate the values of the
needed invariant polynomials directly for every canonical system
we shall define here a family of $T$-comitants expressed
through $C_i$ $(i=0,1,2)$ and $D_j$ $(j=1,2)$:
\begin{gather*}
\hat A= (C_1,T_8-2T_9+D_2^2)^{(2)}/144,\\
\begin{aligned}
\widehat D&= \big[2C_0(T_8-8T_9-2D_2^2)+C_1(6T_7-T_6
 -(C_1,T_5)^{(1)}\\
&\quad +6D_1C_1D_2  -T_5)-9D_1^2C_2\big]/36,
\end{aligned} \\
\widehat E=[D_1(2T_9-T_8)-3(C_1,T_9)^{(1)}
 -D_2(3T_7+D_1D_2)]/72, \\
\begin{aligned}
\widehat F&=[6D_1^2(D_2^2-4T_9)+4D_1D_2(T_6+6T_7)
 + 48C_0(D_2,T_9)^{(1)} \\
&\quad -9D_2^2T_4 + 288 D_1\widehat E
 -24(C_2,\widehat D)^{(2)} + 120(D_2,\widehat  D)^{(1)}\\
 &\quad - 36C_1(D_2,T_7)^{(1)} + 8D_1(D_2,T_5)^{(1)}]/144,
\end{aligned}
\end{gather*}
\begin{align*}
\widehat B
&= \Big\{16D_1(D_2,T_8)^{(1)}(3C_1D_1-2C_0D_2+4T_2)\\
&\quad +32C_0(D_2, T_9)^{(1)}(3D_1D_2-5T_6+9T_7)\\
 &+2(D_2, T_9)^{(1)}(27C_1T_4-18C_1D_1^2
 . -32D_1T_2+32(C_0, T_5)^{(1)}\big)\\
 &+6(D_2, T_7)^{(1)}\Big[8C_0(T_8-12T_9)
 -12C_1(D_1D_2+T_7)+ D_1(26C_2D_1+32T_5)\\
 &\quad + C_2(9T_4+96T_3)\Big]
 +6(D_2, T_6)^{(1)}\Big[32C_0T_9-C_1(12T_7+52D_1D_2)\\
 &\quad -32C_2D_1^2\Big]
 +48D_2(D_2, T_1)^{(1)}(2D_2^2-T_8)\\
 &-32D_1T_8(D_2, T_2)^{(1)}+9D_2^2T_4(T_6-2T_7)
 -16D_1(C_2, T_8)^{(1)}(D_1^2+4T_3 )\\
 &+12D_1(C_1, T_8)^{(2)}(C_1D_2-2C_2D_1 )
 +6D_1D_2T_4(T_8-7D_2^2-42T_9)\\
 &+12D_1(C_1,T_8)^{(1)}(T_7+2D_1D_2)
 +96D_2^2[D_1(C_1, T_6)^{(1)}+D_2(C_0, T_6)^{(1)}]-\\
 &-16D_1D_2T_3(2D_2^2+3T_8)
 -4D_1^3D_2(D_2^2+3T_8+6T_9)
 +6D_1^2D_2^2(7T_6+2T_7)\\
 &-252D_1D_2T_4T_9\Big\} /(2^{8}3^3),
\end{align*}
\[
\widehat K= (T_8+4T_9+4D_2^2)/72, \quad
\widehat H= (8T_9-T_8+2D_2^2)/72.
\]
These polynomials in addition to \eqref{expr:Ci,Di} and
\eqref{expr:Ti} will serve as bricks in constructing affine
invariant polynomials for systems \eqref{sys:QSgen}.

The following 42 affine invariants $A_1,\ldots,A_{42}$ form the
minimal polynomial basis of affine invariants up to degree 12.
This fact was proved in \cite{Bul_Tim} by constructing
$A_1,\ldots,A_{42}$ using the above bricks.
\begin{gather*}
A_1= \hat A,\quad A_2= (C_2, \widehat D)^{(3)}/12,\quad
A_3= \big[ C_2, D_2)^{(1)}, D_2\big)^{(1)},D_2\big)^{(1)}/48, \\
A_4= (\widehat H, \widehat H)^{(2)},\quad
A_5= (\widehat H, \widehat K)^{(2)}/2, \quad
A_6= (\widehat E, \widehat H)^{(2)}/2, \\
A_7= \big[ C_2, \widehat E)^{(2)}, D_2\big)^{(1)}/8,\quad
A_8= \big[\widehat D, \widehat H)^{(2)},D_2\big)^{(1)}/48,\quad
A_9 = \big[\widehat D, D_2)^{(1)}, D_2\big)^{(1)}, \\
A_{10}= \big[\widehat D, \widehat K)^{(2)}, D_2\big)^{(1)}/8, \quad
A_{11}= (\widehat F, \widehat K)^{(2)}/4, \quad
A_{12}= (\widehat F, \widehat H)^{(2)}/4, \\
A_{13}= \big[C_2, \widehat H)^{(1)}, \widehat H\big)^{(2)},
D_2\big)^{(1)}/24, \quad
A_{14}= (\widehat B, C_2)^{(3)}/36,\quad
A_{15}= (\widehat E, \widehat F)^{(2)}/4,\\
A_{16}= \big[\widehat E, D_2)^{(1)}, C_2\big)^{(1)}, \widehat K\big)^{(2)}/16,\quad
A_{17}= \big[\widehat D,\widehat D)^{(2)},D_2\big)^{(1)},D_2\big)^{(1)}/64, \\
A_{18}= \big[\widehat D,\widehat F)^{(2)}, D_2\big)^{(1)}/16, \quad
A_{19}= \big[\widehat D,\widehat D)^{(2)},\widehat H\big)^{(2)}/16,\\
A_{20}= \big[C_2,\widehat D)^{(2)}, \widehat F\big)^{(2)}/16, \quad
A_{21}= \big[\widehat D, \widehat D)^{(2)}, \widehat K\big)^{(2)}/16,\\
A_{22}= \frac{1}{1152}\big[C_2, \widehat D)^{(1)}, D_2\big)^{(1)},
  D_2\big)^{(1)}, D_2\big)^{(1)} D_2\big)^{(1)},\quad
A_{23}= \big[\widehat F, \widehat H)^{(1)}, \widehat K\big)^{(2)}/8,\\
A_{24}= \big[C_2, \widehat D)^{(2)}, \widehat K\big)^{(1)}, \widehat H\big)^{(2)}/32,\quad
A_{25}= \big[\widehat D, \widehat D)^{(2)}, \widehat E\big)^{(2)}/16,\\
A_{26}= (\widehat B, \widehat D)^{(3)}/36,\quad
A_{27}= \big[\widehat B, D_2)^{(1)}, \widehat H\big)^{(2)}/24,\\
A_{28}= \big[C_2,\widehat K)^{(2)},\widehat D\big)^{(1)},\widehat E\big)^{(2)}/16,\quad
A_{29}= \big[\widehat D, \widehat F)^{(1)}, \widehat D\big)^{(3)}/96,\\
A_{30}= \big[C_2,\widehat D)^{(2)},\widehat D\big)^{(1)},\widehat D\big)^{(3)}/288,\quad
A_{31}= \big[\widehat D,\widehat D)^{(2)},\widehat K\big)^{(1)},\widehat H\big)^{(2)}/64,\\
A_{32}= \big[\widehat D, \widehat D)^{(2)}, D_2\big)^{(1)}, \widehat H\big)^{(1)},
 D_2\big)^{(1)}/64,\\
A_{33}= \big[\widehat D, D_2)^{(1)}, \widehat F\big)^{(1)}, D_2\big)^{(1)}, D_2\big)^{(1)}/128,\\
A_{34}= \big[\widehat D, \widehat D)^{(2)}, D_2\big)^{(1)}, \widehat K\big)^{(1)},
D_2\big)^{(1)}/64,\\
A_{35}= \big[\widehat D, \widehat D)^{(2)}, \widehat E\big)^{(1)}, D_2\big)^{(1)}, D_2\big)^{(1)}/128,\\
A_{36}= \big[\widehat D,\widehat E)^{(2)},\widehat D\big)^{(1)},
 \widehat H\big)^{(2)}/16,\quad
A_{37}= \big[\widehat D,\widehat D)^{(2)},\widehat D\big)^{(1)},\widehat D\big)^{(3)}/576,\\
A_{38}= \big[C_2,\widehat D)^{(2)}, \widehat D\big)^{(2)}, \widehat D\big)^{(1)},
  \widehat H\big)^{(2)}/64,\quad
A_{39}= \big[\widehat D,\widehat D)^{(2)},\widehat F\big)^{(1)},\widehat H\big)^{(2)}/64,\\
A_{40}= \big[\widehat D,\widehat D)^{(2)},\widehat F\big)^{(1)},
 \widehat K\big)^{(2)}/64,\quad
A_{41}= \big[C_2,\widehat D)^{(2)}, \widehat D\big)^{(2)}, \widehat F\big)^{(1)}, D_2\big)^{(1)}/64,\\
A_{42}= \big[\widehat D,\widehat F)^{(2)},\widehat F\big)^{(1)},D_2\big)^{(1)}/16.
\end{gather*}

In the above list, the bracket ``$[$'' is used in order to avoid
placing the otherwise necessary up to five parentheses ``$($''.

Using the elements of the minimal polynomial basis given above we
construct the affine invariant polynomials
\begin{gather*}
\gamma_1(\tilde a)=  A_1^2 (3A_6 + 2 A_7) - 2 A_6( A_8 + A_{12}),\\
\begin{aligned}
\gamma_2(\tilde a)
&=9 A_1^2 A_2 (23252 A_3 + 23689 A_4) -
 1440 A_2 A_5 (3 A_{10} + 13 A_{11}) \\
&\quad - 1280 A_{13} (2 A_{17} + A_{18} + 23 A_{19} - 4 A_{20})
 - 320 A_{24} (50 A_8 + 3 A_{10} \\
&\quad+ 45 A_{11} - 18 A_{12}) + 120 A_1 A_6 (6718 A_8
+ 4033 A_9 + 3542 A_{11}\\
&\quad  + 2786 A_{12}) + 30 A_1 A_{15}(14980 A_3 - 2029 A_4 - 48266 A_5)\\
&\quad -  30 A_1 A_7 (76626 A_1^2 - 15173 A_8 + 11797 A_{10}
 + 16427 A_{11} - 30153 A_{12})\\
&\quad  + 8 A_2 A_7 (75515 A_6 - 32954 A_7) +
 2 A_2 A_3 (33057 A_8 - 98759 A_{12}) \\
&\quad - 60480 A_1^2 A_{24}
 +  A_2 A_4 (68605 A_8 - 131816 A_9 + 131073 A_{10} + 129953 A_{11})\\
&\quad - 2 A_2 (141267 A_6^2
 - 208741 A_5 A_{12} + 3200 A_2 A_{13}),
\end{aligned}
\\
\begin{aligned}
\gamma_3(\tilde a)
&=843696 A_5 A_6 A_{10} +
 A_1 (-27 (689078 A_8 + 419172 A_9 - 2907149 A_{10} \\
&\quad- 2621619 A_{11}) A_{13}  -  26 (21057 A_3 A_{23} + 49005 A_4 A_{23}
- 166774 A_3 A_{24} \\
&\quad + 115641 A_4 A_{24})).
\end{aligned}
\\
\begin{aligned}
\gamma_4(\tilde a)&=-9 A_4^2 (14 A_{17} + A_{21}) +
 A_5^2 (-560 A_{17} - 518 A_{18} + 881 A_{19} - 28 A_{20} \\
&\quad + 509 A_{21})
 -  A_4 (171 A_8^2 + 3 A_8 (367 A_9 - 107 A_{10}) +
 4 (99 A_9^2 + 93 A_9 A_{11}\\
&\quad + A_5 (-63 A_{18} - 69 A_{19}
 + 7 A_{20} + 24 A_{21}))) + 72 A_{23} A_{24},
\end{aligned}
\\
\begin{aligned}
\gamma_5(\tilde a)
&=-488 A_2^3 A_4 +
 A_2 (12 (4468 A_8^2 + 32 A_9^2 - 915 A_{10}^2 + 320 A_9 A_{11} -
 3898 A_{10} A_{11} \\
&\quad - 3331 A_{11}^2  +  2 A_8 (78 A_9 + 199 A_{10} + 2433 A_{11})) +
 2 A_5 (25488 A_{18} \\
&\quad - 60259 A_{19} - 16824 A_{21})
 + 779 A_4 A_{21}) + 4 (7380 A_{10} A_{31}\\
&\quad - 24 (A_{10} + 41 A_{11}) A_{33} +
 A_8 (33453 A_{31} + 19588 A_{32} - 468 A_{33} - 19120 A_{34})\\
&\quad + 96 A_9 (-A_{33} + A_{34}) + 556 A_4 A_{41} -
 A_5 (27773 A_{38} + 41538 A_{39}\\
&\quad - 2304 A_{41} + 5544 A_{42})),
\end{aligned}\\
\gamma_6(\tilde a)=2 A_{20} - 33 A_{21},
\\
\begin{aligned}
\gamma_7(\tilde a)
&=A_1 (64 A_3 - 541 A_4) A_7 + 86 A_8 A_{13} + 128 A_9 A_{13}
 - 54 A_{10} A_{13} \\
&\quad - 128 A_3 A_{22} + 256 A_5 A_{22} + 101 A_3 A_{24} - 27 A_4 A_{24},
\end{aligned}
\\
\begin{aligned}
\gamma_8(\tilde a)
&= 3063 A_4 A_9^2 - 42 A_7^2 (304 A_8 + 43 (A_9 - 11 A_{10}))
 - 6 A_3 A_9 (159 A_8 \\
&\quad + 28 A_9 + 409 A_{10})
+ 2100 A_2 A_9 A_{13} + 3150 A_2 A_7 A_{16} \\
&\quad + 24 A_3^2 (34 A_{19} - 11 A_{20})
 + 840 A_5^2 A_{21} - 932 A_2 A_3 A_{22}
 + 525 A_2 A_4 A_{22} \\
&\quad + 844 A_{22}^2 - 630 A_{13} A_{33},
\end{aligned}
\\
\gamma_9(\tilde a)=2 A_8 - 6 A_9 + A_{10},\quad
\gamma_{10}(\tilde a)=3 A_8 + A_{11},\\
\gamma_{11}(\tilde a)=-5 A_7 A_8 + A_7 A_9 + 10 A_3 A_{14},\quad
\gamma_{12}(\tilde a)=25 A_2^2 A_3 + 18 A_{12}^2,\\
\gamma_{13}(\tilde a)=A_2,\quad
\gamma_{14}(\tilde a)=A_2 A_4 + 18 A_2 A_5 - 236 A_{23} + 188 A_{24},\\
\begin{aligned}
\gamma_{15}(\tilde a,x,y)
&= 144 T_1 T_7^2 - T_1^3 (T_{12} + 2
T_{13}) - 4 (T_9 T_{11} + 4 T_7 T_{15} + 50 T_3 T_{23} \\
&\quad + 2 T_4 T_{23}  + 2 T_3 T_{24} + 4 T_4 T_{24}),
\end{aligned}\\
\gamma_{16}(\tilde a,x,y)=T_{15}, \quad
\gamma_{17}(\tilde a,x,y)= -(T_{11}+12\,T_{13}),\\
\tilde\gamma_{18}(\tilde a,x,y)= C_1 (C_2,C_2)^{(2)} - 2C_2(C_1,C_2)^{(2)},\\
\tilde\gamma_{19}(\tilde a,x,y)= D_1 (C_1,C_2)^{(2)}
- ((C_2,C_2)^{(2)},C_0)^{(1)},
\end{gather*}
\begin{gather*}
\delta_1(\tilde a)= 9 A_8 + 31 A_9 + 6 A_{10},
\delta_2(\tilde a)= 41 A_8 + 44 A_9 + 32 A_{10},\\
\delta_3(\tilde a)= 3 A_{19} - 4 A_{17},
\delta_4(\tilde a)= -5 A_2 A_3 + 3 A_2 A_4 + A_{22},\\
\delta_5(\tilde a)= 62 A_8 + 102 A_9 - 125 A_{10},
\delta_6(\tilde a)= 2 T_3 + 3 T_4,\\
\beta_1(\tilde a)= 3 A_1^2 - 2 A_8 - 2 A_{12},
\beta_2(\tilde a)= 2 A_7 - 9 A_6,\\
\beta_3(\tilde a)= A_6,
\beta_4(\tilde a)= -5 A_4 + 8 A_5,\\
\beta_5(\tilde a)= A_4,
\beta_6(\tilde a)= A_1,\\
\beta_7(\tilde a)= 8 A_3 - 3 A_4 - 4 A_5,
\beta_8(\tilde a)= 24 A_3 + 11 A_4 + 20 A_5,\\
\beta_9(\tilde a)= -8 A_3 + 11 A_4 + 4 A_5,
\beta_{10}(\tilde a)= 8 A_3 + 27 A_4 - 54 A_5,\\
\beta_{11}(\tilde a,x,y)= T_1^2 - 20 T_3 - 8 T_4,
\beta_{12}(\tilde a,x,y)= T_1,\\
\beta_{13}(\tilde a,x,y)= T_3,
\end{gather*}
\begin{gather*}
\begin{aligned}
\mathcal{R}_1(\tilde a)
&= -2 A_7 (12 A_1^2 + A_8 + A_{12}) + 5 A_6 (A_{10}
 + A_{11}) - 2 A_1 (A_{23} - A_{24}) \\
&\quad + 2 A_5 (A_{14} + A_{15}) + A_6 (9 A_8 + 7 A_{12}),
\end{aligned}\\
\mathcal{R}_2(\tilde a)= A_8 + A_9 - 2 A_{10},\quad
\mathcal{R}_3(\tilde a)= A_9,\\
\mathcal{R}_4(\tilde a)= -3 A_1^2 A_{11} + 4 A_4 A_{19},\\
\begin{aligned}
\mathcal{R}_5(\tilde a,x,y)
&= (2 C_0 (T_8 - 8 T_9 - 2 D_2^2) +
 C_1 (6 T_7 - T_6) - (C_1,T_5)^{(1)} \\
&\quad + 6 D_1 (C_1 D_2 - T_5) - 9 D_1^2 C_2),
\end{aligned}\\
\begin{aligned}
\mathcal{R}_6(\tilde a)
&= -213 A_2 A_6 + A_1 (2057 A_8 - 1264 A_9 + 677 A_{10}
 + 1107 A_{12}) \\
&\quad + 746 (A_{27} - A_{28}),
\end{aligned}\\
\mathcal{R}_7(\tilde a)= -6 A_7^2 - A_4 A_8 + 2 A_3 A_9 - 5 A_4 A_9 + 4 A_4 A_{10}
 - 2 A_2 A_{13},\\
\mathcal{R}_8(\tilde a)= A_{10},\quad
\mathcal{R}_9(\tilde a)= -5 A_8 + 3 A_9,\\
\mathcal{R}_{10}(\tilde a)= 7 A_8 + 5 A_{10} + 11 A_{11},\quad
\mathcal{R}_{11}(\tilde a,x,y)= T_{16}.
\end{gather*}
\begin{gather*}
H_{12}(\tilde a,x,y)= (\widehat D,\widehat D)^{(2)},\\
N_7(\tilde a)= 12 D_1(C_0,D_2)^{(1)}+2D_1^3+9D_1(C_1,C_2)^{(2)}+
 36\big[C_0,C_1)^{(1)},D_2)^{(1)}.
\end{gather*}

We remark the the last two invariant polynomials $H_{12}(\tilde
a,x,y)$ and $N_7(\tilde a)$ are constructed in \cite{SchVul08-JDDE}.


 \subsection{Preliminary results involving  polynomial
invariants}\label{subs: Prelim-results}

Considering the $GL$-comitant $C_2(\tilde a,x,y)=yp_2(\tilde
a,x,y)-xq_2(\tilde a,x,y)$ as a cubic binary form of $x$ and $y$
we calculate
$$
\eta(\tilde a)=\operatorname{Discrim}[C_2,\xi],\quad
M(\tilde a,x,y)=\operatorname{Hessian}[C_2],
$$
where $\xi=y/x$ or $\xi=x/y$.
According to \cite{Dana-Vlp-JFPT-2010} we have the next result.


\begin{lemma}[\cite{Dana-Vlp-JFPT-2010}] \label{lem:S1-S5}
The number of infinite singularities (real and imaginary)
of a quadratic system in QS is determined by the following
conditions:
\begin{itemize}
\item[(i)] $3$ real if $\eta>0$;

\item[(ii)] $1$ real and $2$ imaginary if $\eta<0$;

\item[(iii)] $2$ real if $\eta=0$ and $M\ne0$;

\item[(iv)] $1$ real if $\eta= M=0$ and $C_2\ne0$;

\item[(v)] $\infty$ if $\eta= M= C_2=0$.
\end{itemize}
Moreover, for each one of these cases the quadratic systems
\eqref{sys:QSgen} can be brought via a linear transformation to
one of the following canonical systems:
\begin{align*}
(\mathbf{S}_I)&\quad \begin{cases}
 \dot x= a+cx+dy+gx^2+(h-1)xy,\\
 \dot y=  b+ex+fy+(g-1)xy+hy^2;
\end{cases} \\
(\mathbf{S}_{II})&\quad \begin{cases}
 \dot x= a+cx+dy+gx^2+(h+1)xy,\\
 \dot y=  b+ex+fy-x^2+gxy+hy^2;
\end{cases} \\
(\mathbf{S}_{III})&\quad \begin{cases}
 \dot x= a+cx+dy+gx^2+hxy,\\
 \dot y= b+ex+fy+(g-1)xy+hy^2;
\end{cases} \\
(\mathbf{S}_{IV})&\quad \begin{cases}
 \dot x= a+cx+dy+gx^2+hxy,\\
 \dot y= b+ex+fy-x^2+gxy+hy^2;
\end{cases} \\
(\mathbf{S}_{V})&\quad \begin{cases}
 \dot x= a+cx+dy+x^2,\\
 \dot y=  b+ex+fy+xy.
\end{cases}
\end{align*}
\end{lemma}

\begin{lemma} \label{lem:H,E-irred-gamma1,2=0}
If a quadratic system \eqref{sys:QSgenCoef} possesses a
non-parabolic irreducible conic then the conditions
$\gamma_1=\gamma_2=0$ hold.
\end{lemma}

\begin{proof}
 According to \cite{Chr1989} a system \eqref{sys:QSgenCoef}
possessing a second order non-parabolic irreducible curve as an
algebraic particular integral can be written in the form
$$
 \dot x=a\Phi(x,y) + \Phi'_y (gx +hy +k),\quad
 \dot y=b\Phi(x,y) - \Phi'_x (gx +hy +k),
$$
where $a,b,g,h,k$ are real parameters and $\Phi(x,y)$ is the conic
\begin{equation} \label{con:Phi(x,y)}
 \Phi(x,y)\equiv p + q x + r y + s x^2 + 2 t x y + u y^2=0.
\end{equation}
 A straightforward calculation gives $\gamma_1=\gamma_2=0$ for
the above systems and this completes the proof.
\end{proof}

Assume that a conic \eqref{con:Phi(x,y)} is an affine algebraic
invariant curve for quadratic systems
\eqref{sys:QSgen}, which we rewrite in the form:
\begin{equation} \label{sys:QSgenCoef}
\begin{gathered}
 \frac {dx}{dt}= a+cx+dy+gx^2+2hxy+ky^2\equiv P(x,y),\\
 \frac {dy}{dt}= b+ex+fy+lx^2+2mxy+ny^2\equiv Q(x,y).
\end{gathered}
\end{equation}

\begin{remark} \label{rem:Delta-ne0=>irred} \rm
Following \cite{Lawrence1972} we construct the determinant
$$
\Delta=\begin{vmatrix} s & t &q/2\\ t & u & r/2\\ q/2 & r/2 &
p\end{vmatrix} ,
$$
associated to the conic \eqref{con:Phi(x,y)}. By
\cite{Lawrence1972} this conic is irreducible (i.e. the polynomial $\Phi$
defining the conic is irreducible over $\mathbb{C}$) if and only if $\Delta\ne0$.
\end{remark}

To detect if an invariant conic \eqref{con:Phi(x,y)} of a
system \eqref{sys:QSgenCoef} has the multiplicity greater than
one, we shall use the notion of \emph{$k$-th extactic curve
$\mathscr{E}_k(X)$} of the vector field $X$ (see
\eqref{vectFild:X}), associated to systems \eqref{sys:QSgenCoef}.
This curve is defined in the paper \cite[Definition5.1]{ChrLliPer2007} as follows:
$$
\mathscr{E}_k(X)=\det\begin{pmatrix}
 v_1 & v_2 & \ldots & v_l\\
 X(v_1) & X(v_2) & \ldots & X(v_l)\\
 \vdots  &  & & \vdots \\
 X^{l-1}(v_1) & X^{l-1}(v_2) & \ldots & X^{l-1}(v_l)
 \end{pmatrix},
$$
where $v_1,v_2,\ldots,v_l$ is the basis of $\mathbb{C}_n[x,y]$, the
$\mathbb{C}$-vector space of polynomials in $\mathbb{C}_n[x,y]$ and
$l=(k+1)(k+2)/2$. Here $ X^0(v_i)=v_i$ and
$X^{j}(v_1)=X(X^{j-1}(v_1))$.

Considering the Definition \ref{def:multipl} of a multiplicity of
an invariant curve, according to \cite{ChrLliPer2007} the
following statement holds:

\begin{lemma} \label{lem:Ek}
If an invariant curve $\Phi(x,y)=0$ of degree $k$ has multiplicity
$m$, then $\Phi(x,y)^m$ divides $\mathscr{E}_k(X)$.
\end{lemma}

We shall apply this lemma in order to detect additional
conditions for a conic to be multiple.
According to definition of an invariant curve (see page
\pageref{ref:page}) considering the cofactor $K= Ux+Vy+W
\in\mathbb{C}[x,y]$ the following identity holds:
$$
 \frac{\partial \Phi}{\partial x}P(x,y)+
 \frac{\partial \Phi}{\partial y}Q(x,y)=\Phi(x,y)(Ux+Vy+W).
$$
This identity yields a system of~10 equations for determining the
9 unknown parameters $p$, $q$, $r$, $s$, $t$, $u$, $U$, $V$, $W$:
\begin{equation} \label{Eqs:gen}
\begin{aligned}
 Eq_1\ \equiv\ & s(2g-U) +2lt=0, \\
 Eq_2\ \equiv\ & 2t(g+2m-U)+s(4h-V)+2lu=0,\\
 Eq_3\ \equiv\ & 2t(2h+n-V)+u(4m-U)+2ks=0, \\
 Eq_4\ \equiv\ & u(2n-V)+2kt=0, \\
 Eq_5\ \equiv\ & q(g-U)+s(2c-W)+2et+lr=0, \\
 Eq_6\ \equiv\ & r (2m - U) + q (2 h - V)+2 t (c + f - W)+2(ds+eu)=0, \\
 Eq_7\ \equiv\ & r(n-V)+u(2f-W)+2dt+kq=0, \\
 Eq_8\ \equiv\ & q(c-W) +2(as+bt)+er-pU=0, \\
 Eq_9\ \equiv\ & r(f-W) +2(bu+at)+dq-pV=0, \\
 Eq_{10}\ \equiv\ & aq+br-pW=0. \\
 \end{aligned}
\end{equation}

\section{Proof of the main theorem}\label{sec:the proof}

Assuming that a quadratic system \eqref{sys:QSgenCoef} in $QS$ has
an invariant hyperbola \eqref{con:Phi(x,y)}, we conclude
that this system must possess at least two real distinct infinite
singularities. So according to Lemmas \ref{lem:S1-S5} and
\ref{lem:H,E-irred-gamma1,2=0} the conditions
$\gamma_1=\gamma_2=0$ and either $\eta\ge0$ and $M\ne0$ or
$C_2=0$ have to be fulfilled.

In what follows, supposing that the conditions
$\gamma_1=\gamma_2=0$ hold, we shall examine three families of
quadratic systems
\eqref{sys:QSgenCoef}: systems with three real distinct infinite
singularities (corresponding to the condition $\eta>0$); systems
with two real distinct infinite singularities (corresponding to
the conditions $\eta=0$ and $M\ne0$) and systems with infinite
number of singularities at infinity, i.e. with degenerate infinity
 (corresponding to the condition $C_2=0$).


\subsection{Systems with three real infinite singularities and $\theta\ne0$}
In this case according to Lemma \ref{lem:S1-S5} systems \eqref{sys:QSgenCoef}
via a linear transformation could be brought to the following
family of systems
\begin{equation} \label{sys:eta-poz}
\begin{gathered}
 \frac {dx}{dt}= a+cx+dy+gx^2+(h-1)xy,\\
 \frac {dy}{dt}= b+ex+fy+(g-1)xy+hy^2.
\end{gathered}
\end{equation}
For this systems we calculate
\begin{equation} \label{val:C2,theta}
C_2(x,y)=xy(x-y),\quad \theta=-(g-1)(h-1)(g+h)/2
\end{equation}
and we shall prove the next lemma.

\begin{lemma} \label{lem:normal form-gamma1=0}
Assume that for a system \eqref{sys:eta-poz} the conditions
$\theta\ne0$ and $\gamma_1=0$ hold. Then this system via an affine
transformation could be brought to the form
\begin{equation} \label{sys:eta-poz-theta-ne0-gamma1=0}
\frac {dx}{dt}= a+cx+gx^2+(h-1)xy,\quad
 \frac {dy}{dt}= b-cy+(g-1)xy+hy^2.
\end{equation}
\end{lemma}

\begin{proof}
Since $\theta\ne0$ the condition $(g-1)(h-1)(g+h)\ne0$
holds and by  a translation we may assume $d=e=0$ for systems
\eqref{sys:eta-poz}. Then we calculate
$$
\gamma_1=\frac{1}{64}(g-1)^2(h-1)^2\mathcal{D}_1
\mathcal{D}_2\mathcal{D}_3,
$$
where
\begin{gather*}
\mathcal{D}_1= c+f, \quad
 \mathcal{D}_2= c (g + 4 h-1)+f (1 + g - 2 h),\\
\mathcal{D}_3= c (1 - 2 g + h) + f ( 4 g + h-1).
\end{gather*}
Since $\theta\ne0$ (i.e. $(g-1)(h-1)\ne0$) the condition
$\gamma_1=0$ is equivalent to $\mathcal{D}_1
\mathcal{D}_2\mathcal{D}_3=0$. We claim that without loss of
generality we may assume $\mathcal{D}_1=c+f=0$, as other cases
could be brought to this one via an affine transformation.

Indeed, assume first $\mathcal{D}_1\ne0$ and
$\mathcal{D}_2=0$. Then as $g+h\ne0$ (due to $\theta\ne0$) we
apply to systems
\eqref{sys:eta-poz} with $d=e=0$ the affine transformation
\begin{equation} \label{transf:1}
x'=y-x-(c-f)/(g+h),\quad y'=-x
\end{equation}
and we obtain the systems
\begin{equation} \label{trabsf-system-1}
 \frac {dx'}{dt}= a'+c'x'+g'x'^2+(h'-1)x'y',\quad
 \frac {dy'}{dt}= b'+f'y'+(g'-1)x'y'+h'y'^2.
\end{equation}
These systems have the following new parameters:
\begin{equation} \label{transf:coef-1}
\begin{gathered}
a'=\big[ c^2 h-f^2 g + c f (g - h) - (a - b) (g + h)^2\big]/(g + h)^2,\\
b'=-a,\quad c'= (c g - 2 f g - c h) /(g + h),\\
f'=( c - f - c g + 2 f g + f h) /(g + h), \quad g'=h, \quad
h'=1-g-h.
\end{gathered}
\end{equation}
A straightforward computation gives
 $$
 \mathcal{D}_1'=c'+f'=\big[c (g + 4 h-1)+f (1 + g - 2
h)\big]/(g+h)=\mathcal{D}_2/(g+h)=0
$$
and hence, the condition $\mathcal{D}_2=0$ is replaced with
$\mathcal{D}_1=0$ via an affine transformation.

Suppose now $\mathcal{D}_1\ne0$ and $\mathcal{D}_3=0$.
Then we apply to systems \eqref{sys:eta-poz} the affine
transformation
$$
x''=-y,\quad y''= x -y+(c-f)/(g+h)
$$
and we obtain the systems
$$
 \frac {dx''}{dt}= a''+c''x''+g''x''^2+(h''-1)x''y'',\quad
 \frac {dy''}{dt}= b''+f''y''+(g''-1)x''y''+h''y''^2,
$$
having the following new parameters:
$$
\begin{gathered}
a''= -b,\quad
b''=\big[f^2 g - c^2 h + c f (-g + h) + (a - b) (g + h)^2\big]/(g + h)^2,\\
c''= ( c - f - c g + 2 f g + f h) /(g + h),\\
f''= (c g - 2 f g - c h) /(g + h), \quad g''=1-g-h, \quad
h''=g.\\
\end{gathered}
$$
We calculate
 $$
 \mathcal{D}_1''=c''+f''=\big[c (1 - 2 g + h) + f ( 4 g + h-1)\big]/(g+h)=
 \mathcal{D}_3/(g+h)=0.
$$
Thus our claim is proved and this completes the proof of the lemma.
\end{proof}

\begin{lemma} \label{lem:theta-ne0-Conics}
A system \eqref{sys:eta-poz-theta-ne0-gamma1=0} possesses an
invariant hyperbola of the indicated form if and only if the corresponding
 conditions indicated on the right hand side are satisfied:
\begin{itemize}
 \item[(I)] $\Phi(x,y)= p+qx+ry+2xy$ \quad $\Leftrightarrow$ \quad
 $\mathcal{B}_1\equiv b(2h-1)- a(2g-1)=0$, $(2h-1)^2+(2g-1)^2\ne0$, $a^2+b^2\ne0$;

 \item[(II)] $\Phi(x,y)= p+qx+ry+2x(x-y)$ \quad $\Leftrightarrow$
 \quad either
\begin{itemize}
 \item[(i)] $c=0$, $\mathcal{B}_2\equiv b(1-2h)+ 2a(g+2h-1)=0$, 
$(2h-1)^2+( g+2h-1)^2\ne0$, $a^2+b^2\ne0$, or 
 \item[(ii)] $h=1/3$, $\mathcal{B}_2'\equiv(1 + 3 g)^2 (b - 2 a + 6 a g) + 6 c^2 (1 - 3 g)=0$, $a\ne0$;
\end{itemize}


 \item[(III)] $\Phi(x,y)= p+qx+ry+2y(x-y)$ \quad $\Leftrightarrow$ \quad either
\begin{itemize}
 \item[(i)] $c=0$, $\mathcal{B}_3\equiv a(1-2g)+ 2b(2g+ h-1)=0$, 
$(2g-1)^2+(2g+ h-1)^2\ne0$, $a^2+b^2\ne0$, or
 \item[(ii)]$g=1/3$, $\mathcal{B}_3'\equiv 
(1 + 3 h)^2 (a - 2 b + 6 b h) + 6 c^2 (1 - 3 h)=0$, $b\ne0$.
\end{itemize}
\end{itemize}
\end{lemma}


\begin{proof}
Since for systems \eqref{sys:eta-poz-theta-ne0-gamma1=0}
we have $C_2=xy(x-y)$ (i.e. the infinite singularities are located
at the ``ends'' of the lines $x=0$, $y=0$ and $x-y=0$) it is clear
that if a hyperbola is invariant for these systems, then its
homogeneous quadratic part has one of the following forms:
(i) $kxy$, (ii) $kx(x-y)$,  (iii) $ky(x-y)$, \label{pageref:1} where
$k$ is a real nonzero constant. Obviously we may assume $k=2$
(otherwise instead of hyperbola \eqref{con:Phi(x,y)} we could
consider $ 2\Phi(x,y)/k=0$).

Considering the equations \eqref{Eqs:gen} we examine each one of
the above mentioned possibilities.

(i) $\Phi(x,y)= p+qx+ry+2xy$; in this case we obtain
$$
\begin{gathered}
 t=1,\ q=r=s=u=0,\ U=2g-1,\ V=2h-1 ,\ W=0,\\
 Eq_8= p(1-2g)+ 2b,\ \ Eq_9= p(1-2h)+ 2a,\\
 Eq_1=Eq_2=Eq_3=Eq_4=Eq_5=Eq_6=Eq_7=Eq_{10}=0.
\end{gathered}
$$
Calculating the resultant of the non-vanishing equations with
respect to the parameter $p$ we obtain
$$
\operatorname{Res}_p (Eq_8,Eq_9)= a(1-2g)+b(2h-1)= \mathcal{B}_1.
$$
So if $(2h-1)^2+(2g-1)^2\ne0$ then the hyperbola exists if and
only if $\mathcal{B}_1=0$. We may assume $2h-1\ne0$, otherwise the
change $(x,y,a,b,c,g,h)\mapsto(y,x,b,a,-c,h,g)$ (which preserves
systems
\eqref{sys:eta-poz-theta-ne0-gamma1=0}) could be applied. Then we obtain
$$
p=2a/(2h-1),\quad b=a(2g-1)/(2h-1),\quad
\Phi(x,y)=\frac{2a}{2h-1}+2xy=0
$$
and clearly for the irreducibility of the hyperbola the condition
$a^2+b^2\ne0$ must hold. This completes the proof of the statement
(I) of the lemma.



(ii) $\Phi(x,y)= p+qx+ry+2x(x-y)$; since $g+h\ne0$ (because
$\theta\ne0$) we obtain
$$
\begin{gathered}
s=2,\ t=-1,\ r=u=0,\ q=4c/(g+h),\ U=2g ,\ V=2h-1,\ W=-hq/2,\\
 Eq_8= 4 a - 2 b - 2 g p + 4 c^2 (g - h)/(g + h)^2,\\
 Eq_9= p(1-2h)-2a,\quad Eq_{10}= 2 c (2 a - h p) /(g + h ),\\
 Eq_1=Eq_2=Eq_3=Eq_4=Eq_5=Eq_6=Eq_7=0.
\end{gathered}
$$


 (1) Assume first $c\ne0$. Then considering the equations
 $Eq_9=0$ and $Eq_{10}=0$ we obtain $p(3h-1)=0$. Taking into account the
 relations above we obtain the hyperbola
 $$
\Phi(x,y)= p+4cx/(g+h) +2x(x-y)=0
$$
which evidently is reducible if $p=0$. So $p\ne0$ and this
implies $h=1/3$. Then from the equation $Eq_9=0$ we obtain
$p=6a$. Since $\theta=(g-1)(3g+1)/9\ne0$ we have $\
Eq_9=Eq_{10}=0$, $ Eq_8=-2\mathcal{B}_2'/(3g+1)^2. $\ So the
equation $Eq_8=0$ gives $\mathcal{B}_2'=0$ and then systems
\eqref{sys:eta-poz-theta-ne0-gamma1=0} with $h=1/3$ possess the
hyperbola
 $$
\Phi(x,y)= 6a+\frac{12c}{3g+1}x +2x(x-y)=0,
$$
which is irreducible if and only if $a\ne0$.


 (2) Suppose now $c=0$. In this case there remain only two
 non--vanishing equations:
$$
 Eq_8= 4 a - 2 b - 2 g p=0,\quad
 Eq_9= p(1-2h)-2a=0.\\
$$
Calculating the resultant of these equations with respect to the
parameter $p$ we obtain
$$
\operatorname{Res}_p (Eq_8,Eq_9)= b(1-2h)+ 2a(g+2h-1)= \mathcal{B}_2.
$$
If $(1-2h)^2+ (g+2h-1)^2\ne0$ (which is equivalent to $(1-2h)^2+
g^2\ne0$) then the condition
 $\mathcal{B}_2'=0$ is necessary and sufficient for a
system \eqref{sys:eta-poz-theta-ne0-gamma1=0} with $c=0$ to
possess the invariant hyperbola
 $$
\Phi(x,y)= p +2x(x-y)=0,
$$
where $p$ is the parameter determined from the equation $Eq_9=0$
(if $2h-1\ne0$), or $Eq_8=0$ (if $g\ne0$). We observe that the
hyperbola is irreducible if and only if $p\ne0$ which due to the
mentioned equations is equivalent to $a^2+b^2\ne0$.

 Thus the
statement II of the lemma is proved.

(iii) $\Phi(x,y)= p+qx+ry+2y(x-y)$; we observe that due
to the change $(x,y,a,b,c,g,h)\mapsto(y,x,b,a,-c,h,g)$ (which
preserves systems
\eqref{sys:eta-poz-theta-ne0-gamma1=0}) this case could be
brought to the previous one and hence, the conditions could be
constructed directly applying this change. This completes the
proof of Lemma \ref{lem:theta-ne0-Conics}. \end{proof}

In what follows the next remark will be useful.

\begin{remark} \label{rem:g,h,1-g-h} \rm
Consider systems \eqref{sys:eta-poz-theta-ne0-gamma1=0}.
\begin{itemize}
\item[(i)] The change $(x,y,a,b,c,g,h)\mapsto(y,x,b,a,-c,h,g)$ which
preserves these systems replaces the parameter $g$ by $h$ and $h$
by $g$.
\item[(ii)] Moreover if $c=0$ then having the relation
$(2h-1)(2g-1)(1-2g-2h)=0$ (respectively $(4h-1)(4g-1)(3-4g-4h)=0$)
due to a change we may assume $2h-1=0$ (respectively $ 4h-1 =0$).
\end{itemize}
\end{remark}

To prove the statement (ii) it is sufficient to observe that in
the
 case $2g-1=0$ (respectively $ 4g-1 =0$) we could apply the
 change given in the statement $(i)$ (with $c=0$), whereas in the
 case $1-2g-2h=0$ (respectively $ 3-4g-4h =0$) we apply the change
$(x,y,a,b,g,h)\mapsto(y-x,-x,b-a,-a,h,1-g-h)$, which conserves
systems \eqref{sys:eta-poz-theta-ne0-gamma1=0} with $c=0$.

Next we determine the invariant criteria which are equivalent to
the conditions given by Lemma~\ref{lem:theta-ne0-Conics}.

\begin{lemma} \label{lem:main-eta>0,theta-ne0}
Assume that for a quadratic system \eqref{sys:QSgenCoef} the
conditions $\eta>0$, $\theta\ne0$ and $\gamma_1=\gamma_2=0$ hold.
Then this system possesses at least one invariant hyperbola if
and only if one of the following sets of the conditions are
satisfied:
\begin{itemize}

\item [(i)] If $\beta_1\ne0$ then either

 \begin{itemize}

 \item[(i.1)] $\beta_2\ne0$, $\mathcal{R}_1\ne0$, or

 \item[(i.2)] $\beta_2=0$, $\beta_3\ne0$, $\gamma_3=0$,
 $\mathcal{R}_1\ne0$, or

 \item[(i.3)] $\beta_2=\beta_3=0$,
 $\beta_4\beta_5\mathcal{R}_2\ne0$, or

 \item[(i.4)] $\beta_2=\beta_3=\beta_4=0$, $\gamma_3=0$,
 $\mathcal{R}_2\ne0$;
 \end{itemize}

 \item [(ii)] If $\beta_1=0$ then either
 \begin{itemize}

 \item[(ii.1)] $\beta_6\ne0$, $\beta_2\ne0$, $\gamma_4=0$, $\mathcal{R}_3\ne0$, or

 \item[(ii.2)] $\beta_6\ne0$, $\beta_2=0$, $\gamma_5=0$, $\mathcal{R}_4\ne0$, or

 \item[(ii.3)] $\beta_6=0$, $\beta_7\ne0$, $\gamma_5=0$, $\mathcal{R}_5\ne0$, or

 \item[(ii.4)] $\beta_6=0$, $\beta_7=0$, $\beta_9\ne0$, $\gamma_5=0$,
 $\mathcal{R}_5\ne0$, or

 \item[(ii.5)] $\beta_6=0$, $\beta_7=0$, $\beta_9=0$, $\gamma_6=0$,
 $\mathcal{R}_5\ne0$.

 \end{itemize}
\end{itemize}
\end{lemma}

\begin{proof}
Assume that for a quadratic system \eqref{sys:QSgenCoef}
the conditions $\eta>0$, $\theta\ne0$ and $\gamma_1=0$ are
fulfilled. According to Lemma \ref{lem:normal form-gamma1=0} due
to an affine transformation and time rescaling this system could
be brought to the canonical form
\eqref{sys:eta-poz-theta-ne0-gamma1=0}, for which we calculate
\begin{equation} \label{val:gamma1=0}
\begin{gathered}
\gamma_2 = -1575 c^2 (g-1)^2 (h-1)^2 (g + h)(3g-1) (3h-1) (3 g + 3
h-4) \mathcal{B}_1,\\
\beta_1= - c^2 (g-1) (h-1) (3g-1)(3h-1)/4,\\
\beta_2= - c (g - h) ( 3 g + 3 h-4)/2.
\end{gathered}
\end{equation}

\subsubsection{Case $\beta_1\ne0$}
According to Lemma \ref{lem:H,E-irred-gamma1,2=0} the condition $\gamma_2=0$ is
necessary for the existence of a hyperbola. Since
$\theta\beta_1\ne0$ in this case the condition $\gamma_2=0$ is
equivalent to $(3 g + 3 h-4)\mathcal{B}_1=0$.

\subsubsection*{Subcase $\beta_2\ne0$.}\label{subs:beta1beta2ne0}
 Then $( 3 g + 3 h-4)\ne0$ and the condition $\gamma_2=0$ gives $\mathcal{B}_1=0$.
Moreover the condition $\beta_2\ne0$ yields $g - h\ne0$ and this
implies $(2h-1)^2+(2g-1)^2\ne0$. According to Lemma
\ref{lem:theta-ne0-Conics} systems
\eqref{sys:eta-poz-theta-ne0-gamma1=0} possess an invariant
hyperbola, which is irreducible if and only if $a^2+b^2\ne0$.

On the other hand for these systems we calculate
$$
\mathcal{R}_1= -3 c(a-b) (g-1)^2(h-1)^2 (g + h) (3g-1) (3h-1)/8
$$
and we claim that for $\mathcal{B}_1=0$ the condition
$\mathcal{R}_1=0$ is equivalent to $a=b=0$. Indeed, as the
equation $\mathcal{B}_1=0$ is linear homogeneous in $a$ and $b$,
as well as the second equation $a-b=0$, calculating the respective
determinant we obtain $-2(g+h)\ne0$ due to $\theta\ne0$. This
proves our claim and hence the statement $(i.1)$ of Lemma
\ref{lem:main-eta>0,theta-ne0} is proved.


\subsubsection*{Subcase $\beta_2=0$.} Since
$\beta_1\ne0$ (i.e. $c\ne0$) we obtain $(g-h)(3 g + 3 h-4)=0$. On
the other hand for systems \eqref{sys:eta-poz-theta-ne0-gamma1=0}
we have
$$
\beta_3=- c (g - h) (g-1)(h-1)/4
$$
and we consider two possibilities: $\beta_3\ne0$ and $\beta_3=0$.


\subsubsection*{Possibility $\beta_3\ne0$.} In this case we
have $g-h\ne0$ and the condition $\beta_2=0$ implies $3 g + 3
h-4=0$, i.e. $g=4/3-h$. So the condition $(2h-1)^2+(2g-1)^2\ne0$
for systems \eqref{sys:eta-poz-theta-ne0-gamma1=0} becomes
$(2h-1)^2+(6h-5)^2\ne0$ and obviously this condition is satisfied.

For systems
\eqref{sys:eta-poz-theta-ne0-gamma1=0} with $g=4/3-h$ we calculate
\begin{gather*}
\gamma_3=  22971 c (h-1)^3 (3h-1)^3 \mathcal{B}_1,\quad
\mathcal{R}_1= (a - b) c (h-1)^3 (3h-1)^3/6,\\
\beta_1=  -c^2 (h-1)^2 (3h-1)^2/4 ,\quad
\beta_3= - c (h-1) (3h-2) (3h-1)/18.
\end{gather*}
So because $\beta_1\ne0$ the condition $\gamma_3=0$ is equivalent
to $\mathcal{B}_1=0$. Moreover if in addition $\mathcal{R}_1=0$
(i.e. $a-b=0$) we obtain $a=b=0$, because the determinant of the
systems of linear equations
$$
3\mathcal{B}_1=a(5-6h)-3b(2h-1)=0, \quad a-b=0
$$
with respect to the parameters $a$ and $b$ equals $4(3h-2)\ne0$
due to the condition $\beta_3\ne0$. So the statement (i.2) of
the lemma is proved.



\subsubsection*{Possibility $\beta_3=0$.} Since
$\beta_1\ne0$ (i.e. $c(g-1)(h-1)\ne0$) we obtain $g=h$ and
 for systems
\eqref{sys:eta-poz-theta-ne0-gamma1=0} we calculate
\begin{gather*}
\gamma_2=  6300 c^2 h(h-1 )^4 (3h-2) (3h-1)^2\mathcal{B}_1,
\quad \theta= -h(h-1)^2,\\
\beta_1=  - c^2 (h-1)^2 (3h-1)^2/4,\quad
\beta_4= 2 h (3h-2),\quad
\beta_5= -2 h^2 (2h-1).
\end{gather*}
So by the condition $\theta\beta_1\ne0$ we obtain that the
necessary condition $\gamma_2=0$ is equivalent to
$\mathcal{B}_1(3h-2)=0$ and we shall consider two cases: $\beta_4
\ne0$ and $\beta_4=0$.


(1) \emph{Case $\beta_4\ne0$.}
Therefore $ 3h-2 \ne0$ and this implies $\mathcal{B}_1=0$. Considering
Lemma~\ref{lem:theta-ne0-Conics} the condition
$(2h-1)^2+(2g-1)^2\ne0$ for $g=h$ becomes $ 2h-1 \ne0$. So for the
existence of a invariant hyperbola the condition $\beta_5\ne0$ is
necessary. Moreover this hyperbola is irreducible if and only if
$a^2+b^2\ne0$. Since for these systems we have
$$
\mathcal{R}_2=(a+b)(h-1)^2(3h-1)/2,\quad \mathcal{B}_1=-(2h-1)(a-b)
$$
we conclude, that when $\mathcal{B}_1=0$ the condition
$\mathcal{R}_2\ne0$ is equivalent to $a^2+b^2\ne0$ and this
completes the proof of the statement (i.3) of the lemma.


(2) \emph{Case $\beta_4=0$.} Then by $\theta\ne0$ we obtain
$h=2/3$ and arrive at the 3-parameter family of systems
 \begin{equation} \label{sys:eta-poz-theta-ne0-gamma1=0a}
\frac {dx}{dt}= a+cx+2x^2/3-xy/3,\quad
 \frac {dy}{dt}=b-cy-xy/3+2y^2/3,
\end{equation}
For these systems we calculate
$ \gamma_3= 7657 c\mathcal{B}_1/9$,
$\beta_1= -c^2/36 $,
$\mathcal{R}_2= (a + b)/18$,
where $\mathcal{B}_1=(b-a)/3$. Since for these systems the
condition $(2h-1)^2+(2g-1)^2=2/9\ne0$ holds, according to Lemma
\ref{lem:theta-ne0-Conics} we conclude that the statement (i.4)
of the lemma is proved.


\subsubsection{Case $\beta_1=0$} Considering \eqref{val:gamma1=0}
and the condition $\theta\ne0$ we obtain $c(3g-1)(3h-1)=0$. On the
other hand for systems \eqref{sys:eta-poz-theta-ne0-gamma1=0} we
calculate
$$
\beta_6=-c (g-1) (h-1)/2
$$
and we shall consider two subcases: $\beta_6\ne0$ and $\beta_6=0$.

\subsubsection*{Subcase $\beta_6\ne0$.} Then $c\ne0$ and the
condition $\beta_1=0$ implies $ (3g-1)(3h-1)=0$. Therefore by
Remark \ref{rem:g,h,1-g-h} we may assume $h=1/3$ and this leads to
the following 4-parameter family of systems
 \begin{equation} \label{sys:eta-poz-theta-ne0-gamma1=0c}
\frac {dx}{dt}= a+cx+ gx^2 -2xy/3,\quad
 \frac {dy}{dt}=b-cy+ (g-1)xy+ y^2/3,
\end{equation}
which is a subfamily of \eqref{sys:eta-poz-theta-ne0-gamma1=0}.
According to Lemma \ref{lem:theta-ne0-Conics} the above systems
possess a hyperbola if and only if either $\mathcal{B}_1=
a(1-2g)- b/3=0$ and $a^2+b^2\ne0$ (the statement I), or
$\mathcal{B}_2'=(1 + 3 g)^2 (b - 2 a + 6 a g) + 6 c^2 (1 - 3 g)=0$
and $a\ne0$ (the statement II). We observe that in the first
case, when $a(1-2g)- b/3=0$ the condition $ a^2+b^2 \ne0$ is
equivalent to $a\ne0$.

On the other hand for these systems we calculate
\begin{gather*}
\gamma_4=  -16 (g-1)^2 (3g-1)^2 \mathcal{B}_1 \mathcal{B}_2'/81,\quad
\beta_6= c (g-1)/3,\\
\beta_2=  c (g-1) (3g-1)/2,\quad
\mathcal{R}_3= a (3g-1)^3/18.
\end{gather*}
So we consider two possibilities: $\beta_2\ne0$ and $\beta_2=0$.


\subsubsection*{Possibility $\beta_2\ne0$.} In this case
$(g-1)(3g-1)\ne0$ and the conditions $\gamma_4=0$ and
$\mathcal{R}_3\ne0$ are equivalent to $\mathcal{B}_1
\mathcal{B}_2'=0$ and $a\ne0$, respectively. This completes the
proof of the statement (ii.1).



\subsubsection*{Possibility $\beta_2=0$.} From the
condition $\beta_6\ne0$ we obtain $g=1/3$ and this leads to the
following 3-parameter family of systems:
 \begin{equation} \label{sys:eta-poz-theta-ne0-gamma1=0d}
 \frac {dx}{dt}= a+cx+ x^2/3-2xy/3,\quad
 \frac {dy}{dt}=b-cy-2xy/3+ y^2/3.
\end{equation}
Since $c\ne0$ (because $\beta_6\ne0$) according to Lemma
\ref{lem:theta-ne0-Conics} these systems possess an invariant
hyperbola if and only if one of the following sets conditions are
fulfilled:
\begin{gather*}
\mathcal{B}_1= (a-b)/3=0,\quad a^2+b^2\ne0;\\
\mathcal{B}_2'= 4b=0,\quad a\ne0;\quad
\mathcal{B}_3'= 4a=0,\quad b\ne0.
\end{gather*}
We observe that the last two conditions are equivalent to $ab=0$
and $a^2+b^2\ne0$.

On the other hand for systems
\eqref{sys:eta-poz-theta-ne0-gamma1=0d} we calculate
\[
\gamma_5=  16 \mathcal{B}_1 \mathcal{B}_2'\mathcal{B}_3'/27,\quad
\mathcal{R}_4= 128 (a^2 - a b + b^2) /6561.
\]
It is clear that the condition $\mathcal{R}_4=0$ is equivalent to
$a^2+b^2=0$. So the statement (ii.2) is proved.


\subsubsection*{Subcase $\beta_6=0$.} Since
 $\theta\ne0$ (i.e. $(g-1)(h-1)\ne0$) the condition
$\beta_6=0$ yields $c=0$. Therefore according to Lemma
\ref{lem:theta-ne0-Conics} systems
\eqref{sys:eta-poz-theta-ne0-gamma1=0} with $c=0$ possess an
invariant hyperbola if and only if one of the following sets of
conditions holds:
\begin{gather*}
\mathcal{B}_1\equiv b(2h-1)- a(2g-1)= 0,\quad (2h-1)^2+(2g-1)^2\ne0 ,\; a^2+b^2\ne0;\\
\mathcal{B}_2\equiv b(1-2h)+ 2a(g+2h-1) =0,\quad (2h-1)^2+(g+2h-1)^2\ne0,\;
 a^2+b^2\ne0;\\
\mathcal{B}_3 \equiv a(1-2g)+ 2b(2g+ h-1)=0,\quad (2g-1)^2+(2g+h -1)^2\ne0,\;
 a^2+b^2\ne0.
\end{gather*}
Considering the following three expressions
$$
\alpha_1=2g-1,\quad \alpha_2=2h-1,\quad \alpha_3=1-2g-2h
$$
we observe that the condition $(2h-1)^2+(2g-1)^2\ne0$
(respectively $(2h-1)^2+(g+2h-1)^2\ne0$; $(2g-1)^2+(2g+
h-1)^2\ne0$) is equivalent to $\alpha_1^2+\alpha_2^2\ne0$
(respectively $\alpha_2^2+\alpha_3^2\ne0$;
$\alpha_1^2+\alpha_3^2\ne0$).

 On the other hand for these systems we calculate
\begin{gather*}
\gamma_5= -288 (g-1) (h-1) (g + h) \mathcal{B}_1 \mathcal{B}_2 \mathcal{B}_3,\\
\theta= -(g-1) (h-1) (g + h)/2,\\
\beta_7=  2\alpha_1\alpha_2\alpha_3 ,\quad
\beta_9= 2(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3),\\
\mathcal{R}_5= 36 (b x - a y)\big[(g-1 )^2 x^2 + 2 (g + h + g h-1
) x y + (h-1)^2 y^2\big].
\end{gather*}
We observe that if $\alpha_1=\alpha_2=0$ (respectively
$\alpha_2=\alpha_3=0$;\ $\alpha_1=\alpha_3=0$) then the factor
$\mathcal{B}_1$ (respectively $\mathcal{B}_2$; $\mathcal{B}_3$)
vanishes identically. Considering the values of the invariant
polynomials $\beta_7$ and $\beta_9$ we conclude that two of the
factors $\alpha_i$ ($i=1,2,3$) vanish if and only if
$\beta_7=\beta_9=0$. So we have to consider two subcases:
$\beta_7^2+\beta_9^2\ne0$ and $\beta_7^2+\beta_9^2=0$.

 \subsubsection*{Possibility $\beta_7^2+\beta_9^2\ne0$.} In
this case by  $\theta\ne0$ the conditions $\gamma_5=0$ and
$\mathcal{R}_5\ne0$ are equivalent to $\mathcal{B}_1
\mathcal{B}_2 \mathcal{B}_3=0$ and $a^2+b^2\ne0$, respectively. So
by Lemma \ref{lem:theta-ne0-Conics} there exists at least one
hyperbola and hence the statements $(ii.3)$ and $(ii.4)$ are
valid.

 \subsubsection*{Possibility $\beta_7^2+\beta_9^2=0$.} As it
was mentioned above, in this case two of the factors $\alpha_i$
($i=1,2,3$) vanish. Considering Remark \ref{rem:g,h,1-g-h}, without
loss of generality we may assume $\alpha_1=\alpha_2=0$.

Thus we have $g=h=1/2$ and we obtain the family of systems
 \begin{equation} \label{sys:eta-poz-theta-ne0-gamma1=0=c}
\frac {dx}{dt}= a+ x^2/2- xy/2,\quad
 \frac {dy}{dt}=b - xy/2+ y^2/2.
\end{equation}
Since $c=0$ and the conditions of the statement I of Lemma
\ref{lem:theta-ne0-Conics} are not satisfied for these systems,
according to Lemma \ref{lem:theta-ne0-Conics} the above systems
possess an invariant hyperbola if and only if $a^2+b^2\ne0$ and
either $\mathcal{B}_2=a=0$ or $\mathcal{B}_3=b=0$. For systems
\eqref{sys:eta-poz-theta-ne0-gamma1=0=c} we calculate
$$
\gamma_6= -9 \mathcal{B}_2 \mathcal{B}_3,\quad
\mathcal{R}_5= 9(bx - a y)(x+y )^2
$$
and we conclude that the statement (ii.5) of the lemma holds.

As all the cases are examined, Lemma
\ref{lem:main-eta>0,theta-ne0} is proved.
\end{proof}

The next lemma is related to the number of the invariant
hyperbolas that quadratic systems with $\eta>0$ and $\theta\ne0$
could have.


\begin{lemma} \label{lem:main-eta>0,theta-ne0-2H}
Assume that for a quadratic system \eqref{sys:QSgenCoef} the
conditions $\eta>0$, $\theta\ne0$ and $\gamma_1=\gamma_2=0$ are
satisfied. Then this system possesses:
\begin{itemize}

\item[(A)] two invariant hyperbolas if and only if either
\begin{itemize}

 \item[(A1)]  $\beta_1=0$, $\beta_6\ne0$, $\beta_2\ne0$,
 $\gamma_4=0$, $\mathcal{R}_3\ne0$ and $\delta_1=0$, or

\item[(A2)]  $\beta_1=0$, $\beta_6=0$, $\beta_7\ne0$,
 $\gamma_5=0$, $\mathcal{R}_5\ne0$ and $\beta_8=\delta_2=0$,
 or

 \item[(A3)]  $\beta_1=0$, $\beta_6=\beta_7=0$,
 $\beta_9\ne0$, $\gamma_5=0$, $\mathcal{R}_5\ne0$ and
 $\delta_3=0$, $\beta_8\ne0$;
\end{itemize}

\item[(B)] three invariant hyperbolas if and only
if $\beta_1=0$, $\beta_6=\beta_7=0$, $\beta_9\ne0$,
$\gamma_5=0$, $\mathcal{R}_5\ne0$ and
 $\delta_3= \beta_8=0$.
\end{itemize}
\end{lemma}

\begin{proof}
For systems \eqref{sys:eta-poz-theta-ne0-gamma1=0} we have
\begin{gather*}
\beta_6= - c (g-1) (h-1)/2,\quad
\theta= - (g-1) (h-1) (g +h)/2,\\
\beta_1= -c^2 (g-1)(h-1)(3g-1)(3h-1)/4.
\end{gather*}


\subsubsection{Case $\beta_6\ne0$} Then $c\ne0$ and
according to Lemma \ref{lem:theta-ne0-Conics} we could have at
least two hyperbolas only if the conditions given either by the
statements I and  II; (ii) (i.e.
$\mathcal{B}_1=\mathcal{B}_2'=0$ and $h=1/3$), or by the statements  I and
 III; (ii) (i.e. $\mathcal{B}_1 =\mathcal{B}_3'=0$ and
$g=1/3$) are satisfied. Therefore the condition $(3g-1)(3h-1)=0$
is necessary. This condition is governed by the invariant
polynomial $\beta_1$. So we assume $\beta_1=0$ and due to Remark
\ref{rem:g,h,1-g-h} we may consider $h=1/3$. Then we calculate
\begin{gather*}
\gamma_4=  -16 (g-1)^2 (3g-1)^2
\mathcal{B}_1\mathcal{B}_2'/81,\quad
\beta_1= 0,\\
\theta =  (g-1) (1 + 3g)/9\ne0,\quad
\beta_2= c (g-1) (3g-1)/2.
\end{gather*}
Solving the systems of equations
$\mathcal{B}_1\big|_{h=1/3}=\mathcal{B}_2'=0$ with respect to $a$
and $b$ we obtain
$$
a=\frac{6 c^2 (3g-1)}{(1 + 3 g)^2}\equiv A_0,\quad b= -\frac{
 18 c^2 (2g-1) (3g-1)}{(1 + 3 g)^2}\equiv B_0.
$$

In this case we obtain the family of systems
\begin{equation} \label{sys:2hyp-theta-ne0-3}
 \frac {dx}{dt}= A_0+cx+ gx^2 -2xy/3,\quad
 \frac {dy}{dt}=B_0-cy+ (g-1)xy+ y^2/3,
\end{equation}
which possess two invariant hyperbolas:
\begin{gather*}
\Phi_1(x,y)= -\frac{36 c^2 (3g-1)}{(1 + 3 g)^2} + 2 x y=0,\\
\Phi_2(x,y)= -\frac{36 c^2 (3g-1)}{(1 + 3 g)^2} +\frac{12 c}{1 +
3 g} x + 2 x(x- y)=0,
\end{gather*}
where $c (3g-1)\ne0$ due to $a\ne0$. Thus for the irreducibility
of the hyperbolas above, the condition $c(3g-1)\ne0$ (i.e.
$\beta_2\ne0$) is necessary.


Since the condition $\gamma_4=0$ gives $\mathcal{B}_1
\mathcal{B}_2'=0$ it remains to find out the invariant polynomial
which in addition to $\gamma_4$ is responsible for the relation
$\mathcal{B}_1 =\mathcal{B}_2'=0$. We observe that in the case
$\mathcal{B}_1=0$ (i.e. $b= 3 a(1-2g)$) we have
$$
\delta_1= (3g-1) \big[a (1 + 3 g)^2-6 c^2
(3g-1)\big]/18=(3g-1)\mathcal{B}_2'/18.
$$
It remains to observe that in the case considered we have
$\mathcal{R}_3=a(3g-1)^3/18\ne0$ and that due to the condition
$\beta_2\ne0$ (i.e. $c(3g-1)\ne0$) by Lemma
\ref{lem:theta-ne0-Conics} we could not have a third hyperbola of
the form $\Phi(x,y)=p+qx+ry+2 y(x- y)=0$. This completes the proof
of the statement $(\mathcal{A}_1)$ of the lemma.


\subsubsection{Case $\beta_6=0$} Then $c=0$ and we
calculate for systems \eqref{sys:eta-poz-theta-ne0-gamma1=0}
$$
\beta_7=  2\alpha_1\alpha_2\alpha_3 ,\quad
\beta_9= 2(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3),\quad
\beta_8= 2(4g-1)(4h-1)(3-4g-4h),
$$
where $\alpha_1=2g-1$, $\alpha_2=2h-1$ and $\alpha_3=1-2g-2h$.

\subsubsection*{Subcase $\beta_7\ne0$.} Then $\alpha_1\alpha_2\alpha_3\ne0$
and we consider two possibilities: $\beta_8\ne0$ and $\beta_8=0$.

\subsubsection*{Possibility $\beta_8\ne0$.} We claim that in
this case we could not have more than one hyperbola. Indeed, as
$c=0$ we observe that all five polynomials $\mathcal{B}_i$
$(i=1,2,3)$, $\mathcal{B}_2'$ and $\mathcal{B}_3'$ are linear (and
homogeneous) with respect to $a$ and $b$ and the condition
$a^2+b^2\ne0$ must hold. So in order to have nonzero solutions
in $(a,b)$ of the equations
$$
\mathcal{U}=\mathcal{V}=0, \quad
\mathcal{U},\mathcal{V}\in\{\mathcal{B}_1,\mathcal{B}_2,
\mathcal{B}_3,\mathcal{B}_2',\mathcal{B}_3'\},\quad
\mathcal{U}\ne\mathcal{V}
$$
it is necessary that the corresponding determinants
$\det(\mathcal{U},\mathcal{V})=0$. We have for each couple,
respectively:
\begin{equation} \label{det:(Bi,Bj)}
\begin{aligned}
(\omega_1)&\ \det(\mathcal{B}_1,\mathcal{B}_2) =\ \ -(2h-1)(4h-1)=0;\\
(\omega_2)&\ \det(\mathcal{B}_1,\mathcal{B}_3) =\ \ -(2g-1)(4g-1)=0;\\
(\omega_3)&\ \det(\mathcal{B}_2,\mathcal{B}_3) =\ \ (1-2g-2h)(3-4g-4h)=0;\\
(\omega_4)&\ \det(\mathcal{B}_1,\mathcal{B}_2')\big|_{h=1/3} =\ \ (3g+1)^2/3;\\
(\omega_5)&\ \det(\mathcal{B}_1,\mathcal{B}_3')\big|_{g=1/3} =\ \ (3h+1)^2/3;\\
(\omega_6)&\ \det(\mathcal{B}_2',\mathcal{B}_3)\big|_{\{c=0,\,h=1/3\}} =\ \ (1 + 3 g)^2 (6g-1) (12g-5)/3=0;\\
(\omega_7)&\ \det(\mathcal{B}_2,\mathcal{B}_3')\big|_{\{c=0,\,g=1/3\}} =\ \ (1 + 3 h)^2 (6h-1) (12h-5)/3=0;\\
(\omega_8)&\ \det(\mathcal{B}_2',\mathcal{B}_3')\big|_{\{h=1/3,\,g=1/3\}} =\ \ -16\ne0.\\
\end{aligned}
\end{equation}
We observe that the determinant $(\omega_8)$ is not zero. Moreover
since $\beta_7\ne0$ and $\beta_8\ne0$ we deduce that none of the
determinants $(\omega_i)$ ($i=1,2,3$) could vanish.

On the other hand for systems
\eqref{sys:eta-poz-theta-ne0-gamma1=0} with $c=0$ we have
$\theta=(g-1)(3g+1)/9$ in the case $h=1/3$ and
$\theta=(h-1)(3h+1)/9$ in the case $g=1/3$. Therefore due to
$\theta\ne0$ in the cases $(\omega_4)$ and $(\omega_5)$ we also
could not have zero determinants.

Thus it remains to consider the cases $(\omega_6)$ and
$(\omega_7)$. Considering Remark \ref{rem:g,h,1-g-h} we observe
that the case $(\omega_7)$ could be brought to the case
$(\omega_6)$. So assuming $h=1/3$ we calculate
$$
\beta_7=2(2g-1)(6g-1)/9,\quad \beta_8=-2(4g-1)(12g-5)/9,\quad
\theta=(g-1)(3g+1)/9
$$
and hence the determinant corresponding to the case $(\omega_6)$
could not be zero due to $\theta\beta_7\beta_8\ne0$. This
completes the proof of our claim.

\subsubsection*{Possibility $\beta_8=0$.}
In this case we obtain $(4g-1)(4h-1)(3-4g-4h)=0$ and due to Remark
\ref{rem:g,h,1-g-h} we may assume $h=1/4$. Then
$\det(\mathcal{B}_1,\mathcal{B}_2)=0$ (see the case $(\omega_1)$)
and we obtain $\mathcal{B}_1= (2 a - b - 4 a
g)/2=-\mathcal{B}_2=0$. Since in this case we have
$$
\delta_2= 2 (2g-1) (4g-1) ( b-2 a + 4 a g),\quad
\beta_7= (2g-1) (4g-1)/2 \\
$$
we conclude that due $\beta_7\ne0$ the condition $2 a - b - 4 a
g=0$ is equivalent to $\delta_2=0$. So setting $b=2a(1-2g)$ we
arrive at the family of systems
\begin{equation} \label{sys:2hyp-theta-ne0-2}
\frac {dx}{dt}= a +gx^2 - 3xy/4,\quad
 \frac {dy}{dt}=2a(1-2g) +(g-1)xy+ y^2/4.
\end{equation}
These systems possess the invariant hyperbolas
$$
\Phi_1''(x,y)=-4a + 2 xy=0,\quad \Phi_2''(x,y)=4 a + 2x(x -y)=0,
$$
which are irreducible if and only if $a\ne0$. Since for these
systems we have
$$
\mathcal{R}_5= 9a(2x-4gx-y)\big[16(g-1)^2x^2+8(5g-3)xy+9y^2\big]/4
$$
the condition $a\ne0$ is equivalent to $\mathcal{R}_5\ne0$. On
the other hand for these systems we calculate
$$
\mathcal{B}_3= -2 a (2g-1) (4g-1),\quad
\mathcal{B}_3'\big|_{h=1/4}= 49 a /24
$$
and because $\beta_7\mathcal{R}_5\ne0$ we obtain
$\mathcal{B}_3\mathcal{B}_3'\ne0$, i.e. systems
\eqref{sys:2hyp-theta-ne0-2} could not possess a third hyperbola.
This completes the proof of the statement (A2).

\subsubsection*{Subcase $\beta_7=0$.} Then
$(2g-1)(2h-1)(1-2g-2h)=0$ and due to Remark \ref{rem:g,h,1-g-h} we
may assume $h=1/2$. Then by Lemma \ref{lem:theta-ne0-Conics} we
must have $g(2g-1)\ne0$ and this is equivalent to
$\beta_9=-4g(2g-1)\ne0$. Herein we have
$\det(\mathcal{B}_1,\mathcal{B}_2)=0$ and we obtain
$\mathcal{B}_1= a ( 1 - 2 g)=0$ and $\mathcal{B}_2= 2 a g=0$. This
implies $a=0$, which due to $\beta_9\ne0$ is equivalent to
$\delta_3=16a^2g^2(2g-1)^2 =0$. So we obtain the family of systems
\begin{equation} \label{sys:2hyp-theta-ne0-1}
\frac {dx}{dt}= gx^2-xy/2,\quad
 \frac {dy}{dt}= b +(g-1)xy+ y^2/2
\end{equation}
which possess the following two hyperbolas
$$
\Phi_1(x,y)=-\frac{2 b}{2g-1}+2xy=0,\quad
\Phi_2(x,y)=-\frac{b}{g}+2x(x-y)=0.
$$
These hyperbolas are irreducible if and only if $b\ne0$ which
is equivalent to
$\mathcal{R}_5=9bx\big[4(g-1)^2x^2+4(3g-1)xy+y^2\big]\ne0$.


For the above systems we have $\mathcal{B}_3= b (4g- 1)$ and
$\mathcal{B}_3'= 25 b /4$. Since $b\ne0$ only the condition
$\mathcal{B}_3= 0$ could be satisfied and this implies $g=1/4$.
It is not too hard to find out that in this case we obtain the third
hyperbola:
$$
\Phi_3(x,y)=-4b +2y(x-y)=0.
$$
We observe that for the systems above $\beta_8=-2(4g-1)^2$ and
hence the third hyperbola exists if and only if $\beta_8=0$. So
the statements $(\mathcal{A}_3)$ and $(\mathcal{B})$ are proved.

Since all the possibilities are examined, Lemma
\ref{lem:main-eta>0,theta-ne0-2H} is proved.
 \end{proof}


\subsection{Systems with three real infinite singularities and
$\theta=0$}
Considering \eqref{val:C2,theta} for systems
\eqref{sys:eta-poz} we obtain $(g-1)(h-1)(g+h)=0$ and we may assume
$g=-h$, otherwise in the case $g=1$ (respectively $h=1$) we apply
the change $(x,y,g,h)\mapsto(-y,x-y,1-g- h,g)$ (respectively
$(x,y,g,h)\mapsto(y-x,-x,h,1-g-h)$) which preserves the quadratic
parts of systems~\eqref{sys:eta-poz}.

So $g=-h$ and for systems \eqref{sys:eta-poz} we calculate $
N=9(h^2-1)(x-y)^2$. We consider two cases: $N\ne0$ and $N=0$.

\subsubsection{Case $N\ne0$} Then $(h-1)(h+1)\ne0 $ and due
to a translation we may assume $d=e=0$ and this leads to the
family of systems
\begin{equation} \label{sys:eta-poz-theta=0-Nne0}
\frac {dx}{dt}= a+cx-hx^2+(h-1)xy,\quad
 \frac {dy}{dt}= b+fy-(h+1)xy+hy^2.
\end{equation}

\begin{remark} \label{rem:h->-h} \rm
We observe that by  changing
$(x,y,a,b,c,f,h)\mapsto(y,x,b,a,f,c,-h)$ which conserves systems
\eqref{sys:eta-poz-theta=0-Nne0} we can change the sign of the
parameter $h$.
\end{remark}

\begin{lemma} \label{lem:theta=0-Nne0-Conics}
A system \eqref{sys:eta-poz-theta=0-Nne0} with $(h-1)(h+1)\ne0$
possesses at least one invariant hyperbola of the indicated form
if and only if the following conditions are satisfied,
respectively:
\begin{itemize}
 \item[(I)] $\Phi(x,y)= p+qr+ry+2xy\; \Leftrightarrow\;c+f=0$,
$\mathcal{E}_1\equiv a(2h+1)+b(2h-1)=0$, $a^2+b^2\ne0$;
\item[(II)] $\Phi(x,y)= p+qr+ry+2x(x-y)\; \Leftrightarrow\;
c-f=0$ and either
\begin{enumerate}
\item[(i)] $(2h-1)(3h-1)\ne0$, $\mathcal{E}_2\equiv 2 c^2 (h-1) (2h-1) +
(3h-1)^2 (b-2 a + 2 a h - 2 b h)=0$, $a\ne0$, or
\item[(ii)] $h=1/3$, $c=0$, $a\ne0$, or
\item[(iii)] $h=1/2$, $a=0$, $b+4c^2\ne0$;
\end{enumerate}
\item[(III)] $\Phi(x,y)= p+qr+ry+2y(x-y)\; \Leftrightarrow\;
c-f=0$ and either
\begin{enumerate}
\item[(i)] $(2h+1)(3h+1)\ne0$, $\mathcal{E}_3\equiv 2 c^2 (h+1) (2h+1) +
(3h+1)^2 (a-2 b - 2 b h + 2 a h)=0$, $b\ne0$, or
\item[(ii)] $h=-1/3$, $c=0$, $b\ne0$, or
\item[(iii)] $h=-1/2$, $b=0$, $a+4c^2\ne0$.
\end{enumerate}
\end{itemize}
\end{lemma}

\begin{proof} As it was mentioned in the proof of Lemma
\ref{lem:theta-ne0-Conics} (see page \pageref{pageref:1}) we may
assume that the quadratic part of an invariant hyperbola has one
of the following forms: (i) $2xy$, (ii)  $2x(x-y)$,
 (iii) $2y(x-y)$. Considering the equations \eqref{Eqs:gen} we examine
each one of these possibilities.

(i) $\Phi(x,y)= p+qx+ry+2xy$; in this case because
$N\ne0$ (i.e. $(h-1)(h+1)\ne0$) we obtain
\begin{gather*}
 t=1,\quad q=r=s=u=0,\quad U=-2h-1,\quad V=2h-1 ,\quad W=c+f,\\
 Eq_8= p(1+2h)+ 2b,\quad Eq_9= p(1-2h)+ 2a,\quad  Eq_{10}=-p(c+f),\\
 Eq_1=Eq_2=Eq_3=Eq_4=Eq_5=Eq_6=Eq_7=0.
\end{gather*}
Since in this case the hyperbola has the form $\Phi(x,y)= p+2xy$
it is clear that $p\ne0$, otherwise we obtain a reducible hyperbola.
So the condition $c+f=0$ is necessary.

Calculating the resultant of the non-vanishing equations with
respect to the parameter $p$ we obtain
$$
\operatorname{Res}_p(Eq_8,Eq_9)= 2[a(2h+1)+b(2h-1)]= 2\mathcal{E}_1.
$$

Since $(2h-1)^2+(2h+1)^2\ne0$ we conclude that an invariant
hyperbola exists if and only if $\mathcal{E}_1=0$. Due to Remark
\ref{rem:h->-h} we may assume $2h-1\ne0$. Then we obtain
$$
p=2a/(2h-1),\quad b=a(2h+1)/(2h-1),\quad
\Phi(x,y)=\frac{2a}{2h-1}+2xy=0
$$
and clearly for the irreducibility of the hyperbola the condition
$a\ne0$ must hold.

This completes the proof of the statement I of the lemma.

(ii) $\Phi(x,y)= p+qx+ry+2x(x-y)$; since
$(h-1)(h+1)\ne0$ (because $N\ne0$) we obtain
\begin{equation} \label{eqs:x(x-y)}
\begin{gathered}
s=2,\quad t=-1,\quad r=u=0,\quad U=-2h ,\quad V=2h-1,\quad W=(4 c + h q)/2,\\
 Eq_6=2(c-f),\quad Eq_8= 4 a - 2 b + 2 h p -cg- hq^2/2,\\
 Eq_9= p(1-2h)-2a,\quad Eq_{10}= -2 cp+aq-hpq/2,\\
 Eq_1=Eq_2=Eq_3=Eq_4=Eq_5=Eq_7=0.
\end{gathered}
\end{equation}
We observe that the equation $Eq_6=0$ implies the condition
$c-f=0$.


(1) Assume first $(2h-1)(3h-1)\ne0$. Then considering
the equation $Eq_9=0$ we obtain $p= 2 a /(1 - 2 h)$. As the
hyperbola $\Phi(x,y)= p+qx+2x(x-y)=0$ has to be irreducible the
condition $p\ne0$ holds and this implies $a\ne0$. Therefore from
 $$
 Eq_{10}= \frac{a (4 c - q + 3 h q)}{2h-1}=0
 $$
 From $3h-1\ne0$ we obtain $q= 4 c /(1 - 3 h)$ and then we obtain
$$
 Eq_8=\frac{2\mathcal{E}_2}{(2h-1) (3h-1)^2}=0.
$$
So we deduce that the conditions $c-f=0$, $\mathcal{E}_2=0$ and
$a\ne0$ are necessary and sufficient for the existence of a
hyperbola of systems \eqref{sys:eta-poz-theta=0-Nne0} in the case
$(2h-1)(3h-1)\ne0$.

(2) Suppose now $h=1/3$. Then considering
\eqref{eqs:x(x-y)} we have $ Eq_9=(p-6a)/3=0$, i.e. $p=6a\ne0$
(otherwise we obtain a reducible hyperbola). Therefore the equation
$Eq_{10}= -12ac=0$ yields $c=0$. Herein the equation $Eq_8=0$
becomes $Eq_8=[12(4a - b) - q^2]/6=0$, i.e.
$q=\pm2\sqrt{3(4a-b)}$ and obviously we obtain at leas one real
hyperbola if $4a-b\ge0$ and two complex if $4a-b<0$.

Thus in the case $h=1/3$ we have at least one hyperbola if and
only if the conditions $f=c=0$ and $a\ne0$ hold.


(3) Assume finally $h=1/2$. In this case we obtain
$Eq_9=-2a=0$, i.e. $a=0$ and we have
\begin{gather*}
 Eq_8= -2b+p-cq-q^2/4=0,\quad
 Eq_{10}= -p(8c+q)/4=0,\\ \Phi(x,y)= p+qx +2x(x-y).
\end{gather*}
Therefore $p\ne0$ and we obtain $q=-8c$ and $p=2(b+4c^2)\ne0$.
This completes the proof of the statement II of the lemma.

(iii) $\Phi(x,y)= p+qx+ry+2y(x-y)$; we observe that due
to the change $(x,y,a,b,c,f,h)\mapsto(y,x,b,a, c,f, -h)$ (which
preserves systems
\eqref{sys:eta-poz-theta=0-Nne0}) this case could be
brought to the previous one and hence, the conditions could be
constructed directly applying this change.

Thus Lemma \ref{lem:theta=0-Nne0-Conics} is proved.
\end{proof}

We shall construct now the affine invariant conditions for the
existence of at least one invariant hyperbola for quadratic
systems in the considered family.

\begin{lemma} \label{lem:main-eta>0,theta0,Nne0}
Assume that for a quadratic system \eqref{sys:QSgenCoef} the
conditions $\eta>0$, $\theta=0$, $N\ne0$, and
$\gamma_1=\gamma_2=0$ hold. Then this system possesses at least
one invariant hyperbola if and only if one of the following
sets of the conditions is satisfied:
\begin{itemize}
\item [(i)] If $\beta_6\ne0$ then either

 \begin{itemize}

 \item[(i.1)] $\beta_{10}\ne0$, $\gamma_7=0$, $\mathcal{R}_6\ne0$, or

 \item[(i.2)] $\beta_{10}=0$, $\gamma_4=0$, $\beta_2\mathcal{R}_3\ne0$;
 \end{itemize}

 \item [(ii)] If $\beta_6=0$ then either
 \begin{itemize}

 \item[(ii.1)] $\beta_2\ne0$, $\beta_7\ne0$, $\gamma_8=0$, 
$\beta_{10}\mathcal{R}_7\ne0$, or

 \item[(ii.2)] $\beta_2\ne0$, $\beta_7=0$, $\gamma_9=0$, $\mathcal{R}_8\ne0$, or

 \item[(ii.3)] $\beta_2=0$, $\beta_7\ne0$, $\beta_{10}\ne0$, $\gamma_7\gamma_8=0$,
 $\mathcal{R}_5\ne0$, or

 \item[(ii.4)] $\beta_2=0$, $\beta_7\ne0$, $\beta_{10}=0$, $\mathcal{R}_3\ne0$,
 $\gamma_7\ne0$, or

 \item[(ii.5)] $\beta_2=0$, $\beta_7\ne0$, $\beta_{10}=0$, 
$\mathcal{R}_3\ne0$, $\gamma_7=0$, or

 \item[(ii.6)] $\beta_2=0$, $\beta_7=0$, $\gamma_7=0$,
 $\mathcal{R}_3\ne0$.
 \end{itemize}
\end{itemize}
\end{lemma}

\begin{proof}
Assume that for a quadratic system
\eqref{sys:QSgenCoef} the conditions $\eta>0$, $\theta=0$ and
$N\ne0$ are fulfilled. As it was mentioned earlier due to an
affine transformation and time rescaling this system could be
brought to the canonical form
\eqref{sys:eta-poz-theta=0-Nne0}, for which we calculate
\begin{gather*}
\gamma_1 =  (c - f)^2 (c + f) (h-1)^2 (h+1)^2 (3h-1) (3h+1)/64,\\
\beta_6=(c - f) (h-1)(h+1)/4,\quad
\beta_{10}=-2 (3h-1) (3h+1).
\end{gather*}

\subsubsection*{Subcase $\beta_6\ne0$.} By Lemma
\ref{lem:H,E-irred-gamma1,2=0} for the existence of an invariant
hyperbola of systems \eqref{sys:eta-poz-theta=0-Nne0} the
condition $\gamma_1=0$ is necessary and this condition is
equivalent to $(c + f)(3h-1) (3h+1)=0$. We examine two
possibilities: $\beta_{10}\ne0$ and $\beta_{10}=0$.

\subsubsection*{Possibility $\beta_{10}\ne0$.} Then we obtain
$f=-c$ (this implies $\gamma_2=0$) and we have
$$
\gamma_7=  8 (h-1) (h+1) \, \mathcal{E}_1. \\
$$
Therefore becuase $\beta_6\ne0$ the condition $\gamma_7=0$ is
equivalent to $\mathcal{E}_1=0$. So we have $a=\lambda(2h-1)$,
$b=-\lambda(2h+1)$ (where $\lambda\ne0$ is an arbitrary parameter)
and then we calculate
$$
\mathcal{R}_6=-632\lambda c (h-1) (h+1).
$$
Since $\beta_6\ne0$ we deduce that the condition
$\mathcal{R}_6\ne0$ is equivalent to $a^2+b^2\ne0$. This completes
the proof of the statement (i.1) of the lemma.

\subsubsection*{Possibility $\beta_{10}=0$.} Then we have
$(3h-1) (3h+1)=0$ and by Remark \ref{rem:h->-h} we may assume
$h=1/3$. Then we obtain the 4-parameter family of systems
\begin{equation} \label{sys:eta-poz-theta=0-Nne0-h=1/3}
\frac {dx}{dt}= a+cx- x^2/3-2xy/3,\quad
 \frac {dy}{dt}= b+fy-4xy/3+ y^2/3,
\end{equation}
for which we calculate $\gamma_1=0$ and
\begin{gather*}
\gamma_2=  44800(c - f)^2 (c + f)(2 c - f) /243 , \\
\beta_6= -2(c - f)/9 ,\quad
\beta_2= -4(2 c - f)/9.
\end{gather*}
Since $\beta_6\ne0$ (i.e. $c-f\ne0$) by Lemma
\ref{lem:H,E-irred-gamma1,2=0} the necessary condition
$\gamma_2=0$ gives $(c + f)(2 c - f)=0$. We claim that for the
existence of an invariant hyperbola the condition $2c-f\ne0$
(i.e. $\beta_2\ne0$) must be satisfied. Indeed, setting $f=2c$ we
obtain $\beta_6=2c/9\ne0$. However, according to the Lemma
\ref{lem:theta=0-Nne0-Conics}, for the existence of a hyperbola of
systems
\eqref{sys:eta-poz-theta=0-Nne0-h=1/3}, the condition
$(c + f) (c -f)=0$ is necessary , which for $f=2c$ becomes $-3c^2=0$.
The contradiction obtained proves our claim.

Thus the condition $\beta_2\ne0$ is necessary and then we have
$f=-c$. By Lemma \ref{lem:theta=0-Nne0-Conics} in the case $h=1/3$
we have an invariant hyperbola (which is of the form $\Phi(x,y)=
p+qx+ry+2xy=0$) if and only if $\mathcal{E}_1=(5a-b)/3=0$ and
$a^2+b^2\ne0$.

On the other hand for systems
\eqref{sys:eta-poz-theta=0-Nne0-h=1/3} with $f=-c$ we calculate
$$
\gamma_4=  -4096 c^2\mathcal{E}_1/243,\quad
\beta_6= - 4 c /9 ,\quad
\mathcal{R}_3= - 4 a /9.
$$
So the statement (i.2) of the lemma is proved.

\subsubsection*{Subcase $\beta_6=0$.} Then $f=c$ (this implies
 $\gamma_2=0$) and we calculate
\begin{gather*}
\gamma_8=  42(h-1)(h+1)\mathcal{E}_2\mathcal{E}_3, \quad
\beta_2= c(h-1)(h+1)/2,\\
\beta_7= -2 (2h-1) (2h+1),\quad
\beta_{10}= -2 (3h-1) (3h+1),\\
\mathcal{R}_7= - (h-1)(h+1) U(a,b,c,h)/4,
\end{gather*}
where $U(a,b,c,h)= 2 c^2 (h-1)(h+1) - b (h+1) (3h-1)^2 + a (h-1)
(3h+1)^2$.

\subsubsection*{Possibility $\beta_2\ne0$.} Then $c\ne0$ and we
shall consider two cases: $\beta_7\ne0$ and $\beta_7=0$.

(1) \emph{Case $\beta_7\ne0$.} We observe that in this
case for the existence of a hyperbola the condition
$\beta_{10}\ne0$ is necessary. Indeed, since $f=c\ne0$ and $(2h-1)
(2h+1)\ne0$, according to Lemma \ref{lem:theta=0-Nne0-Conics} (see
the statements  II and  III for the existence of at
least one invariant hyperbola it is necessary and sufficient
$(3h-1) (3h+1)\ne0$ and either $\mathcal{E}_2=0$ and $a\ne0$, or
$\mathcal{E}_3=0$ and $b\ne0$.

We claim that the condition $a\ne0$ (when $\mathcal{E}_2=0$) as
well as the condition $b\ne0$ (when $\mathcal{E}_3=0$) is
equivalent to $U(a,b,c,h)\ne0 $. Indeed, as $\mathcal{E}_2$ as
well as $\mathcal{E}_3$ and $U(a,b,c,h)$ are linear polynomials in
$a$ and $b$, then the equations $\mathcal{E}_2=U(a,b,c,h)=0$
(respectively $\mathcal{E}_2=U(a,b,c,h)=0$) with respect to $a$
and $b$ gives $a=0$ and $b=2c^2(h-1)/(3h-1)^2$ (respectively $b=0$
and $a=-2c^2(h+1)/(3h+1)^2$). This proves our claim.

It remains to observe that the condition
$\mathcal{E}_2\mathcal{E}_3=0$ is equivalent to $\gamma_8=0$. So
this completes the proof of the statement \emph{(ii.1)} of the
lemma.

(2) \emph{Case $\beta_7=0$.} Then by Remark
\ref{rem:h->-h} we may assume
 $h=1/2$ and since $f=c$, by Lemma
\ref{lem:theta=0-Nne0-Conics} for the existence of a hyperbola of
systems \eqref{sys:eta-poz-theta=0-Nne0} (with $h=1/2$ and $f=c$)
the conditions $a=0$ and $b+4c^2\ne0$. On the other hand we
calculate
$$
\gamma_9=  3a/2, \quad
\mathcal{R}_8= (7 a + b + 4 c^2)/8
$$
and clearly these invariant polynomials govern the above
conditions. So the statement (ii.2) of the lemma is proved.


\subsubsection*{Possibility $\beta_2=0$.} In this case we
have $f=c=0$.


(1) \emph{Case $\beta_7\ne0$.} Then $(2h-1)(2h+1)\ne0$.

(a) \emph{Subcase $\beta_{10}\ne0$.} In this case
$(3h-1)(3h+1)\ne0$. By Lemma \ref{lem:theta=0-Nne0-Conics} we
could have an invariant hyperbola if and only if
$\mathcal{E}_1\mathcal{E}_2\mathcal{E}_3=0$. On the other hand for
systems \eqref{sys:eta-poz-theta=0-Nne0} with $f=c=0$ we have
\begin{gather*}
\gamma_7\gamma_8= -336 (h-1)^2 (1 + h)^2
\mathcal{E}_1\mathcal{E}_2\mathcal{E}_3,\\
\mathcal{R}_5= 36 (b x - a y)(x - y) \big[(1 + h)^2 x - (h-1)^2
y\big]
\end{gather*}
and therefore the condition $\mathcal{R}_5\ne0$ is equivalent to
$a^2+b^2\ne0$. This completes the proof of the statement
(ii.3) of the lemma.


(b) \emph{Subcase $\beta_{10}=0$.} Then we have
$(3h-1)(3h+1)=0$ and by Remark \ref{rem:h->-h} we may assume
$h=1/3$. By Lemma \ref{lem:theta=0-Nne0-Conics} we could have an
invariant hyperbola if and only if either the conditions I
or II; (ii) of Lemma \ref{lem:theta=0-Nne0-Conics} are
satisfied. In this case we calculate
$$
\gamma_7= -64 \mathcal{E}_1/9,\quad \mathcal{R}_3= -4a/9
$$
and hence, the condition $\mathcal{R}_3\ne0$ implies the
irreducibility of the hyperbola. Therefore in the case
$\gamma_7\ne0$ we arrive at the statement \emph{(ii.4)} of the
lemma, whereas for $\gamma_7=0$ the statement \emph{(ii.5)} of the
lemma holds.




(2) \emph{Case $\beta_7=0$.} Then $(2h-1)(2h+1)=0$ and
 by Remark \ref{rem:h->-h} we may assume $h=1/2$. By Lemma
\ref{lem:theta=0-Nne0-Conics} we could have an invariant hyperbola
if and only if either the conditions $\mathcal{E}_1=2a=0$ and
$b\ne0$ (see statement I) or $a=0$ and $b\ne0$ (see
statement II; (iii) of the lemma) are fulfilled. As we
could see the conditions coincide and hence by this lemma we have
two hyperbolas: the asymptotes of one of them are parallel to the
lines $x=0$ and $y=0$, whereas the asymptotes of the other
hyperbola are parallel to the lines $x=0$ and $y=x$.

On the other hand for systems
\eqref{sys:eta-poz-theta=0-Nne0} (with $h=1/2$ and $f=c=0$)
 we calculate
$$
\gamma_7= -12a,\quad \mathcal{R}_3= (5a-b)/16
$$
and this leads to the statement (ii.6) of the lemma.

Since all the possibilities are considered, Lemma
\ref{lem:main-eta>0,theta0,Nne0} is proved.
 \end{proof}

\begin{lemma} \label{lem:main-eta>0,theta=0-Nne0-2H}
Assume that for a quadratic system \eqref{sys:QSgenCoef} the
conditions $\eta>0$, $\theta=0$, $N\ne0$ and $\gamma_1=\gamma_2=0$
are satisfied. Then this system possesses:
\begin{itemize}

\item[(A)] three distinct invariant hyperbolas
 if and only if $\beta_6=\beta_2=\beta_{10}=\gamma_7=0$,
$\beta_7\mathcal{R}_3\ne0$ and $\gamma_{10}\ne0$; more precisely
all three hyperbolas are real (1 $ \mathcal{H}$ and 2 $
\mathcal{H}^p$) if $\gamma_{10}>0$ and one is real and two are
complex (1 $ \mathcal{H}$ and 2
$\overset{c}{\mathcal{H}^p}$) if $\gamma_{10}<0$;

\item[(B)] two distinct invariant hyperbolas if and only if $\beta_6=0$ and either
\begin{itemize}

 \item[(B1)] $\beta_2\ne0$, $\beta_7\ne0$,
 $\gamma_8=0$, $\beta_{10}\mathcal{R}_7\ne0$ and $\delta_4=0$\ ($\Rightarrow$\ 2 $ \mathcal{H}$), or
 \item[(B2)] $\beta_2\ne0$, $\beta_7=0$,
 $\gamma_9=0$, $\mathcal{R}_8\ne0$ and $\delta_5=0$\ ($\Rightarrow$\ 2 $ \mathcal{H}$), or
 \item[(B3)] $\beta_2=0$, $\beta_7\ne0$, $\beta_{10}\ne0$,
 $\gamma_7\gamma_8=0$, $\mathcal{R}_5\ne0$ and $\beta_8=\delta_2=0$\ ($\Rightarrow$\ 2 $ \mathcal{H}$), or
 \item[(B4)] $\beta_2=0$, $\beta_7\ne0$, $\beta_{10}=0$,
 $\gamma_7\ne0$, $\mathcal{R}_3\ne0$ and $\gamma_{10}<0$\ ($\Rightarrow$\ 2
 $\overset{c}{\,\mathcal{H}^p}$)),
 or
 \item[(B5)] $\beta_2=0$, $\beta_7\ne0$, $\beta_{10}=0$,
 $\gamma_7\ne0$, $\mathcal{R}_3\ne0$ and $\gamma_{10}>0$\ ($\Rightarrow$\ 2 $ \mathcal{H}^p$),
 or
 \item[(B6)] $\beta_2=0$, $\beta_7=0$,
 $\gamma_7=0$, $\mathcal{R}_3\ne0$ ($\Rightarrow$\ 2
 $\mathcal{H}$);
\end{itemize}

\item[(C)] one double ($\mathcal{H}^p_2$) invariant hyperbola if and only if
 $\beta_6=\beta_2=0$, $\beta_7\ne0$, $\beta_{10}=0$,
 $\gamma_7\ne0$, $\mathcal{R}_3\ne0$ and $\gamma_{10}=0$.
\end{itemize}
\end{lemma}

\begin{proof} For systems
\eqref{sys:eta-poz-theta=0-Nne0} we calculate
\begin{equation} \label{val:beta6,7,9,10}
\begin{gathered}
\beta_6= (c - f) (h-1)(h+1)/4,\quad
\beta_7= -2(2h+1)(2h-1),\\
\beta_{10}=  -2(3h+1)(3h-1),\quad
\beta_2= \big[(c + f)(h^2 - 1) - 8 (c - f) h)\big]/4.
\end{gathered}
\end{equation}
According to Lemma \ref{lem:theta=0-Nne0-Conics} in order to have
at least two invariant hyperbolas the condition $c-f=0$ must
hold. This condition is governed by the invariant polynomial
$\beta_6$ and in what follows we assume $\beta_6=0$ (i.e. $f=c$).

\subsubsection*{Case $\beta_2\ne0$.} Then we have $c\ne0$ and the
conditions given by the statement I of Lemma
\ref{lem:theta=0-Nne0-Conics} could not be satisfied.


\subsubsection*{Case $\beta_7\ne0$.} We observe that in this
case due to $c\ne0$ we could have two invariant hyperbolas if
and only if $(3h-1) (3h+1)\ne0$ (i.e $\beta_{10}\ne0$),
$\mathcal{E}_2= \mathcal{E}_3=0$ and $ab\ne0$. The system of
equations $\mathcal{E}_2=\mathcal{E}_3=0$ with respect to the
parameters $a$ and $b$ gives the solution
\begin{equation} \label{val:a,b}
a= -\frac{2 c^2 (1 + h)^3 (2h-1 )}{(3h-1 )^2 (1 + 3 h)^2}\equiv a_0,\quad
 b= -\frac{ 2 c^2 (h-1 )^3 (1 + 2 h)}{(3h-1)^2 (1 + 3 h)^2}\equiv b_0,
\end{equation}
which exists and $ab\ne0$ by the condition
$(2h-1)(2h+1)(3h-1)(3h+1)\ne0$.

In this case systems \eqref{sys:eta-poz-theta=0-Nne0} with $a=a_0$
and $b=b_0$ possess the  two hyperbolas
\begin{gather*}
\Phi_1^{(1)} (x,y)=  \frac{4 c^2 (1 + h )^3}{(3h-1)^2 (1 + 3
h )^2}-\frac{4 c}{ 3h-1 }\, x + 2x(x - y)=0,\\
\Phi_2^{(1)} (x,y)=  \frac{4 c^2 (h-1)^3}{(3h-1)^2 (1 + 3 h)^2}
- \frac{4c}{1 + 3 h}\, y + 2 y (x - y)=0.
\end{gather*}
Since $c\ne0$ by Lemma \ref{lem:theta=0-Nne0-Conics} we could not
have a third invariant hyperbola.

Now we need the invariant polynomials which govern the condition
$\mathcal{E}_2=\mathcal{E}_3=0$. First we recall that for these
systems we have $\gamma_8=
42(h-1)(h+1)\mathcal{E}_2\mathcal{E}_3$, and hence the condition
$\gamma_8=0$ is necessary. In order to set $\mathcal{E}_2=0$ we
use the following parametrization:
$$
c=c_1 (3h-1)^2,\quad a=a_1 (2h-1)
$$
and then the condition $\mathcal{E}_2=0$ gives
$b=2(h-1)(a_1+c_1^2)$. Herein for systems
\eqref{sys:eta-poz-theta=0-Nne0} with
$$
f=c=c_1 (3h-1)^2,\quad a=a_1 (2h-1),\quad b=2(h-1)(a_1+c_1^2)
$$
we calculate
$$
\mathcal{E}_3=3\big[2 c_1^2 (1 + h)^3 + a_1 (1 + 3 h)^2\big],\quad
\delta_4=(h-1)(2h-1)\mathcal{E}_3/2
$$
and hence the condition $\mathcal{E}_3=0$ is equivalent to
$\delta_4=0$.

It remains to observe that in this case
$\mathcal{R}_7=-3a_1(h-1)^4(h+1)/4\ne0$, otherwise $a_1=0$ and
then the condition $\delta_4=0$ implies $c_1=0$, i.e. $c=0$ and this
contradicts to $\beta_2\ne0$.
So we arrive at the statement (B1) of the lemma.

\subsubsection*{Case $\beta_7=0$.} Then $(2h-1)(2h+1)=0$
and by Remark \ref{rem:h->-h} we may assume $h=1/2$. In this case
by Lemma \ref{lem:theta=0-Nne0-Conics} in order to have at least
two hyperbolas the conditions II; (iii) and
III (i) have to be satisfied simultaneously. Therefore we
arrive at the conditions
$$
a=0, \quad b+4c^2\ne0,\quad \mathcal{E}_3=(50 a - 75 b + 24 c^2)/4=0
$$
and as $a=0$ we have $b=24c^2/75$ and $b+4c^2= 108 c^2 /25\ne0$
due to $\beta_2\ne0$. So we obtain the family of systems
\begin{equation} \label{sys:2hyp-theta=0-Nne0-1}
\frac {dx}{dt}= cx-x(x+y)/2,\quad
 \frac {dy}{dt}= 8c^2/25+cy-y(3x-y)/2
\end{equation}
which possess the  two invariant hyperbolas
\begin{gather*}
\Phi_1^{(2)}(x,y)= 216 c^2 /25 - 8 c x + 2 x (x - y)=0,\\
\Phi_2^{(2)}(x,y)=- 8 c^2 /25 - 8 c y /5 + 2 y (x - y)=0.
\end{gather*}
These hyperbolas are irreducible due to $\beta_2\ne0$ (i.e.
$c\ne0$).

We need to determine the affine invariant conditions which are
equivalent to $a=\mathcal{E}_3=0$. For systems
\eqref{sys:eta-poz-theta=0-Nne0} with $f=c$ and $h=1/2$ we
calculate
$$
\gamma_9= 3 a /2, \quad \delta_5=-3(25 b - 8 c^2)/2
$$
and obviously these invariant polynomials govern the conditions mentioned before.
 It remains to observe that for systems
\eqref{sys:2hyp-theta=0-Nne0-1} we have
$\mathcal{R}_8=108c^2/25\ne0$ due to $\beta_2\ne0$. This completes
the proof of the statement (B2) of the lemma.

\subsubsection*{Case $\beta_2=0$.} Then $c=0$ and by Lemma
\ref{lem:theta=0-Nne0-Conics} systems
\eqref{sys:eta-poz-theta=0-Nne0} with $f=c=0$ could possess at
least two invariant hyperbolas if and only if one of the
following sets of conditions holds:
\begin{equation} \label{cases:alpha-i}
\begin{aligned}
(\phi_1)\ \ & \mathcal{E}_1=\mathcal{E}_2=0,\ \ (2h-1)(3h-1)\ne0,\ \ a\ne0;\\
(\phi_2)\ \ & \mathcal{E}_1=\mathcal{E}_3=0,\ \ (2h+1)(3h+1)\ne0,\ \ b\ne0;\\
(\phi_3)\ \ & \mathcal{E}_2=\mathcal{E}_3=0,\ \ (2h-1)(2h+1)(3h-1)(3h+1)\ne0,\ \ ab\ne0;\\
(\phi_4)\ \ & \mathcal{E}_1=0,\ \ h=1/3,\ \ a\ne0;\\
(\phi_5)\ \ & \mathcal{E}_1=a=0,\ \ h=1/2,\ \ b\ne0;\\
(\phi_6)\ \ & \mathcal{E}_1=0,\ \ h=-1/3,\ \ b\ne0;\\
(\phi_7)\ \ & \mathcal{E}_1=b=0,\ \ h=-1/2,\ \ a\ne0.
\end{aligned}
\end{equation}
As for systems
\eqref{sys:eta-poz-theta=0-Nne0} with $f=c=0$ we have
$$
\beta_7= -2(2h+1)(2h-1),\quad
\beta_{10}= -2(3h+1)(3h-1)
$$
we consider two subcases: $\beta_7\ne0$ and $\beta_7=0$.

\subsubsection*{Subcase $\beta_7\ne0$.} Then $(2h+1)(2h-1)\ne0$
and we examine two possibilities: $\beta_{10}\ne0$ and
$\beta_{10}=0$.


(1) \emph{Possibility $\beta_{10}\ne0$.} In this case
$(3h+1)(3h-1)\ne0$. We observe that due to $f=c=0$ all tree
polynomials $\mathcal{E}_i$ are linear (homogeneous) with respect
to the parameters $a$ and $b$. So each one of the sets of
conditions $(\phi_1)$--$(\phi_3)$ could be compatible only if the
corresponding determinant vanishes, i.e.
\begin{equation} \label{det:(Ei,Ej)}
\begin{gathered}
\det(\mathcal{E}_1,\mathcal{E}_2) \Rightarrow -(2h-1) (3h-1)^2 (4h-1)=0, \\
\det(\mathcal{E}_1,\mathcal{E}_3) \Rightarrow (2h+1) (3h+1)^2 (4h+1)=0, \\
\det(\mathcal{E}_2,\mathcal{E}_3) \Rightarrow -3 (3h-1)^2 (3h+1)^2=0, \\
\end{gathered}
\end{equation}
otherwise we obtain the trivial solution $a=b=0$. Clearly the third
determinant could not be zero due to the condition
$\beta_{10}\ne0$, i.e. the conditions in the set $(\phi_3)$ are
incompatible in this case. As regard the conditions $(\phi_1)$
(respectively $(\phi_2)$) we observe that they could be compatible
only if $4h-1=0$ (respectively $4h+1=0$).

On the other hand we have $\beta_8=-6(4h-1)(4h+1)$ and we conclude
that for the existence of two hyperbolas in these case the
condition $\beta_8=0$ is necessary.

Assuming $\beta_8=0$ we may consider $h=1/4$ due to Remark
\ref{rem:h->-h} and we obtain
$$
 \mathcal{E}_1= (3 a - b)/2=-16 \mathcal{E}_2=0.
$$
So we obtain $b=3a$ and we arrive at the systems
\begin{equation} \label{sys:2hyp-theta=0-Nne0-a}
\frac {dx}{dt}= a - x^2/4 -3xy/4,\quad
 \frac {dy}{dt}= 3a -5 xy/4+ y^2/4,
\end{equation}
which possess the  two invariant hyperbolas
$$
\Phi_1^{(3)} (x,y)=-4a + 2 xy=0,\quad \Phi_2^{(3)} (x,y)=4 a +
2x(x - y)=0.
$$
Clearly these hyperbolas are irreducible if and only if $a\ne0$.

On the other hand for systems
\eqref{sys:eta-poz-theta=0-Nne0} with $f=c=0$ and $h=1/4$ we have
\begin{gather*}
\gamma_7= -15(3a-b),\quad \gamma_8= 15435
(3 a - 5 b) (3 a - b))/8192,\\
\delta_2= -6 (3 a - b),\quad \mathcal{R}_5=9 (b x - a y)(25 x - 9 y)
(x - y)/4.
\end{gather*}
We observe that the conditions $\mathcal{E}_1=\mathcal{E}_2=0$
and $a\ne0$ are equivalent to $\gamma_7=0$ and
$\mathcal{R}_5\ne0$. However  to insert this possibility in
the generic diagram (see Figure \ref{diagr:eta-poz}) we
remark that these conditions are equivalent to
$\gamma_7\gamma_8=\delta_2=0$ and $ \mathcal{R}_5\ne0$.

It remains to observe that for the systems above we have
$\mathcal{E}_3= 147 a/8\ne0$ and, hence we could not have a
third hyperbola. So the statement (B3) of the lemma
is proved.


(2) \emph{Possibility $\beta_{10}=0$.} In this case
$(3h+1)(3h-1)=0$ and without loss of generality we may assume
$h=1/3$ due to the change $(x,y,a,b,h)\mapsto(y,x,b,a, -h)$,
which conserves systems \eqref{sys:eta-poz-theta=0-Nne0} with
$f=c=0$ and transfers the conditions $(\phi_6)$ to $(\phi_4)$.

So $h=1/3$ and we arrive at the following 2-parameter family of
systems
\begin{equation} \label{sys:2hyp-theta=0-Nne0-b}
\frac {dx}{dt}= a - x^2/3 -2xy/3,\quad
 \frac {dy}{dt}= b -4 xy/3+ y^2/3,
\end{equation}
for which we have $\mathcal{E}_1= (5 a - b)/3$ and we shall
prove the next statements:
\begin{itemize}

 \item if $\mathcal{E}_1\ne0$, $4a-b<0$ and $a\ne0$ we have 2
 complex invariant hyperbolas $\overset{c}{\,\mathcal{H}^p}$;

 \item if $\mathcal{E}_1\ne0$, $4a-b>0$ and $a\ne0$ we have 2
 real invariant hyperbolas $\mathcal{H}^p$;

 \item if $\mathcal{E}_1\ne0$, $4a-b=0$ and $a\ne0$ we have one
 double invariant hyperbola $\mathcal{H}^p_2$.

 \item if $\mathcal{E}_1=0$, $4a-b>0$ and $a\ne0$ we have 3 real
 invariant hyperbolas (two of them being $\mathcal{H}^p$);

 \item if $\mathcal{E}_1=0$, $4a-b<0$ and $a\ne0$ we have 1 real
 and two complex invariant hyperbolas (of the type
 $\overset{c}{\,\mathcal{H}^p}$).
\end{itemize}
So we consider two cases: $\mathcal{E}_1\ne0$ and
$\mathcal{E}_1=0$


(a)  \emph{Case $\mathcal{E}_1\ne0$.} In this case by
Lemma \ref{lem:theta=0-Nne0-Conics} we could not have an invariant
hyperbola with the quadratic part of the form $xy$. However
systems \eqref{sys:2hyp-theta=0-Nne0-b} possess the following two
invariant hyperbola:
$$
 \Phi_{1,2}^{(4)} (x,y)= 3a
\pm \sqrt{3(4a-b)\,}x + x(x - y)=0
$$
and these conics are irreducible if and only if $a\ne0$.
Moreover the above hyperbolas have parallel asymptotes and they
are real if $4a-b>0$ (i.e . we have two $\mathcal{H}^p$) and
complex if $4a-b<0$ (i.e. we have two
$\overset{c}{\,\mathcal{H}^p}$). We observe that in the
case $4a-b=0$ the hyperbola $\Phi_{1,2}^{(4)} (x,y)=0$ collapse
and we obtain a hyperbola of multiplicity two (i.e . we have
$\mathcal{H}^p_2$).


(b)  \emph{Case $\mathcal{E}_1=0$.} Then $b=5a$ and
 we obtain the following 1-parameter family of systems
\begin{equation} \label{sys:2hyp-theta=0-Nne0-b1}
\frac {dx}{dt}= a - x^2/3 -2xy/3,\quad
 \frac {dy}{dt}= 5a -4 xy/3+ y^2/3.
\end{equation}
which possess three invariant hyperbolas
\[
\Phi_{1,2}^{(4)} (x,y)= 3a \pm \sqrt{-3a\,}x + x(x - y)=0,\quad
 \Phi_3^{(4)} (x,y) = 3a -xy=0.
\]
These conics are irreducible if and only if $a\ne0$. Also
the hyperbolas $\Phi_{1,2}^{(4)} (x,y)=0$ have parallel asymptotes
and they are real if $a<0$ and complex if $a>0$.

Thus the above statements are proved and in order to determine
the corresponding invariant conditions, for systems
\eqref{sys:eta-poz-theta=0-Nne0} with $c=f=0$ and $h=1/3$ we
calculate
$$
\gamma_7= -64(5a-b)/27, \quad \gamma_{10}= 8(4a-b)/27,\quad
\mathcal{R}_3= -4a/9.
$$
Considering the conditions given by the above statements it is
easy to observe that the corresponding invariant conditions are
given by the statements (B4), (B5),
(C) and (A) of Lemma
\ref{lem:main-eta>0,theta=0-Nne0-2H}, respectively.

\subsubsection*{Subcase $\beta_7=0$.} Then $(2h+1)(2h-1)=0$ and
by Remark \ref{rem:h->-h} we may assume $h=1/2$. Considering
\eqref{det:(Ei,Ej)} we conclude that only the case $(\phi_5)$
could be satisfied and we obtain the additional conditions:
 $ a=0$, $b\ne0$. Therefore we arrive at the family of systems
\begin{equation} \label{sys:2hyp-theta=0-Nne0-c}
\frac {dx}{dt}= - x^2/2 - xy/2,\quad
 \frac {dy}{dt}= b -3 xy/2+ y^2/2,
\end{equation}
which possess the following two hyperbolas
$$
\Phi_1^{(5)}, (x,y)=  -b + 2xy=0,\quad \Phi_2^{(5)} (x,y)
= 2b + 2x(x - y)=0.
$$
 We observe that the condition $a=0$ is equivalent to
$\gamma_7=-12a=0$. Regarding the condition $b\ne0$, in the case
$a=0$ it is equivalent to $\mathcal{R}_3=- b /16\ne0$. Since for
these systems we have $\mathcal{E}_3=75b/4\ne0$ we deduce that we
could not have a third invariant hyperbola. This completes the
proof of the statement (B6) of the lemma.

Since all the cases are examined, Lemma
\ref{lem:main-eta>0,theta=0-Nne0-2H} is proved.
\end{proof}


\subsubsection{Case $N=0$} As $\theta=-(g-1)(h-1)(g+h)/2=0$ we observe that
the condition $N=0$ implies the vanishing of two factors of
$\theta$. We may assume $g=1=h$, otherwise in the case
$g + h = 0$ and $g - 1 \ne 0$ (respectively $h-1\ne0$) we
apply the change $(x,y,g,h)\mapsto(-y,x-y,1-g- h,g)$ (respectively
$(x,y,g,h)\mapsto(y-x,-x,h,1-g-h)$) which preserves the form of
systems \eqref{sys:eta-poz}.

So $g=h=1$ and from an additional translation, systems \eqref{sys:eta-poz}
become
\begin{equation} \label{sys:eta-poz-N=0}
\frac {dx}{dt}= a+ dy+x^2,\quad
 \frac {dy}{dt}= b+ex +y^2.
\end{equation}

\begin{lemma} \label{lem:N=0-Conics}
A system \eqref{sys:eta-poz-N=0} possesses at least one
 invariant hyperbola of the indicated form if and only
if the corresponding conditions on the right hand side are satisfied:
\begin{itemize}
 \item[I] $\Phi(x,y)= p+qr+ry+2xy\; \Leftrightarrow\;
 d=e=0$ and $a-b=0$;
 \item[II] $\Phi(x,y)= p+qr+ry+2x(x-y)\; \Leftrightarrow\;
 d=0$, $\mathcal{M}_1\equiv 64 a - 16 b - e^2=0$, $ 16 a + e^2 \ne0$;
 \item[III] $\Phi(x,y)= p+qr+ry+2y(x-y)\;\Leftrightarrow\;
 e=0$, $\mathcal{M}_2\equiv 64 b - 16 a - d^2=0$, $ 16 b + d^2 \ne0$.
\end{itemize}
\end{lemma}

\begin{proof}
As it was mentioned in the proof of Lemma
\ref{lem:theta-ne0-Conics} (see page \pageref{pageref:1}) we may
assume that the quadratic part of an invariant hyperbola has one
of the following forms: (i) $2xy$, (ii) $2x(x-y)$, (iii)
$2y(x-y)$. Considering the equations \eqref{Eqs:gen} we examine
each one of these possibilities.

(i) $\Phi(x,y)= p+qx+ry+2xy$; in this case we obtain
\begin{gather*}
 t=1,\quad s=u=0,\quad p=(4b+q^2+qr)/2,\quad U=1,\quad V=1 ,\quad W=-(q+r)/2,\\
 Eq_9= (4 a - 4 b - q^2 + r^2)/2,\quad Eq_{10}=4 a q + 4 b (q + 2 r)+ q (q + r)^2,\\
 Eq_1=Eq_2=Eq_3=Eq_4=Eq_5=Eq_6=Eq_7=Eq_8=0.
\end{gather*}
Calculating the resultant of the non-vanishing equations with
respect to the parameter $r$ we obtain
$$
\operatorname{Res}_r (Eq_9,Eq_{10})= (a - b) (4 b + q^2)^2/4.
$$
If $ b = -q^2/4$ then we obtain the hyperbola $\Phi(x,y)= (r + 2
x) (q + 2 y)/2=0$, which is reducible.

Thus $b=a$ and we obtain
$$
Eq_9= -(q - r) (q + r)/2=0,\quad
Eq_{10}=(q + r) (8 a + q^2 + q r)/4=0.
$$
It is not too difficult to observe that the case $q+r\ne0$ (then
$q=r$) leads to reducible hyperbola (as we obtain $b=a=-q^2/4$,
see the case above). So $q=-r$ and the above equations are
satisfied. This leads to the invariant hyperbola $\Phi(x,y)= 2
a - r x + r y + 2 x y=0$. Considering Remark
\ref{rem:Delta-ne0=>irred} we calculate $\Delta=-(4a+r^2)/2$. So
the hyperbola above is irreducible if and only if $4a+r^2\ne0$.
Thus any system belonging to the family
\begin{equation} \label{sys:Famil-F1,2,3}
\frac {dx}{dt}= a+ x^2,\quad
 \frac {dy}{dt}= a +y^2
\end{equation}
possesses one-parameter family of invariant hyperbolas
$\Phi(x,y)= 2 a - r (x - y) + 2 x y=0$, where $r\in \mathbb{R}$ is a
parameter satisfying the relation $4a+r^2\ne0$. This completes
the proof of the statement {\bf I } of the lemma.


(ii) $\Phi(x,y)= p+qx+ry+2x(x-y)$; in this case we obtain
\begin{gather*}
s=2,\quad t=-1,\quad u=0,\quad p=(8 a - 4 b + 4 d e - 2 e^2 + q^2)/4,\\
r=2 d - e - q,\quad U=2 ,\quad V=1,\quad W=-(2e+ q)/2,\quad  Eq_7=-2d
\end{gather*}
and hence the condition $d=0$ is necessary. Then we calculate
\begin{gather*}
 Eq_1=Eq_2=Eq_3=Eq_4=Eq_5=Eq_6=Eq_7=Eq_8=0,\\
 Eq_9= -4 a + b - ( 2 e^2 + 6 e q + 3 q^2)/4,\\
 Eq_{10}= \big[ 16 a (e + q)-4 b (4 e + 3 q) + (2 e + q) (q^2- 2 e^2 )\big]/8,\\
\operatorname{Res}_q (Eq_9,Eq_{10})= - (64 a - 16 b - e^2) (4 a - 4 b -
e^2)^2/256.
\end{gather*}


(1) Assume first $64 a - 16 b - e^2=0$. Then $b=4 a - e^2/16$ and we obtain
$$
 Eq_9= -3(e + 2 q) (3 e + 2 q)/16 =0,\quad
 Eq_{10}= - (3 e + 2 q) (64 a + 4 e^2 - e q - 2 q^2)/32=0.
$$


(1a) If $q=-3e/2$ all the equations vanish and we
arrive at the invariant hyperbola
$$
\Phi(x,y)= -2 a + e^2/8 + e (-3 x + y)/2+ 2 x (x - y)=0
$$
for which we calculate $\Delta=(16 a + e^2)/8$. Therefore this
hyperbola is irreducible if and only if $ 16 a + e^2 \ne0$.


(1b) In the case $3 e + 2 q\ne0$ we have $q=-e/2\ne0$ and
the equation $Eq_{10}=0$ implies $e(16a+e^2)=0$. Therefore because
$e\ne0$ we obtain $ 16a+e^2 =0$. However in this case we have the
hyperbola
$$
\Phi(x,y)=- (16 a + 3 e^2)/8 - e (x + y)/2 + 2 x (x - y)=0,
$$
the determinant of which equals $(16a+e^2)/8$ and hence the
condition above leads to an irreducible hyperbola.


(2) Suppose now $4 a - 4 b - e^2=0$, i.e. $b= a -e^2/4$.
Herein we obtain
$$
 Eq_9= -3\big[4 a + (e+q)^2\big]/4 =0,\quad
 Eq_{10}= q\big[4 a + (e+q)^2\big]/8 =0
$$
and the hyperbola
$$
 \Phi(x,y)= 2 x (x - y) + q x - (e + q) y + (4 a - e^2 + q^2)/4=0,
$$
for which we calculate $\Delta=-[4 a + (e+q)^2\big]/4$. Obviously
the condition $Eq_9=0$ implies $\Delta=0$ and hence the invariant
hyperbola is reducible. So in the case $d=0$ and $4 a - 4 b -
e^2=0$ systems \eqref{sys:eta-poz-N=0} could not possess an
 invariant hyperbola and the statement II of the
lemma is proved.

(iii) $\Phi(x,y)= p+qx+ry+2y(x-y)$; we observe that because
the change $(x,y,a,b,d,e)\mapsto(y,x,b,a, e,d)$ (which
preserves systems
\eqref{sys:eta-poz-N=0}) this case could be
brought to the previous one and hence, the conditions could be
constructed directly applying this change.
Thus Lemma \ref{lem:N=0-Conics} is proved.
\end{proof}

\begin{lemma} \label{lem:main-eta>0,theta0,N=0}
Assume that for a quadratic system \eqref{sys:QSgenCoef} the
conditions $\eta>0$ and $\theta=N=0$ hold. Then this system could
possess either a single invariant hyperbola or a family of invariant hyperbolas.
More precisely, it possesses:
\begin{itemize}

\item [(i)] one invariant hyperbola if and only if $\beta_1=0$, $\mathcal{R}_9\ne0$
and either $(i.1)$ $\beta_2\ne0$ and $\gamma_{11}=0$, or $(i.2)$\
$\beta_2=\gamma_{12}=0$;
 \item [(ii)]\ a family of such hyperbolas if and only if
 $\beta_1=\beta_2=\gamma_{13}=0$.
\end{itemize}
\end{lemma}

\begin{proof} For systems
\eqref{sys:eta-poz-N=0} we calculate
\begin{gather*}
\beta_1=  4 d e,\quad
\beta_2= -2(d+ e),\quad
\gamma_{11}= 19 d e (d + e) + e \mathcal{M}_1 + d \mathcal{M}_2,\\
\mathcal{R}_9\big|_{d=0}= \big[5 (16 a + e^2) - \mathcal{M}_1\big]/2,\quad
\mathcal{R}_9\big|_{e=0}= \big[5 (16 b + d^2) - \mathcal{M}_2\big]/2.
\end{gather*}
By Lemma \ref{lem:N=0-Conics} the condition $de=0$ (i.e.
$\beta_1=0$) is necessary for a system \eqref{sys:eta-poz-N=0} to
possess an invariant hyperbola.

\subsubsection*{Subcase $\beta_2\ne0$.} Then $d^2+e^2\ne0$ and
considering the values of the above invariant polynomials by Lemma
\ref{lem:N=0-Conics} we deduce that the statement $ (i.1)$ of the
lemma is proved.

\subsubsection*{Subcase $\beta_2=0$.} In this case we obtain
$d=e=0$ and we calculate
$$
\gamma_{13}=4(a-b),\quad \mathcal{R}_9=8(a+b), \quad
 \gamma_{12}=-128(a-4b)(4a-b)=\mathcal{M}_1\mathcal{M}_2/2.
$$
 Therefore by Lemma
\ref{lem:N=0-Conics} in the case $\gamma_{12}=0$ we arrive at the
statement $ (i.2)$, whereas for $\gamma_{13}=0$ we arrive at the
statement $ (ii)$ of the lemma.

It remains to observe that if the systems \eqref{sys:eta-poz-N=0}
possess the family we mentioned of invariant hyperbolas, then they
have the form \eqref{sys:Famil-F1,2,3}, depending on the parameter
$a$. We may assume $a\in\{-1,0,1\}$ due to the rescaling
$(x,y,t)\mapsto (|a|^{1/2} x, |a|^{1/2} y,|a|^{-1/2}t)$.

\subsection{Systems with two real distinct infinite singularities
and $\theta\ne0$} For this family of systems by Lemma
\ref{lem:S1-S5} the conditions $\eta=0$ and $M\ne0$ are satisfied
and then via a linear transformation and time rescaling systems
\eqref{sys:QSgenCoef} could be brought to the
following family of systems:
\begin{equation} \label{sys:eta=0-Mne0}
\begin{gathered}
 \frac {dx}{dt}= a+cx+dy+gx^2+hxy,\\
 \frac {dy}{dt}= b+ex+fy+(g-1)xy+hy^2.
\end{gathered}
\end{equation}
For this systems we calculate
\begin{equation} \label{val:C2,theta-eta0}
C_2(x,y)=x^2y,\quad \theta= -h^2(g-1)/2
\end{equation}
 and since $\theta\ne0$ due to a translation we may assume
 $d=e=0$. So in what follows we consider the family of systems
\begin{equation} \label{sys:eta=0-Mne0-theta-ne0}
\begin{gathered}
 \frac {dx}{dt}= a+cx+ gx^2+hxy,\\
 \frac {dy}{dt}= b+ fy+(g-1)xy+hy^2.
\end{gathered}
\end{equation}
\end{proof}

\begin{lemma} \label{lem:eta=0-theta-ne0-Conics}
A system \eqref{sys:eta=0-Mne0-theta-ne0} could not posses more
than one invariant hyperbola. And it possesses one such
hyperbola if and only if
 $c+f=0$,\ $\mathcal{G}_1\equiv a(1-2g)+2bh=0$ and $a\ne0$.
\end{lemma}

\begin{proof} Since $C_2= x^2y$ we may assume that the quadratic part
of an invariant hyperbola has the form $ 2xy$. Considering the
equations
\eqref{Eqs:gen} and the condition $\theta\ne0$ (i.e. $h(g-1)\ne0$)
for systems
\eqref{sys:eta=0-Mne0-theta-ne0} we obtain
\begin{gather*}
 t=1,\quad s=u=q=r= 0,\quad p=a/h, \quad U=2g-1,\quad V=2h ,\quad W=c+f,\\
 Eq_8=(a-2ag + 2bh)/h=\mathcal{G}_1/h,\quad Eq_{10}=-a(c + f)/h ,\\
 Eq_1=Eq_2=Eq_3=Eq_4=Eq_5=Eq_6=Eq_7=Eq_9=0.
\end{gather*}
Since the hyperbola \eqref{con:Phi(x,y)} in this case becomes
$\Phi(x,y)=a/h+2xy=0$ the condition $a\ne0$ is necessary in order
to have an invariant hyperbola. Then the equation $Eq_{10}=0$
implies $c+f=0$ and the condition $Eq_8/h= 0$ yields
$\mathcal{G}_1=0$. Since $h\ne0$ we set $b=a(2g-1)/(2h)$ and this
leads to the family of systems
\begin{equation} \label{sys:eta=0-Mne0-theta-ne0-hyp-1}
\begin{gathered}
 \frac {dx}{dt}= a+cx +gx^2+hxy,\\
 \frac {dy}{dt}= \frac{a(2g-1)}{2h}-cy+(g-1)xy+hy^2,
\end{gathered}
\end{equation}
which possess the  invariant hyperbola
$$
\Phi(x,y)= \frac{a}{h}+ 2xy=0.
$$
This completes the proof of the lemma.
\end{proof}

Next we determine the corresponding affine invariant conditions.

\begin{lemma} \label{lem:main-eta=0,theta-ne0}
Assume that for a quadratic system \eqref{sys:QSgenCoef} the
conditions $\eta=0$, $M\ne0$ and $\theta\ne0$ hold. Then this
system possesses a single invariant hyperbola (which could be
simple or double) if and only if one of the following sets of the
conditions hold, respectively:
\begin{itemize}
\item[(i)] $\beta_2\beta_1\ne0$, $\gamma_1=\gamma_2=0$,
 $\mathcal{R}_1\ne0$:\ simple;

 \item[(ii)] $\beta_2\ne0$, $\beta_1= \gamma_1=\gamma_4=0$,
 $\mathcal{R}_3\ne0$: \ simple if $\delta_1\ne0$ and
 double if $\delta_1=0$;

 \item[(iii)] $\beta_2= \beta_1 =\gamma_{14}=0$,
$\mathcal{R}_{10}\ne0$:\ simple if $\beta_7\beta_8\ne0$ and double
if $\beta_7\beta_8=0$.
\end{itemize}
\end{lemma}

\begin{proof}
For systems \eqref{sys:eta=0-Mne0-theta-ne0} we calculate
$$
\gamma_1=  (2c - f) (c + f)^2 h^4 (g-1)^2/32,\quad
\beta_2= h^2(2c - f)/2.
$$
According to Lemma \ref{lem:H,E-irred-gamma1,2=0} for the
existence of an invariant hyperbola the condition $\gamma_1=0$
is necessary and therefore we consider two cases: $\beta_2\ne0$
and $\beta_2=0$.

\subsubsection{Case $\beta_2\ne0$} Then $2c-f\ne0$ and the
condition $\gamma_1=0$ implies $f=-c$. Then we calculate
\begin{gather*}
\gamma_2=  14175 c^2 h^5 (g-1)^2 (3g-1) \mathcal{G}_1,\quad
\beta_2= 3ch^2/2,\\
\beta_1= -3c^2 h^2 (g-1) (3g-1)/4 ,\ \
\mathcal{R}_1= -9a c h^4 (g-1)^2 (3g-1)/8
\end{gather*}
and we examine two subcases $\beta_1\ne0$ and $\beta_1=0$.

\subsubsection*{Subcase $\beta_1\ne0$.} Then the necessary
condition $\gamma_2=0$ (see Lemma \ref{lem:H,E-irred-gamma1,2=0})
gives $\mathcal{G}_1=0$ and by Lemma
\ref{lem:eta=0-theta-ne0-Conics} systems
\eqref{sys:eta=0-Mne0-theta-ne0} possess an invariant hyperbola.
We claim that this hyperbola could not be double. Indeed, since
the condition $\theta\ne0$ holds we apply Lemma
\ref{lem:main-eta>0,theta-ne0-2H} which provides necessary and
sufficient conditions in order to have at least two hyperbolas.
According to this lemma the condition $\beta_1=0$ is necessary for
the existence of at least two hyperbolas. So it is clear that in
this case the hyperbola of systems
\eqref{sys:eta=0-Mne0-theta-ne0-hyp-1} could not be double due to
$\beta_1\ne0$. This completes the proof of the statement $(i)$ of
the lemma.

\subsubsection*{Subcase $\beta_1=0$.} Because $\beta_2\ne0$ (i.e.
$c\ne0$) this implies $g=1/3$ and then $\gamma_2=0$ and
$$
\gamma_4= 16 h^6 (a + 6 b h)^2/3=48h^6\mathcal{G}_1^2,\quad
\mathcal{R}_3= 3 b h^3 /2.
$$
Therefore the condition $\gamma_4=0$ is equivalent to
$\mathcal{G}_1=0$ and in this case $\mathcal{R}_3\ne0$ gives
$b\ne0$ which is equivalent to $a\ne0$. By Lemma
\ref{lem:eta=0-theta-ne0-Conics} systems
\eqref{sys:eta=0-Mne0-theta-ne0} possess a hyperbola. We claim
that this hyperbola is double if and only if the condition
$a=-12c^2$ holds.

Indeed, as we would like after some perturbation to have two
hyperbolas, then the respective conditions provided by Lemma
\ref{lem:main-eta>0,theta-ne0-2H} must hold. We calculate:
$$
\beta_1=0,\quad \beta_2=3ch^2/2,\quad \beta_6=ch/3,\quad \gamma_4=0,\quad
\delta_1=-(a+12c^2)h^2/4
$$
and since $\beta_6\ne0$ (because $\beta_2\ne0$) we could have a
double hyperbola only if the identities provided by the statement
(A1) are satisfied. Therefore the condition
$\delta_1=0$ is necessary and due to $\theta\ne0$ (i.e. $h\ne0$)
we obtain $a=-12c^2$.

So our claim is proved and we obtain the family of systems
\begin{equation} \label{sys:eta=0-Mne0-theta-ne0-hyp-double}
\frac {dx}{dt}=-12c^2+cx + x^2/3+hxy,\quad
 \frac {dy}{dt}= 2c^2/h-cy- 2xy/3+hy^2,
\end{equation}
which possess the hyperbola $\Phi(x,y)=-12c^2/h+2xy=0$. The
perturbed systems
\begin{equation} \label{sys:eta=0-Mne0-theta-ne0-hyp-pert}
\begin{gathered}
\frac {dx}{dt}= - \frac{18 c^2 (2h+\varepsilon)(3 h +
 \varepsilon)}{(3h-\varepsilon)^2}+cx + x^2/3+(h+ \varepsilon)xy,\\
 \frac {dy}{dt}= \frac{6 c^2 (3h+\varepsilon)}{(3h-\varepsilon)^2}-cy-
 2xy/3+hy^2,\quad |\varepsilon|\ll1
\end{gathered}
\end{equation}
possess the  two distinct invariant hyperbolas:
\begin{gather*}
\Phi_1^{\varepsilon}(x,y)=-
\frac{36c^2(3h+\varepsilon)}{(3h-\varepsilon)^2} + 2 x y=0,\\
\Phi_2^{\varepsilon}(x,y)=-
\frac{36c^2(3h+\varepsilon)}{(3h-\varepsilon)^2} -
\frac{12c\varepsilon}{3h-\varepsilon} y +2y(x+\varepsilon y)=0.
\end{gather*}
It remains to observe that the hyperbola
$\Phi(x,y)=-12c^2/h+2xy=0$ could not be triple, because in this
case for systems \eqref{sys:eta=0-Mne0-theta-ne0-hyp-double} the
necessary conditions provided by the statement $(\mathcal{B})$ of
Lemma \ref{lem:main-eta>0,theta-ne0-2H} to have three invariant
hyperbolas are not satisfied: we have $\beta_6\ne0$.

Thus the statement (ii) of the lemma is proved.

\subsubsection{Case $\beta_2=0$} Then $f=2c$ and this
implies $\gamma_1=0$. On the other hand we calculate
$$
\gamma_2=  -14175 a c^2 (g-1)^3 (1 + 3 g) h^5,\quad
\beta_1= -9c^2 (g-1)^2 h^2/16
$$
and since $f=2c$, according to Lemma
\ref{lem:eta=0-theta-ne0-Conics} the condition $c=0$ is necessary
in order to have an invariant hyperbola. The condition $c=0$ is
equivalent to $\beta_1=0$ and this implies $\gamma_2=0$. It
remains to detect invariant polynomials which govern the
conditions $\mathcal{G}_1=0$ and $a\ne0$. For $c=0$ we have
$$
\gamma_{14}=  80 h^3\big[ a(1-2g)+2bh\big]= 80
h^3\mathcal{G}_1,\quad \mathcal{R}_{10}= -4ah^2.
$$
So for $\beta_1= \beta_2=0$, $\gamma_{14}=0$ and $\mathcal{R}_{10}\ne0$
systems \eqref{sys:eta=0-Mne0-theta-ne0-hyp-1} (with $c=0$)
possess the invariant hyperbola $ \Phi(x,y)= {a}/{h}+ 2xy=0$.

Next we shall determine the conditions under which this hyperbola
is simple or double. In accordance with Lemma
\ref{lem:main-eta>0,theta-ne0-2H} we calculate:
$$
\beta_1= \beta_6=0, \beta_7=-8(2g-1)h^2.
$$
We examine two possibilities: $\beta_7\ne0$ and $\beta_7=0$.

\subsubsection*{Possibility $\beta_7\ne0$.} According to
Lemma \ref{lem:main-eta>0,theta-ne0-2H} for systems
\eqref{sys:eta=0-Mne0-theta-ne0-hyp-1} with $c=0$ could be
satisfied only the identities given by the statement
(A2). So we have to impose the following conditions:
$$
\gamma_5=\beta_8=\delta_2=0.
$$
We have $\beta_8=-32(4g-1)h^2=0$ which implies $g=1/4$. Then we
obtain $\gamma_5=\delta_2=0$ and we obtain the family of systems
\begin{equation} \label{sys:eta=0-Mne0-theta-ne0-hyp-2}
\frac {dx}{dt}=a+ x^2/4+hxy,\quad
 \frac {dy}{dt}= -a/(4h) - 3xy/4+hy^2,
\end{equation}
which possess the hyperbola $\Phi(x,y)=a/h+2xy=0$. On the other
hand we observe that the perturbed systems
\begin{equation} \label{sys:eta=0-Mne0-theta-ne0-hyp-pert-2}
\frac {dx}{dt}=a+\frac{\varepsilon}{2h} + x^2/4+(h+\varepsilon)xy,\quad
 \frac {dy}{dt}= -a/(4h) - 3xy/4+hy^2,
\end{equation}
which possess the  two distinct invariant hyperbolas:
$$
\Phi_1^{\varepsilon}(x,y)=a/h+2xy=0,\quad
\Phi_2^{\varepsilon}(x,y)=a/h+ 2y(x+\varepsilon y)=0.
$$
Since $\beta_7\ne0$, according to Lemma
\ref{lem:main-eta>0,theta-ne0-2H} the hyperbola
$\Phi(x,y)=a/h+2xy=0$ could not be triple.

\subsubsection*{Possibility $\beta_7=0$.} In this case we obtain
$g=1/2$ and this implies $\gamma_8=\delta_3=0$. Hence the
identities given by the statement (A3) of Lemma
\ref{lem:main-eta>0,theta-ne0-2H} are satisfied. In this case we
obtain the family of systems
\begin{equation} \label{sys:eta=0-Mne0-theta-ne0-hyp-3}
\frac {dx}{dt}=a+ x^2/2+hxy,\quad
 \frac {dy}{dt}= - xy/2+hy^2,
\end{equation}
which possess the hyperbola $\Phi(x,y)=a/h+2xy=0$. On the other
hand we observe that the perturbed systems
\begin{equation} \label{sys:eta=0-Mne0-theta-ne0-hyp-pert-3}
\frac {dx}{dt}=a+ x^2/2+(h+\varepsilon)xy,\quad
 \frac {dy}{dt}= - xy/2+hy^2,
\end{equation}
possess the  two distinct invariant hyperbolas:
$$
\Phi_1^{\varepsilon}(x,y)=\frac{2 a}{2 h +
\varepsilon}+2xy=0,\quad \Phi_2^{\varepsilon}(x,y)=a/h+
2y(x+\varepsilon y)=0.
$$
Since for systems
\eqref{sys:eta=0-Mne0-theta-ne0-hyp-3} we have $\beta_8=-32h^2\ne0$, according to
Lemma \ref{lem:main-eta>0,theta-ne0-2H} the hyperbola
$\Phi(x,y)=a/h+2xy=0$ could not be triple.

It remains to observe that the conditions of the statement
(B) of Lemma \ref{lem:main-eta>0,theta-ne0-2H} in
order to have three invariant hyperbolas could not be satisfied
for systems
\eqref{sys:eta=0-Mne0-theta-ne0-hyp-1} (i.e. the necessary conditions for these
systems to possess a triple hyperbola). Indeed for systems
\eqref{sys:eta=0-Mne0-theta-ne0-hyp-1} we have
$$
\beta_7=-8(2g-1)h^2,\quad \beta_8=-32(4g-1)h^2,\quad
\theta=-(g-1)h^2/2
$$
and hence the conditions $\beta_7=0$ and $\beta_8=0$ are
incompatible due to $\theta\ne0$.
As all the cases are examined we deduce that Lemma
\ref{lem:main-eta=0,theta-ne0} is proved.
\end{proof}


\subsection{Systems with two real distinct infinite singularities
and $\theta=0$}
By Lemma \ref{lem:S1-S5} systems
\eqref{sys:QSgenCoef} via a linear transformation could be brought
to the systems \eqref{sys:eta=0-Mne0} for which we have
\begin{equation} \label{val:theta,mu_0}
 \theta= -h^2(g-1)/2, \quad  \beta_4=2h^2,\quad
 N=(g^2-1)^2x^2+2h(g-1)xy+h^2y^2.
\end{equation}
We shall consider to cases: $N\ne0$ and $N=0$.

\subsubsection{Case $N\ne0$} Since $\theta=0$ we obtain $h(g-1)=0$
and $(g^2-1)^2+h^2 \ne0$. So we examine two subcases:
$\beta_4\ne0$ and $\beta_4=0$.

\subsubsection*{Subcase $\beta_4\ne0$.} Then $h\ne0$ (this implies $N\ne0$) and
we obtain $g=1$. Applying a translation and the additional rescaling
$y\to y/h$ we may assume $c=f=0$ and $h=1$. So in what follows we
consider the family of systems
\begin{equation} \label{sys:eta=0-theta=0-mu-ne0}
\frac {dx}{dt}= a+dy+ x^2+xy,\quad
 \frac {dy}{dt}= b+ ex+ y^2.
\end{equation}

\begin{lemma} \label{lem:eta=0-theta=0-mu-ne0-Conics}
A system \eqref{sys:eta=0-theta=0-mu-ne0} possesses an invariant
hyperbola if and only if
 $e=0$,\ $\mathcal{L}_1\equiv 9 a - 18 b + d^2=0$ and $a+d^2\ne0$.
\end{lemma}

\begin{proof}
Since $C_2= x^2y$ we determine that the quadratic part of
an invariant hyperbola has the form $ 2xy$. Considering the
equations \eqref{Eqs:gen} for systems
\eqref{sys:eta=0-theta=0-mu-ne0} we obtain
\begin{gather*}
 t=1,\quad s=u=0,\quad r= 2d,\quad p=2 b + 2 d e + d q + q^2/2, \\
 U= 1,\quad V=2,\quad W=-(q+r)/2,\quad  Eq_5=e,\\
Eq_1=Eq_2=Eq_3=Eq_4= Eq_6=Eq_7=Eq_8=0.
\end{gather*}
Therefore the condition $Eq_5=0$ yields $e=0$ and then we have
$$
Eq_9=2 a - 4 b + 2 d^2 - q^2,\quad
Eq_{10}=a q+ b (4 d + q) + q (2 d + q)^2/4.
$$
Clearly in order to have a common solution of the equations
$Eq_9=Eq_{10}=0$ with respect to the parameter $q$ the condition
$$
\operatorname{Res}_q(Eq_9,Eq_{10})= (a + d^2)^2 (9 a - 18 b + d^2)/2=0
$$
is necessary. We claim that the condition $a + d^2=0$ leads to a hyperbola.
Indeed, setting $a=-d^2$ we obtain
$Eq_9=-(4b+q^2)=0$. On the other hand we obtain the hyperbola
$$
\Phi(x,y)= 2 b + d q + q^2/2 +qx+2dy+2xy=0
$$
for which by considering Remark \ref{rem:Delta-ne0=>irred} we
calculate $\Delta=-(4b + q^2)/2$. Therefore the equation
$Eq_9=-(4b+q^2)=0$ leads to an invariant hyperbola. This
proves our claim.

So $a + d^2\ne0$ and we set $b=(9a+d^2)/18$. Then $Eq_9=0$ gives
$(4 d - 3 q) (4 d + 3 q)=0$ and we examine two subcases: $q=4d/3$
and $q=-4d/3$.


(1) Assuming $q=4d/3$ we obtain $Eq_{10}=4 d (a + d^2)=0$.
Since $a + d^2\ne0$ we have $d=0$ and this leads to the family
of systems
\begin{equation} \label{sys:eta=0-theta=0-mu-ne0-hyp-a}
\frac {dx}{dt}= a+ x^2+xy,\quad
 \frac {dy}{dt}= a/2+ y^2.
\end{equation}
These systems possess the invariant hyperbola $\Phi(x,y)=a+2xy=0$.

(2) Suppose now $q=-4d/3$. This implies $Eq_{10}=0$
and we obtain the systems
\begin{equation} \label{sys:eta=0-theta=0-mu-ne0-hyp-b}
\frac {dx}{dt}= a+ dy+ x^2+xy,\quad
 \frac {dy}{dt}= (9a+d^2)/18 + y^2,
\end{equation}
which possess the invariant hyperbola
$$
\Phi_1(x,y)= (3 a - d^2)/3 - 2d (2 x - 3 y)/3 + 2 x y=0.
$$
Its determinant $\Delta$ equals $-(a+d^2)$ and hence, the
conic is irreducible if and only if $a+d^2\ne0$.

It remains to observe that the family of systems
\eqref{sys:eta=0-theta=0-mu-ne0-hyp-a} is a subfamily of the
family \eqref{sys:eta=0-theta=0-mu-ne0-hyp-b} (corresponding to
$d=0$) and this complete the proof of the lemma.
\end{proof}


\subsubsection*{Subcase $\beta_4=0$.} This implies $h=0$ and the condition
$N\ne0$ gives $g^2-1\ne0$. Using a translation we may assume
$e=f=0$ and we arrive at the family of systems
\begin{equation} \label{sys:eta=0-theta=0-mu=0-N-ne0}
\frac {dx}{dt}= a+cx+dy+ gx^2,\quad
 \frac {dy}{dt}= b+ (g-1)xy.
\end{equation}

\begin{lemma} \label{lem:eta=0-theta=0-mu=0-N-ne0-Conics}
A system \eqref{sys:eta=0-theta=0-mu=0-N-ne0} possesses at least
one invariant hyperbola if and only if $d=0$, $2g-1\ne0$ and
either
\begin{itemize}

 \item[(i)] $3g-1\ne0$,  $\mathcal{K}_1\equiv c^2 (1 - 2 g) + a
(3g-1)^2 =0$ and $b\ne0$, or

 \item[(ii)] $g=1/3$,  $c=0$ and $b\ne0$.
\end{itemize}
Moreover in the second case we have two real hyperbolas
$(\mathcal{H}^p)$ if $a<0$; two complex hyperbolas
($\overset{c}{\,\mathcal{H}^p}$) if $a>0$ and these
hyperbolas coincide if $a=0$.
\end{lemma}

\begin{proof}
As earlier we assumed that the quadratic part of an
invariant hyperbola has the form $ 2xy$ and considering the
equations
\eqref{Eqs:gen} for systems
\eqref{sys:eta=0-theta=0-mu=0-N-ne0} we obtain
\begin{gather*}
 t=1,\quad s=u=q=0,\quad
 U= 2g-1,\quad V=0,\quad W=c- gr/2,\\
 Eq_7=2d,\quad Eq_8=2 b + p(1 - 2 g),\quad Eq_9=2a-cr+gr^2/2,\\
 Eq_{10}=br-cp+gpr/2,\quad Eq_1=Eq_2=Eq_3=Eq_4=Eq_5= Eq_6=0.
\end{gather*}
Therefore the condition $Eq_7=0$ yields $d=0$ and we claim that
the condition $2g-1\ne0$ must hold. Indeed, supposing $g=1/2$ the
equation $Eq_8=0$ yields $b=0$ and then
$$
Eq_9= 2a+r(r-4c)/4=0,\quad Eq_{10}= p(r-4c)/4=0.
$$
Since $p\ne0$ (otherwise we obtain a reducible hyperbola) we obtain
$r=4c$, however in this case $Eq_9=0$ implies $a=0$ and we arrive
at degenerate systems. This completes the proof of our claim.

Thus we have $2g-1\ne0$ and then the equation $Eq_8=0$ gives
$p=2b/(2g-1)$ and we obtain:
$$
Eq_{10}= b ( 2 c + r - 3 g r))/(1 - 2 g).
$$
Since in this case the hyperbola is of the form
$$
\Phi(x,y)=\frac{2b}{2g-1 } +ry + 2 x y=0
$$
it is clear that the condition $b\ne0$ must hold and, therefore
we obtain $2 c + r(1 - 3 g)=0$.


(1) Assume first $3g-1\ne0$. Then we obtain $r=2c/(3g-1)$
and the equation $Eq_9=0$ becomes
$$
Eq_9=\frac{2}{(3g-1)^2} \big[c^2 (1 - 2 g) + a
(3g-1)^2\big]=\frac{2}{(3g-1)^2}\mathcal{K}_1=0.
$$
The condition $\mathcal{K}_1=0$ implies $a=c^2(2g-1)/(3g-1)^2$ and
we arrive at the family of systems
\begin{equation} \label{sys:eta=0-theta=0-mu=0-N-ne0-hyp-a}
\frac {dx}{dt}= \frac{c^2(2g-1)}{(3g-1)^2}+cx+ gx^2,\quad
 \frac {dy}{dt}= b+ (g-1)xy,
\end{equation}
possessing the invariant hyperbola
$$
\Phi(x,y)=\frac{2b}{2g-1} +\frac{2c}{3g-1} y + 2 x y=0,
$$
which is irreducible if and only if $b\ne0$.


(2) Suppose now $g=1/3$. In this case the equation
$Eq_{10}=0$ yields $c=0$ and then we obtain $p=-6b$ and the equation
 $Eq_9=0$ becomes $Eq_9= (12 a + r^2)/6=0$. Therefore for the
 existence of an invariant hyperbola the condition $a\le0$ is
 necessary. In this case setting $a=-3z^2\le0$ we arrive at
 the family of systems
\begin{equation} \label{sys:eta=0-theta=0-mu=0-N-ne0-hyp-b}
\frac {dx}{dt}=a+ x^2/3,\quad
 \frac {dy}{dt}= b- 2xy/3,
\end{equation}
possessing the  two invariant conics
$$
\Phi_{1,2}(x,y)=3b \pm \sqrt{-3a\,} y - x y=0,
$$
which are irreducible if and only if $b\ne0$. Clearly these
hyperbolas are real for $a<0$, they are complex for $a>0$ and
coincide (and we obtain a double one) if $a=0$.\end{proof}

\begin{lemma} \label{lem:main-eta=0,theta=0,Nne0}
Assume that for a quadratic system \eqref{sys:QSgenCoef} the
conditions $\eta=0$, $M\ne0$, $\theta=0$ and $N\ne0$ are
satisfied. Then this system could possess either a single
 invariant hyperbola, or two distinct $(\mathcal{H}^p)$
such hyperbolas, or one triple invariant hyperbola. More
precisely, it possesses:
\begin{itemize}

\item [(i)] one invariant hyperbola if and only if
either
\begin{itemize}

\item [(i.1)] $\beta_4\ne0$, $\beta_3= \gamma_8=0$ and
$\mathcal{R}_7\ne0$ \ (simple if $\delta_4\ne0$ and double if
$\delta_4=0$), or
 \item [(i.2)] \ $\beta_4=\beta_6=0$, $\beta_{11}\mathcal{R}_{11}\ne0$, $\beta_{12}\ne0$
 and $\gamma_{15}=0$\ (simple if $\gamma_{16}^2+\delta_6^2\ne0$ and double if
$\gamma_{16}=\delta_6=0$);
\end{itemize}

 \item [(ii)]\ two distinct invariant hyperbolas (both simple)
if and only if $\beta_4=\beta_6=0$,
$\beta_{11}\mathcal{R}_{11}\ne0$, $\beta_{12}=\gamma_{16}=0$ and
$\gamma_{17}\ne0$. Moreover these hyperbolas are real
$(\mathcal{H}^p)$ if $\gamma_{17}<0$ and they are complex
($\overset{c}{\,\mathcal{H}^p}$) if $\gamma_{17}>0$;

 \item [(iii)]\ one triple invariant hyperbola (which splits into three distinct hyperbolas,
 two of them being $(\mathcal{H}^p)$)
 if and only if $\beta_4=\beta_6=0$, $\beta_{11}\mathcal{R}_{11}\ne0$,
 $\beta_{12}=\gamma_{16}=0$ and $\gamma_{17}=0$.
\end{itemize}
\end{lemma}

 \begin{proof} Assume that for a quadratic system
\eqref{sys:QSgenCoef} the conditions $\eta=0$, $M\ne0$, $\theta=0$
and $N\ne0$.

\subsubsection*{Case $\beta_4\ne0$.} As it was shown earlier in this
case via an affine transformation and time rescaling the system
could be brought to the form \eqref{sys:eta=0-theta=0-mu-ne0}, for
which we calculate
$$
\gamma_1=  -9de^2/8,\quad \beta_3=-e/4,
$$
and by Lemma \ref{lem:eta=0-theta=0-mu-ne0-Conics} the condition
$\beta_3=0$ is necessary in order to have an invariant hyperbola.
In this case we obtain
$$
\gamma_8=42(9a-18b+d^2)^2=42\mathcal{L}_1^2,\quad
\mathcal{R}_7=-\mathcal{L}_1/8-(a+d^2)/3
$$
and considering Lemma \ref{lem:eta=0-theta=0-mu-ne0-Conics} for
$\beta_3=\gamma_8=0$ we obtain systems
\eqref{sys:eta=0-theta=0-mu-ne0-hyp-b} possessing the hyperbola
$\Phi(x,y)= (3 a - d^2)/3 - 2d (2 x - 3 y)/3 + 2 x y=0$. To detect
its multiplicity we apply Lemma \ref{lem:Ek} setting $k=2$. So in
order to have the polynomial $\Phi(x,y)$ as a double factor in
$\mathscr{E}_k$, we force its cofactor in $\mathscr{E}_2$ to be
zero along the curve $\Phi(x,y)=0$ (i.e we set $y=(-3 a + d^2 + 4
d x)/(6 (d + x))$). We obtain
$$
\frac{\mathscr{E}_2}{\Phi(x,y)}= \frac{(a + d^2)^4 (81 a + 17
d^2)}{2^{11}3^{12} (d + x)^{10}} (7 d + 15 x) (3 a + d^2 + 4 d x +
6 x^2)^{10} =0
$$
and since $a+d^2\ne0$ (see Lemma
\ref{lem:eta=0-theta=0-mu-ne0-Conics}) we obtain $81a+17d^2=0$. So we
obtain the family of systems
\begin{equation} \label{sys:eta=0-theta=0-mu-ne0-doule}
 \frac {dx}{dt}= - 17 d^2 /81+dy+ x^2+xy,\quad
 \frac {dy}{dt}= - 4 d^2 /81 + y^2,
\end{equation}
which possess the invariant hyperbola: $\Phi(x,y)
=-44d^2/81-4dx/3+2dy+2xy=0$. The perturbed systems
\begin{equation} \label{sys:eta=0-theta=0-mu-ne0-doule-pert}
\frac {dx}{dt}= - \frac{d^2(17-2\varepsilon+\varepsilon^2)}{(\varepsilon^2-9)^2}+dy+
 x^2+(1+\varepsilon)xy,\quad
 \frac {dy}{dt}= - \frac{4d^2}{(\varepsilon^2-9)^2} + y^2,
\end{equation}
possess the two hyperbolas:
\begin{gather*}
\Phi_1^\varepsilon(x,y) =-
\frac{4d^2(11-4\varepsilon+\varepsilon^2}{(\varepsilon^2-9)^2(1+\varepsilon)}-
 \frac{4d}{(1+\varepsilon)(3+\varepsilon)}x+\frac{2d}{1+\varepsilon}y+2xy=0,
\\
\Phi_2^\varepsilon(x,y) =
\frac{4d^2(11+4\varepsilon+\varepsilon^2}{(\varepsilon^2-9)^2(
\varepsilon-1)}-
 \frac{4d}{(1-\varepsilon)(3-\varepsilon)} x-\frac{6d}{\varepsilon-3} y
+2y(x+\varepsilon y)=0,
\end{gather*}
We observe that for systems
\eqref{sys:eta=0-theta=0-mu-ne0-hyp-b} we have $\delta_4=(81 a + 17
d^2)/6$ and $\beta_7=-8$. Therefore if $\delta_4=0$ the invariant
hyperbola is double and by Lemma \ref{lem:main-eta>0,theta-ne0-2H}
it could not be triple due to $\beta_7\ne0$. This completes the
proof of the statement $(i.1)$ of the lemma.

\subsubsection*{Case $\beta_4=0$.} Then we arrive at the family of
systems \eqref{sys:eta=0-theta=0-mu=0-N-ne0}, for which we have
$$
\beta_6=  d(g^2-1)/4,\quad N=4(g^2-1)x^2,\quad
 \beta_{11}= 4(2g-1)^2x^2,\quad
 \beta_{12}= (3g-1) x,
$$
So from  $N\ne0$ the necessary conditions $d=0$ and $2g-1\ne0$
(see Lemma \ref{lem:eta=0-theta=0-mu=0-N-ne0-Conics}) are
equivalent to $\beta_6=0$ and $\beta_{11}\ne0$, respectively.

\subsubsection*{Subcase $\beta_{12}\ne0$.} In this case
$3g-1\ne0$ and then by Lemma
\ref{lem:eta=0-theta=0-mu=0-N-ne0-Conics} an invariant hyperbola
exists if and only if $\mathcal{K}_1=0$ and $b\ne0$. On the other
hand for systems \eqref{sys:eta=0-theta=0-mu=0-N-ne0} with $d=0$
we calculate
$$
\gamma_{15}=4(g-1)^2(3g-1)\mathcal{K}_1x^5,\quad
\mathcal{R}_{11}=-3b(g-1)^2x^4
$$
and hence the above conditions are governed by the invariant
polynomials $\gamma_{15}$ and $\mathcal{R}_{11}$. So we obtain
systems \eqref{sys:eta=0-theta=0-mu=0-N-ne0-hyp-a} possessing the
hyperbola $ \Phi(x,y)= {2b}/(2g-1) + {2cy}/(3g-1) + 2 x y=0$.


By to Lemma \ref{lem:Ek} we calculate the polynomial
$\mathscr{E}_2$ and we observe that $\mathscr{E}_2$ contains
the polynomial $\Phi(x,y)$ as a simple factor.

To have this polynomial as a double factor in
$\mathscr{E}_2$, we force its cofactor in $\mathscr{E}_2$ to be
zero along the curve $\Phi(x,y)=0$ (i.e we set $y=b (3g-1)/((2g-1)
(c - x + 3 g x))$). We obtain
\begin{align*}
\frac{\mathscr{E}_2}{\Phi(x,y)}
&= \frac{288b^3(g-1)[c +(3g-1) x]^3}{(2g-1)^3
(3g-1)^{16}}\big[c (2g-1) + g (3g-1) x\big]^{10}
 \big[c^2 (31 \\
&\quad - 87 g + 62 g^2) 
 +  6 c (3g-2) (3g-1)^2 x + (3g-1)^3 (4g-1) x^2\big]=0
\end{align*}
and since $(2g-1)(3g-1)\ne0$ we obtain $c=0$ and either $g=1/4$ or
$g=0$. However in the second case we obtain degenerate systems. So
$g=1/4$ and we arrive at the family of systems
\begin{equation} \label{sys:eta=0-theta=0-mu=0-N-ne0-hyp-a2-double}
\frac {dx}{dt}= x^2/4,\quad
 \frac {dy}{dt}= b- 3xy/4,
\end{equation}
which possess the hyperbola $\Phi(x,y)=-4b + 2 x y=0$. On the
other hand the perturbed systems
\begin{equation} \label{sys:eta=0-theta=0-mu=0-N-ne0-hyp-a2-double-pert}
\frac {dx}{dt}= -2b\varepsilon + \varepsilon xy+x^2/4,\quad
 \frac {dy}{dt}= b- 3xy/4
\end{equation}
possess the two invariant hyperbolas
$$
\Phi_1^\varepsilon(x,y) =-2b + x y=0,\quad
\Phi_2^\varepsilon(x,y) =
-2b+ y(x+\varepsilon y)=0.
$$
It remains to determine the invariant polynomials which govern the
conditions $c=0$ and $g=1/4$. We observe that for systems
\eqref{sys:eta=0-theta=0-mu=0-N-ne0-hyp-a} we have
$\gamma_{16}=-c(g-1)^2x^3/2$ and $\delta_6=(g-1)(4g-1)x^2/2$.

To deduce that the hyperbola $\Phi(x,y)=-4b + 2 x y=0$ could not
be triple it is sufficient to calculate $\mathscr{E}_2$ for
systems
\eqref{sys:eta=0-theta=0-mu=0-N-ne0-hyp-a2-double}:
$$
\mathscr{E}_2=-\frac{135x^{15}}{65536}\Phi(x,y)^2(5b-3xy)(17b-7xy)
$$
and to observe that the cofactor of $\Phi(x,y)^2$ could not vanish
along the curve $\Phi(x,y)=0$. This leads to the statement $(i.2)$
of the lemma.

\subsubsection*{Subcase $\beta_{12}=0$.} Then $g=1/3$ and by
Lemma \ref{lem:eta=0-theta=0-mu=0-N-ne0-Conics} at least one
 invariant hyperbola exists if and only if $c=0$, $
a\le0$ and $b\ne0$. On the other hand for systems
\eqref{sys:eta=0-theta=0-mu=0-N-ne0} with $d=0$ and $g=1/3$ we
calculate
$$
\gamma_{16}=-2cx^3/9,\quad \gamma_{17}=32ax^2/9, \quad
\mathcal{R}_{11}= - 4 b x^4 /3
$$
Therefore the condition $c=0$ (respectively $b\ne0$) is
equivalent to $\gamma_{16}=0$ (respectively
$\mathcal{R}_{11}\ne0$). Considering the statement $(ii)$ of Lemma
\ref{lem:eta=0-theta=0-mu=0-N-ne0-Conics} we examine two
possibilities: $\gamma_{17}\ne0$ (i.e. $a\ne0$) and
$\gamma_{17}=0$ (i.e. $a=0$).


(1) \emph{Possibility $\gamma_{17}\ne0$.} By Lemma
\ref{lem:eta=0-theta=0-mu=0-N-ne0-Conics} in this case we arrive
at systems \eqref{sys:eta=0-theta=0-mu=0-N-ne0-hyp-b} possessing
the two invariant hyperbolas
$\Phi_{1,2}(x,y)=3b \pm \sqrt{-3a\,} y - x y=0$.
 We claim that none of the hyperbolas
could be double. Indeed calculating $\mathscr{E}_2$ (see Lemma
\ref{lem:Ek}) we obtain:
$$
\mathscr{E}_2=-\frac{2560 (x^2\pm 3a)^6}{177147}\Phi_1\Phi_2(2 b x
- x^2 y \pm a y) \big[3 b x^2 - x^3 y \pm 9a (xy-b)\big].
$$
So each hyperbola appears as a factor of degree one and we could
not increase there degree because of $b\ne0$. This proves our
claim and we arrive at the statement $(ii)$ of the lemma.


(2) \emph{Possibility $\gamma_{17}=0$.} In this case we
have $a=0$ and this leads to the systems
\begin{equation} \label{sys:eta=0-theta=0-mu=0-N-ne0-hyp-triple}
\frac {dx}{dt}= x^2/3,\quad
 \frac {dy}{dt}= b- 2xy/3,
\end{equation}
possessing the hyperbola $\Phi(x,y)=-3b + x y=0$. Calculating
$\mathscr{E}_2$ for this systems we obtain that $\Phi(x,y)$ is a
triple factor of $\mathscr{E}_2$. According to Lemma \ref{lem:Ek}
this hyperbola is triple, as it is shown by the following
perturbed systems:
\begin{equation} \label{sys:eta=0-theta=0-mu=0-N-ne0-hyp-triple-prt}
\frac {dx}{dt}= -12b^2\varepsilon^2+ x^2/3,\quad
 \frac {dy}{dt}= b-2xy/3+3b\varepsilon^2y^2,
\end{equation}
possessing the three distinct invariant hyperbolas:
$$
\Phi_{1,2}=-3b\pm 3b\varepsilon y+ xy=0,\quad \Phi_3=-3b +
y(x-3b\varepsilon^2y).
$$
So we arrive at the statement (iii) of Lemma
\ref{lem:main-eta=0,theta=0,Nne0} and this completes the proof
of this lemma.
\end{proof}


\subsubsection{Case $N=0$} Considering \eqref{val:theta,mu_0} the condition
$N=0$ implies $h=0$ and $g=\pm1$. On the other hand for
\eqref{sys:eta=0-Mne0} with $h=0$ we have $\beta_{13}=(g-1)^2x^2/4$ and
we consider two cases: $\beta_{13}\ne0$ and $\beta_{13}=0$.


\subsubsection*{Subcase $\beta_{13}\ne0$.} Then $g-1\ne0$ (this implies
$g=-1$) and due to a translation we may assume $e=f=0$.
So we obtain the  family of systems
\begin{equation} \label{sys:eta=0-N=0-beta15-ne0}
\frac {dx}{dt}= a+cx+dy-x^2,\quad
 \frac {dy}{dt}= b -2xy.
\end{equation}

\begin{lemma} \label{lem:eta=0-N=0-beta15-ne0-Conics}
A system \eqref{sys:eta=0-N=0-beta15-ne0} possesses at least one
 invariant hyperbola if and only if $d=0$,
 $16a+3c^2 =0$ and $b\ne0$.
\end{lemma}

\begin{proof}
We again assume that the quadratic part of an invariant
hyperbola has the form $ 2xy$ and considering the equations
\eqref{Eqs:gen} for systems
\eqref{sys:eta=0-N=0-beta15-ne0} we obtain
\begin{gather*}
 t=1,\quad s=u=q=0,\quad p=-2b/3,\quad r=-c/2,\quad  U= -3,\\
V=0,\quad W= c +r/2,\quad  Eq_7=2d,\quad Eq_9=(16a+3c^2)/8,\\
 Eq_1=Eq_2=Eq_3=Eq_4=Eq_5= Eq_6=Eq_8= Eq_{10}=0.
\end{gather*}
Therefore the conditions $Eq_7=0$ and $Eq_9=0$ yield $d=0$ and
$16a+3c^2=0$. In this case we obtain the systems
\begin{equation} \label{sys:eta=0-N=0-beta15-ne0-hyp-1}
\frac {dx}{dt}= -3c^2/16+cx -x^2,\quad
 \frac {dy}{dt}= b -2xy,
\end{equation}
which possess the invariant hyperbola
$$
\Phi(x,y)=-2b/3-cy/2+2xy=0.
$$
Obviously this conic is irreducible if and only if $b\ne0$. So
Lemma \ref{lem:eta=0-N=0-beta15-ne0-Conics} is proved.
\end{proof}

\subsubsection*{Subcase $\beta_{13}=0$.} Then $g=1$
 and due to a translation we may assume
$c=0$. So we obtain the following family of systems
\begin{equation} \label{sys:eta=0-N=0-beta15=0}
\frac {dx}{dt}= a+ dy+x^2,\quad
 \frac {dy}{dt}= b +ex+fy.
\end{equation}

\begin{lemma} \label{lem:eta=0-N=0-beta15=0-Conics}
A system \eqref{sys:eta=0-N=0-beta15=0} could not possess a finite number
of invariant hyperbolas. And it has 1-parameter family 
of invariant hyperbolas  if and only if $d=e=0$ and $4a+f^2=0$.
\end{lemma}

\begin{proof} Considering the equations
\eqref{Eqs:gen} and the fact that the quadratic part of an invariant
hyperbola has the form $ 2xy$, for systems
\eqref{sys:eta=0-N=0-beta15=0} we calculate
\begin{gather*}
 t=1,\quad s=u= 0,\quad  U=1,\quad V=0,\quad W= f -r/2,\\
 Eq_5=2e,\quad Eq_7=2d,\quad  Eq_1=Eq_2=Eq_3=Eq_4=Eq_6=0.
\end{gather*}
Therefore the conditions $Eq_5=0$ and $Eq_7=0$ yield $d=e=0$ and
then we have
$$
Eq_8=2b-p- fq+qr/2,\quad Eq_9=(4 a + r^2)/2,\quad Eq_{10}=a q + b r -
 p (2 f - r)/2.
$$
The equations $Eq_8= Eq_{10}=0$ have a common solution with
respect to the parameter $q$ only if
$$
 \operatorname{Res}_q(Eq_8,Eq_{10})=-2 a b + p(a + f^2) - f r(b + p) + r^2(2
b + p)/4=0.
$$
On the other hand in order to have a common solution of the above
equations with respect to $r$ the following condition is
necessary:
$$
 \operatorname{Res}_r\big(Eq_9,\operatorname{Res}_q(Eq_8,Eq_{10})\big)
= (4 a + f^2) (4 a b^2 + f^2 p^2)/4=0.
$$
We claim, that the condition $4 a + f^2=0$ is necessary for the
existence of an invariant hyperbola.

Indeed, supposing $4 a + f^2\ne0$ we deduce that the condition
$4 a b^2 + f^2 p^2=0$ must hold.


(1) Assume first $f\ne0$. If $b=0$ then we obtain $p=0$ and
the equation $Eq_{10}=0$ gives $aq=0$. In the case $q=0$ we obtain
a reducible conic. If $a=0$ then the equation $Eq_9=0$ implies
$r=0$ and we again get a reducible conic.

Thus $b\ne0$ and hence $a\le0$. We set $a=-z^2\le0$ and then
$r=\pm2z$ and $p=\pm2bz/f$. It is not too hard to convince
ourselves that all four possibilities lead either to reducible
conics, or to the equality $4 a + f^2=0$, which contradicts
our assumption.

(2) Suppose now $f=0$. This implies $ab=0$ and since
$b\ne0$ (otherwise we obtain degenerate systems) we have $a=0$ and
this again contradicts to $4 a + f^2\ne0$. This completes the
proof of our claim.

Thus $4 a + f^2=0$ and setting $a=-f^2/4$ we arrive at the family
of systems
\begin{equation} \label{sys:eta=0-N=0-beta15=0-hyp-1}
\frac {dx}{dt}= -f^2/4+x^2,\quad
 \frac {dy}{dt}= b +f y,
\end{equation}
which possess the  family of invariant hyperbolas
$$
\Phi (x,y)= (4 b - f q)/2 + q x + f y + 2 x y=0 , \\
$$
depending on the free parameter $q$. Since the corresponding
determinant $\Delta$ (see Remark \ref{rem:Delta-ne0=>irred}) for
this family equals $fq-2 b$, we conclude that all the conics
are irreducible, except the hyperbola, for which the equality
$fq-2 b=0$ holds. Thus the lemma is proved. \end{proof}

We observe that in the above systems we may assume $b=1$. Indeed,
if $b=0$ then $f\ne0$ (otherwise we obtain a degenerate system) and
therefore due to the translation $y\to y+b'/f $ with $b'\ne0$ and
the addition rescaling $y\to b'y$ we obtain $b'=1$. Moreover, in this
case we may assume $f\in\{0,1\}$ due to rescaling
$(x,y,t)\mapsto(fx, fy, t/f)$ in the case $f\ne0$.

\begin{lemma} \label{lem:main-eta=0,theta=0,N=0}
Assume that for a quadratic system \eqref{sys:QSgenCoef} the
conditions $\eta=0$, $M\ne0$, $\theta=0$ and $N=0$ hold. Then
this system could possess either a single invariant hyperbola,
or a family of such hyperbolas. More precisely this system
possesses
\begin{itemize}

\item [(i)] one simple invariant hyperbola if and only if
 $\beta_{13}\ne0$, $ \gamma_{10}=\gamma_{17}=0$ and
$\mathcal{R}_{11}\ne0$;
 \item [(ii)] one family of invariant hyperbolas
 if and only if
 $\beta_{13}=\gamma_9=\tilde\gamma_{18}=\tilde\gamma_{19}=0$.
\end{itemize}
\end{lemma}


\begin{proof}
Assume that for a quadratic system \eqref{sys:QSgenCoef}
the conditions $\eta=0$, $M\ne0$ $\theta=0$ and $N=0$ hold.

\subsubsection*{Subcase $\beta_{13}\ne0$.} In this case we consider systems
 \eqref{sys:eta=0-N=0-beta15-ne0} for which we calculate
\begin{gather*}
 \gamma_{10}=  14d^2,\quad
 \mathcal{R}_{11}= -12bx^4+6dxy^2(cx+dy),\\
 \gamma_{17}= 8 (16 a + 3 c^2) x^2 - 4 d y (14 c x + 9 d y).
\end{gather*}
So for $ \gamma_{10}=\gamma_{17}=0$ and $\mathcal{R}_{11}\ne0$
we obtain systems \eqref{sys:eta=0-N=0-beta15-ne0-hyp-1} possessing the
hyperbola $\Phi(x,y)=-2b/3-cy/2+2xy=0$. We claim that this
hyperbola is a simple one. Indeed calculating $\mathscr{E}_2$ we
obtain that the polynomial $\Phi(x,y)$ is a factor of degree one
in $\mathscr{E}_2$. So setting $y=-4b/(3(c-4x))$ (i.e.
$\Phi(x,y)\equiv0$) we obtain
$$
\frac{\mathscr{E}_2}{\Phi(x,y)}
= -2^{-24} 5 b^3 (c - 4 x)^3 (3 c -4 x)^{12}/3\ne0
$$
because $b\ne0$. So the hyperbola above could not be double and
this proves our claim.

Thus the statement (i) of lemma is proved.

\subsubsection*{Subcase $\beta_{13}=0$.} Then we consider systems
\eqref{sys:eta=0-N=0-beta15=0} and we calculate
$$
\gamma_9=  -6d^2,\quad
 \tilde\gamma_{18}=8e x^4,\quad
 \tilde{\gamma}_{19} = 4(4a+f^2)x.
$$
So the conditions $d=e=0$ are equivalent to
$\gamma_9=\tilde\gamma_{18}=0$ and $4a+f^2=0$ is equivalent to
$\tilde\gamma_{19}=0$.
Considering Lemma \ref{lem:eta=0-N=0-beta15=0-Conics} we arrive at
the statement (ii).

It remains to observe that for systems
\eqref{sys:eta=0-N=0-beta15=0} with $d=e=0$ and $a=-f^2/4$ we have
$\gamma_{17}=8f^2x^2$ and this invariant polynomial governs the
condition $f=0$.
As all the cases are examined, Lemma
\ref{lem:main-eta=0,theta=0,N=0} is proved.
\end{proof}

 To complete the proof of the Main Theorem we remark,
that both generic families of quadratic systems (with three and
with two distinct real infinite singularities) are examined and
now we could compare the obtained results with the statements of
the Main Theorem.

So comparing the statements of Lemmas
\ref{lem:main-eta>0,theta-ne0}, \ref{lem:main-eta>0,theta-ne0-2H},
\ref{lem:main-eta>0,theta0,Nne0},
\ref{lem:main-eta>0,theta=0-Nne0-2H} and
\ref{lem:main-eta>0,theta0,N=0} with the conditions given by
Figure \ref{diagr:eta-poz}, it is not too difficult to
conclude that the statement (B)(1) of the Main
Theorem is valid.

Analogously, comparing the statements of Lemmas
\ref{lem:main-eta=0,theta-ne0}, \ref{lem:main-eta=0,theta=0,Nne0}
and \ref{lem:main-eta=0,theta=0,N=0} with the conditions given by
 Figure \ref{diagr:eta=0} we deduce that the statement
(B2) of the Main Theorem is valid.


\subsection{Systems with infinite number of singularities
at infinity: $C_2=0$}

In this section we construct the conditions for a quadratic
system with $C_2=0$ to possess at least one invariant hyperbola.
So consider the family of quadratic systems \eqref{sys:QSgenCoef}
assuming $C_2=0$ and we prove the next assertion.

\begin{lemma} \label{lem:C20-exist-Hyp}
If for a quadratic
system \eqref{sys:QSgenCoef} the condition $C_2(x,y)=0$ holds,
then this system possesses invariant hyperbola if and only if
$N_7=0$.
\end{lemma}

\begin{proof}
Assume that for a quadratic system \eqref{sys:QSgenCoef}
the condition $C_2(x,y)=0$ is satisfied. Then the line at infinity
is filled up with singularities and according to Lemma
\ref{lem:S1-S5} in this case via an affine transformation and
time rescaling quadratic systems could be brought to the following
systems
\begin{equation} \label{sys:C2=0-Gen}
 \dot x=\hat a+\hat cx+\hat dy+x^2,\quad \dot y=\hat b + xy.
\end{equation}
We observe that for $\hat d=0$ these systems possess two
parallel invariant lines and we consider two subcases: $\hat
d\ne0$ and $\hat d=0$.

\subsubsection{Subcase $\hat d\ne0$}
As it was shown in \cite[page 749]{SchVul08-JDDE} in this case via
some parametrization and using an additional affine transformation
and time rescaling we arrive at the following 2-parameter family
of systems
\begin{equation} \label{sys:C2=0-d-ne0}
 \dot x=a+y+ (x+c)^2,\quad \dot y= xy.
\end{equation}
Considering \eqref{Eqs:gen} for these systems we obtain
$Eq_1=s(2-U)=0$. We claim that $U=2$ due to the condition
$s^2+t^2+u^2\ne0$. Indeed, supposing $U\ne2$ we obtain $s=0$ and then
calculations yield
$$
Eq_2= 2 t ( 2 -U)=0\quad Eq_3= u (2 - U) - 2 t V=0.
$$
Clearly because $U\ne2$ we have $t=u=0$ which contradicts to
$s^2+t^2+u^2\ne0$ and this completes the proof of our claim. So we
assume $U=2$, and  calculations yield $Eq_2=-sV=0$, $ Eq_3=-2tV=0$, 
$Eq_4=-uV=0$. Since $ \Phi(x,y)=0$ must be a conic (i.e.
$s^2+t^2+u^2\ne0$) the above relations imply $V=0$. Then we have
\begin{gather*}
Eq_5=-q + 4 c s - s W=0,\quad Eq_6=-r + 2 s + 4 c t - 2 t W=0,\\
Eq_7=2 t - u W=0,\quad  Eq_8=-2 p + 2 c q + 2 a s + 2 c^2 s - q W=0
\end{gather*}
and this gives
\begin{gather*}
q = s (4 c - W),\quad r = 2 s + 2 c u W - u W^2,\quad t = u W/2,\\
p = s (2 a + 10 c^2 - 6 c W + W^2)/2.
\end{gather*}
Considering the values of the parameters we detected we finally
obtain
\begin{gather*}
Eq_i=0,\quad i=1,2,\ldots,8,\quad Eq_{10}= s (2 c - W) (4 a + 4 c^2 -
4 c W + W^2)/2=0,\\
Eq_9=4 c s + ( a u-3 s + c^2 u) W - 2 c u W^2 + u W^3=0.
\end{gather*}
We observe that $s\ne0$, otherwise we obtain $ \Phi(x,y)= u y (2 c W
- W^2 + W x + y)$, i.e. the conic becomes reducible. So we
consider the two possibilities defined by the equality $ (2 c -
W)\big[4 a + (W-2 c)^2\big]=0. $

\subsubsection*{Possibility $W=2c$.} Then we obtain $Eq_{10}=0$ and
$Eq_9=2 c ( a u-s + c^2 u)=0$.

\subsubsection*{Case $c=0$.} In this case we obtain the 1-parameter
family of systems
\begin{equation} \label{sys:C2=0-d-ne0-c=0}
 \dot x=a+y+ x^2,\quad \dot y= xy
\end{equation}
which possess the  2-parameter family of invariant conics
$ \Phi(x,y)=a s + 2 s y + s x^2 + u y^2=0 $ which will be of
hyperbolic type if and only if the condition $su<0$ holds.
Moreover following Remark \ref{rem:Delta-ne0=>irred} we calculate
$\Delta= s^2 (au-s)$ and this conic is irreducible if and only if
$s(au-s)\ne0$.

Since $su<0$ we may set a new parameter $u=-sm^2$ and this leads
to the 1-parameter family of hyperbolas
\begin{equation} \label{Hyp:7}
\widetilde \Phi(s,x,y)=a + 2 y + x^2 - m^2 y^2=0.
\end{equation}

\subsubsection*{Case $a u-s + a u + c^2 u=0$.}
Then $s=(a + c^2) u$ and systems \eqref{sys:C2=0-d-ne0} possess the following
invariant conic $ \Phi(x,y)=(a + c^2) (a + c^2 + 2 c x + 2 y) + (a
+ c^2) x^2 + 2 c x y + y^2 =0$ for which we calculate $\Delta=0$,
i.e. by Remark \ref{rem:Delta-ne0=>irred} this conic is
reducible.


\subsubsection*{Possibility $4 a + (W-2 c)^2= 0$.}
If $a=0$ then $W=2c$ and as it was shown above for the
existence of a hyperbola it is necessary $c=0$. So we arrive at
the particular case of the family of hyperbolas \eqref{Hyp:7}
defined by the condition $a=0$. Therefore we consider two cases:
$a<0$ and $a>0$.

\subsubsection*{Case $a<0$.} Then we may assume $a=-k^2$ and
after the rescaling $(x,y,t)\mapsto(kx, k^2 y, t/k)$ we obtain the
systems
\begin{equation} \label{sys:C2=0-H10-a<0}
 \dot x=y-1+ (x+c)^2,\quad \dot y= xy,
\end{equation}
for which we have $W=2(c\pm 1)$ and we obtain\ $ Eq_{10}=0$,
$ Eq_9=2(c\pm3)\big[(c\pm 1)^2-s\big]=0$. We consider the two
subcases given by two factors of $Eq_9$.


\textbf{(1)} \emph{Subcase $c\pm3=0$}. We may assume $c>0$ because
of the rescaling $(x,y,t)\mapsto(-x, y, -t)$ in the above
systems. Therefore we set $c=3$ and then systems
\eqref{sys:C2=0-H10-a<0} could be brought to system
\eqref{sys:C2=0-d-ne0} with $c=0$ and $a=-1$ via the transformation
$\ (x,y,t)\mapsto\big(2(x-1),\ 4(y-x-1),\ t/2\big). $\ So we
arrive at the system \eqref{sys:C2=0-d-ne0-c=0} with $a=-1$ and as
it was shown above this system possesses the family of hyperbolas
\eqref{Hyp:7} with $a=-1$.


\textbf{(2)} \emph{Subcase $(c\pm 1)^2-s=0$}. Then $s=(c\pm
1)^2$ and this leads to the reducible conics
$\ \Phi(x,y)= (c^2-1 \pm x + c x + y)^2=0$.

\subsubsection*{Case $a>0$.} Then we may assume $a= k^2$ and
applying the same rescaling as above we arrive at the family
systems $ \dot x=1+y + (x+c)^2$,  $\dot y= xy$.
So we have $W=2(c\pm i)$ and we obtain
$\ Eq_{10}=0\quad Eq_9=2(c\pm3i)\big[(c\pm i)^2-s\big]=0$.
Since $c\in\mathbb{R}$ we obtain $s=(c\pm i)^2$ and this again leads to the reducible
conics $\ \Phi(x,y)= ( c^2-1 \pm i x + c x + y)^2=0$.

Thus we detect that in the case $d\ne0$ a system
\eqref{sys:C2=0-d-ne0} could possesses an invariant
hyperbola if and only if either the conditions $c=0$ or $a<0$
(then $a=-1$) and $c^2-9=0$ hold. On the other hand for these
systems we calculate $N_7= c (9 a + c^2)/2$ and we claim that
the above conditions are equivalent to $N_7=0$. Indeed, if $c=0$
or $a=-1$ and $c^2-9=0$ we obtain $N_7=0$. Conversely, assuming
$N_7=0$ we have either $c=0$ or $9 a + c^2=0$. However in the
second case the condition $a\le0$ must hold. If $a=0$ we obtain $c=0$
and we arrive at the first case. If $a<0$ as it was mentioned
earlier due to a rescaling we may assume $a=-1$ (see systems
\eqref{sys:C2=0-H10-a<0}) and then we obtain $c^2+9a=c^2-9=0$ and
this completes the proof of our claim.


\subsubsection{Subcase $\hat d=0$} In this case systems
\eqref{sys:C2=0-Gen} become as systems
\begin{equation} \label{sys:C2=0-d=0-Gen}
 \dot x= \hat a+\hat cx+ x^2,\quad \dot y=\hat b + xy,
\end{equation}
for which following \cite{SchVul08-JDDE} we calculate the value of
invariant polynomial $H_{12}=-8 \hat a^2 x^2$ and we consider two
possibilities: $\hat a\ne0$ and $\hat a=0$.

\subsubsection*{Possibility $\hat a\ne0$.} As it
was shown in \cite[page 750]{SchVul08-JDDE} in this case via an
affine transformation and time rescaling after some additional
parametrization we arrive at the following 2-parameter family of
systems
\begin{equation} \label{sys:C2=0-d=0-k-ne0}
 \dot x=a+ (x+c)^2,\quad \dot y= xy
\end{equation}
for which the condition $H_{12}=-8 (a + c^2)^2 x^2\ne0$ must hold.


Next, to determine the conditions for the existence of a
hyperbola as earlier we apply the equations
\eqref{Eqs:gen}. Since the quadratic parts of the above
systems coincide with quadratic parts of systems
\eqref{sys:C2=0-d-ne0} by the same reasons from the first four
equations \eqref{Eqs:gen} we determine that $s\ne0$, $U=2$ and
$V=0$ and then calculations yield
\begin{gather*}
Eq_5=-q + 4 c s - s W=0,\quad Eq_6=-r + 4 c t - 2 t W=0,\\
Eq_7=-u W=0,\quad Eq_8=-2 p + 2 c q + 2 a s + 2 c^2 s - q W=0.
\end{gather*}
So we obtain $q = s (4 c - W)$, $r =2 t (2 c - W)$, $p = s (2
a + 10 c^2 - 6 c W + W^2)/2$, $uW=0 $ and we consider two cases:
$u=0$ and $u\ne0$.

\subsubsection*{Case $u=0$.} In this case we have $Eq_i=0$,
$i=1,2,\ldots,8$ and
\begin{gather*}
 Eq_9=2 t (a + c^2 - 2 c W + W^2)=0,\\
 Eq_{10}=s (2 c - W) (4a + 4 c^2 - 4 c W + W^2)=0
\end{gather*}
and we observe that $t\ne0$ otherwise we obtain
$$
\Phi(x,y)= s (2 a + 10 c^2 - 6 c W + W^2 + 8 c x - 2 W x + 2
x^2)/2=0,
$$
i.e. $\Phi(x,y)$ is a product of two parallel lines. It was
mentioned above that the condition $s\ne0$ also must hold, i.e.
$st\ne0$ and we calculate $\operatorname{Res}_W(Eq_9,Eq_{10}) =2 s^2
t^3 (a + c^2)^2 (9 a + c^2) $\quad and clearly for the existence
of a common solution of the equations $Eq_9=Eq_{10}=0$ the
condition $(a + c^2)^2 (9 a + c^2)=0$ is necessary. However the
condition $H_{12}\ne0$ implies $a + c^2\ne0$ and therefore we obtain
$9 a + c^2=0$.


So $a=-c^2/9$ and we detect that in this case the polynomials
$Eq_9$ and $Eq_{10}$ have as a common factor $4 c - 3W$. Therefore
we obtain $W=4c/3$ and we arrive at the systems
$$
 \dot x= (2 c + 3 x) (4 c + 3 x)/9,\quad \dot y= xy,
$$
which possess the following family of hyperbolas $\ \Phi(x,y)= 16
c^2 s + 24 c s x + 9 s x^2 + 12 c t y + 18 t x y=0. $\ In order to
have irreducible invariant conics we determine $\Delta=-324 c^2 s
t^2\ne0$. So $s\ne0$ and setting a new parameter $m=6t/s$ we
arrive at the 1-parameter family of hyperbolas
$$
\Phi(x,y)= 16 c^2 + 24 c x + 2 c m y+ 9 x^2 + 3 m x y=0.
$$

\subsubsection*{Case $u\ne0$.} Then we obtain $W=0$ and we
calculate
$$
Eq_i=0,\quad i=1,2,\ldots,8,\quad Eq_9= 2 (a + c^2) t=0,\quad
 Eq_{10}=4 c (a + c^2) s=0
$$
and since $s\ne0$ and $a + c^2\ne0$ (due to $H_{12}\ne0$) we obtain
$t=c=0$. So we arrive at the 1-parameter family of systems
$ \dot x=a + x^2$, $\dot y= xy$,
which possess the  family of conics
$$
\Phi(x,y)= a s + s x^2 + u y^2=0.
$$
Clearly these conics are of hyperbolic type if $su<0$ and they
are irreducible if in addition we have $a\ne0$. So setting
$u=-m^2s$ we obtain the following 1-parameter family of hyperbolas:
$$
\Phi(x,y)= a + x^2 -m^2 y^2=0.
$$


Thus we detect that in the case $\hat d=0$ and $\hat a\ne0$ a
system \eqref{sys:C2=0-d=0-Gen} could be brought to
\eqref{sys:C2=0-d=0-k-ne0} which possess an invariant hyperbola if
and only if the condition $c (9 a + c^2)=0$ holds. On the other
hand for these systems we have $N_7= c (9 a + c^2)/2$ and we
deduce that in the case under consideration Lemma
\ref{lem:C20-exist-Hyp} is valid.


\subsubsection*{Possibility $\hat a=0$.} This condition
implies $\hat b\ne0$ (otherwise we obtain degenerate systems
\eqref{sys:C2=0-d=0-Gen}). So we may assume $\hat b=1$ due to the
rescaling $y\to \hat by$ and this leads to the 1-parameter family
of systems (we set $\hat c=c$)
\begin{equation} \label{sys:C2=0-d=k=0}
 \dot x= cx +x^2,\quad \dot y=1 + xy,
\end{equation}
And again, since the quadratic parts of the above systems
coincides with quadratic parts of systems
\eqref{sys:C2=0-d-ne0} by the same reasons from the first four
equations \eqref{Eqs:gen} we determine that $s\ne0$, $U=2$ and
$V=0$ and then calculations yield
\begin{gather*}
Eq_5=-q + 2 c s - s W=0,\quad Eq_6=-r + 2 c t - 2 t W=0,\\
Eq_7=-u W=0,\quad Eq_8=-2 p + c q + 2 t - q W=0.
\end{gather*}
So we obtain $q = s (2 c - W)$, $r =2 t (c - W)$,
$p = (2 c^2s + 2 t - 3 c s W + s W^2)/2$, $uW=0 $ and we claim that the
condition $u=0$ must hold. Indeed supposing $u\ne0$ we obtain
$W=0$ and this implies $Eq_9=2u=0$ and this contradiction proves
our claim. So $u=0$ and calculations yield $Eq_i=0$,
$i=1,2,\ldots,8$, $Eq_9=-2 t (c - W) W=0$, and
$ Eq_{10}=(4 ct - 2 c^2 s W - 6 t W + 3 c s W^2 - s W^3)/2=0$.
 We observe that $t\ne0$ otherwise we obtain
$\Phi(x,y)= s (2 c^2 - 3 c W + W^2 + 4
c x - 2 W x + 2 x^2)/2=0$, i.e. $\Phi(x,y)$ is a product of two
parallel lines. So we obtain $W(c - W)=0$ and we have to consider
the two subcases given by these two factors. However we obtain
$Eq_{10}=2 c t=0$ if $W=0$ and $Eq_{10}=-c t=0$ if $W=c$ and
therefore due to $t\ne0$ in both cases we obtain $c=0$. So we
arrive at the system
$$
 \dot x= x^2,\quad \dot y=1 + xy,
$$
which possess the following family of hyperbolas $\ \Phi(x,y)= t +
s x^2 + 2 t x y=0 $\ and for the irreducibility of these conics
the condition $t\ne0$ is necessary. Then setting $m=s/t$ we obtain
the 1-parameter family of hyperbolas
$$
\Phi(x,y)= 1 + m x^2 + 2 x y=0.
$$
Thus in the case $\hat d= \hat a=0$ a system
\eqref{sys:C2=0-d=0-Gen} could be brought to
\eqref{sys:C2=0-d=k=0} which possess an invariant hyperbola if and only
if the condition $c=0$ holds. On the other
hand for these systems we have $N_7=-16 c^3$ and this completes
the proof of Lemma \ref{lem:C20-exist-Hyp}.
\end{proof}

Then, we conclude that the Main Theorem is completely proved.



\subsection*{Acknowledgments}
The first author is supported by CNPq grant ``Projeto Universal''
472796/2013-5. The first and the third authors are partially
supported by FP7-PEOPLE-2012-IRSES-316338. The second author is
supported by CAPES CSF-PVE-88887.068602/2014-00. The third author is
partially supported by the grant 12.839.08.05F from SCSTD of ASM.



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