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\begin{document}

\title{Conic Sections}
\author{David Pierce}
\date{January 3, 2008; edited, \today}


\publishers{Mathematics Department\\
Mimar Sinan Fine Arts University\\
Istanbul\\
\url{http://mat.msgsu.edu.tr/~dpierce/}}
  \maketitle
\tableofcontents

\listoffigures
%\newpage
\chapter{Introduction}

These notes are about the plane curves known as \tech{conic sections.}
The mathematical presentation is mainly in the `analytic' style whose
origins are sometimes said to be the \emph{Geometry}
\cite{Descartes-Geometry} of Ren\'e Descartes.
However, the features of conic sections presented in
\S~\ref{sect:analysis} below
were apparently known to mathematicians of the eastern
Mediterranean in ancient times.  Accordingly,
\S~\ref{sect:background} below contains a review what I
have been able to find out about the ancient knowledge.  I try to give
references to the original
texts (or translations of them).  Meanwhile, I list some relevant
approximate dates; the ancient dates are selected from
\cite[pp.~685~f.]{MR0234791}: 
\begin{compactitem}
\newcommand{\datewidth}{1.2cm}
  \item[]
\makebox[\datewidth][r]{\textsc{b.c.e.}\ 350:} Menaechmus on conic sections;
\item[]
\makebox[\datewidth][r]{300:} Euclid, \emph{Elements;}
\item[]
\makebox[\datewidth][r]{225:} Apollonius, \emph{Conics;}
\item[]
\makebox[\datewidth][r]{212:} death of Archimedes;
\item[]
\makebox[\datewidth][r]{\textsc{c.e.}\ 320:} 
Pappus, \emph{Mathematical Collections;}
\item[]
\makebox[\datewidth][r]{560:} Eutocius, commentaries on Archimedes;
\item[]
\makebox[\datewidth][r]{1637:} Descartes, \emph{Geometry.}
\end{compactitem}

The reader of these notes may agree that the conic sections are worthy
of study, independently of any
application.  However, Isaac Newton
(1643--1727), for example, could not have developed his theory of
gravitation \cite{MR1965137} without knowing what the Ancients knew
about conic sections.\footnote{An inverse-square law of gravitation
  causes planetary orbits to be conic sections.  Newton showed this,
  apparently using such knowledge as can be found in Apollonius.
  It may be that Newton inferred, from ancient secondary sources, that
  the ancient scientists themselves were aware of an inverse-square
  law of gravity \cite[\S~11.7]{MR2038833}.}  

\chapter{Background}\label{sect:background}

\section{Definitions}

A \defn{cone} and its associated \defn{conic surface} are
determined by the following data:  
\begin{compactenum}[1)]
  \item
a circle, called the \defn{base} of the cone;
\item
a point, called the \defn{vertex} of the cone and the conic surface;
the vertex must not lie in the plane of the base.
\end{compactenum}
The conic surface consists of the points on the lines that pass through the
vertex and the circumference of the base.  The cone itself is the solid
figure bounded by the surface and the base.  See Figure~\ref{fig:cone}.

\begin{figure}[ht!]
  \begin{center}
    \begin{pspicture}(-3,-3)(1.6,1.6)
%\psgrid(-3,-3)(2,2)
      \psellipse(-0.5,-2)(1.5,0.5)
\psdots(0,0)(-0.17,-2.48)
\psline(-3,-2.9)(0,0)(1.5,1.45)
\psline(1.4,-2.8)(0,0)(-0.7,1.4)
\psline(-0.2,-3)(0.1,1.5)
%\uput[dl](-1,-2.5){base}
\uput{8pt}[l](0,0){vertex}
    \end{pspicture}
    \begin{pspicture}(-3,-3)(1.6,1.6)
%\psgrid(-3,-3)(2,2)
      \psellipse(-0.5,-2)(1.5,0.5)
\psdots(0,0)
\psline(-1.93,-1.87)(0,0)
\psline(0.98,-1.95)(0,0)
%\psline(-0.2,-3)
\uput[dl](-1,-2.5){base}
%\uput{8pt}[l](0,0){vertex}
    \end{pspicture}
  \end{center}
\caption{A conic surface and cone}\label{fig:cone}
\end{figure}

The definitions of cone and conic surface can be found at the
beginning of the treatise \emph{On Conic Sections}
\cite{Heath-Apollonius, MR0053831, MR1660991, MR13:419b}, by
Apollonius of Perga.\footnote{Perga 
  or Perge was near what is now Antalya; its remains are well worth a
  visit.}  The \defn{axis} of the cone is the line joining the vertex
to the center of the base.  There is no assumption that the axis is
perpendicular to the base; if it is, then the
cone is \defn{right;} otherwise, the cone is \defn{oblique.}

A \defn{conic section} is the intersection of a plane with a conic
surface.  The discovery of conic sections (as objects worthy of study)
is generally\footnote{See for example \cite[p.~283~f.,
    n.~\emph{a}]{MR13:419b} or \cite[p.~1]{MR0245397}.}
attributed to Apollonius's predecessor
Menaechmus.
However, there are three kinds of conic sections: the \defn{ellipse,}
the \defn{parabola,} and the \defn{hyperbola.}  According to
Eutocius \cite[pp.~276--281]{MR13:419b}, Apollonius was the first mathematician
to show that each kind of conic section can be obtained from
\emph{every} conic surface.  Indeed, the names of the three kinds of conic
sections appear \cite[p.~283~f., n.~\emph{a}]{MR13:419b} to be due to
Apollonius as well.  The names are meaningful  
in Greek and reflect the different geometric properties of the
sections, in a way shown in \S~\ref{sect:analysis}.

\section{Motivation}

Menaechmus used conic sections to solve the problem of
\defn{duplicating the cube.}  Suppose a cube is given, with volume
$V$; how can a cube be
constructed with volume $2V$?  We can give a symbolic answer:  If the
side of the original cube has length $s$, then the new cube must have
side of length $s\sqrt[3]2$.  But how can a side of that length be
\emph{constructed?}

The corresponding problem for squares can be solved as follows.
Suppose 
$AB$ is a diameter of a circle, and $C$ is on $AB$, and $D$ is on the
circumference of the circle, and $CD\perp AB$.  Then the square on
$CD$ is equal in area to the rectangle whose sides are $AC$ and $BC$.
More symbolically, if lengths are as in Figure~\ref{fig:circle}, 
\begin{figure}[ht!]
\psset{unit=8mm}
  \begin{center}
    \begin{pspicture}(-4,-0.5)(4,3.2)
      \psline[showpoints=true](-3,0)(3,0)
      \psline[showpoints=true](-1,0)(-1,2.828)
      \psarc[linewidth=1.6pt](0,0){3}{0}{180}
      \uput[dl](-3,0){$A$}
      \uput[d](-1,0){$C$}
      \uput[dr](3,0){$B$}
      \uput[ul](-1,2.828){$D$}
      \uput[d](-2,0){$a$}
      \uput[d](1,0){$b$}
      \uput[r](-1,1.414){$x$}
    \end{pspicture}
  \end{center}
  \caption{The method of finding a mean proportional}\label{fig:circle}
\end{figure}
then
\begin{equation*}
\frac ax=\frac xb,\qquad\text{ or }\qquad  ab=x^2,
\end{equation*}
so that
\begin{equation*}
  \frac{x^2}{a^2}=\frac ba.
\end{equation*}
In particular, if $b/a=2$, then a square with side of length $x$ has
area twice that of a square with side of length $a$.

Suppose instead we have
\begin{equation}\label{eqn:2}
\frac ax=\frac xy=\frac yb.  
\end{equation}
Then
\begin{equation*}
\frac{x^3}{a^3}=\frac xa\cdot\frac yx\cdot\frac by=\frac ba.
\end{equation*}
If $b/a=2$, then a cube with side of length $x$ has volume twice that of a
cube with side of length $a$.  In any case, the several lengths can be
arranged as in Figure~\ref{fig:circle2}.  There, angle $ACB$ is right,
and $BCD$ and $ACE$ are diameters of the indicated circles.  

\begin{figure}[ht!]
  \begin{center}
    \begin{pspicture}(-1.2,-2.4)(2,2)%(-2.5,-4)(2.16,3)
\psset{unit=0.6cm}
      \psline[linewidth=1.6pt](-2,0)(0,0)(0,-4)
      \psline(3.16,0)(0,0)(0,2.52)(3.16,2.52)(3.16,0)
      \pscircle(0.58,0){2.58}%{0}{180}
      \pscircle(0,-0.74){3.26}%{-90}{90}
      \psdots(-2,0)(0,0)(0,-4)(3.16,0)(0,2.52)(3.16,2.52)
      \uput[d](-1,0){$a$}
      \uput[ul](0,-2){$b$}
      \uput[r](0,1.26){$x$}
      \uput[u](1.58,0){$y$}
      \uput[l](-2,0){$A$}
      \uput[dr](0,0){$C$}
      \uput{7pt}[ul](0,-4){$B$}
      \uput[u](0,2.52){$D$}
      \uput[r](3.16,0){$E$}
      \uput[ur](3.16,2.52){$F$}
    \end{pspicture}
  \end{center}
\caption{Two mean proportionals}\label{fig:circle2}
\end{figure}

The problem is, How
can $D$ and $E$ be chosen on the extensions of $BC$ and $AC$ so that
the circles intersect as in Figure~\ref{fig:circle2}?  The solution of
Menaechmus (along with many other solutions) is given 
in the commentary \cite[pp.~288-290]{MR2093668} by Eutocius on the
second volume \emph{On
  the Sphere and the Cylinder} by Archimedes.  In
Figure~\ref{fig:circle2}, if $CDFE$ is a rectangle, then $F$
determines $x$ and $y$.
But by Equations~\eqref{eqn:2}, rearranged, $x$ and $y$ must satisfy
two equations,
\begin{equation*}
  ay=x^2,\qquad bx=y^2.
\end{equation*}
Each of these equations determines a curve, and $F$ is the
intersection of the two curves.  The curves turn out to be conic
sections, namely parabolas.
Points on the curve given by $ay=x^2$ can be
plotted as in Figure~\ref{fig:par-const}.  

\begin{figure}[ht!]
\psset{unit=15mm}
  \begin{center}
    \begin{pspicture}(-2,-3)(4.8,3.5)
      \pscircle(-1,0){1}%{0}{180}
      \pscircle(-0.5,0){1.5}%{0}{180}
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      \psline[linewidth=1.6pt]{->}(-2,0)(4.8,0)
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      \psline(3,-3)(3,3)
      \psline(4,-3)(4,3)
      \psline(0,1.414)(4,1.414)
      \psline(0,2)(4,2)
      \psline(0,2.45)(4,2.45)
      \psline(0,2.82)(4,2.82)
      \psline(0,-1.414)(4,-1.414)
      \psline(0,-2)(4,-2)
      \psline(0,-2.45)(4,-2.45)
      \psline(0,-2.82)(4,-2.82)
      \psdots(0,0)(0,1.414)(0,2)(0,2.45)(0,2.82)
                  (0,-1.414)(0,-2)(0,-2.45)(0,-2.82)
(1,0)(2,0)(3,0)(4,0)
(1,1.414)(2,2)(3,2.45)(4,2.82)
(1,-1.414)(2,-2)(3,-2.45)(4,-2.82)
%\pscurve(0,0)(1,1.414)(2,2)(3,2.45)(4,2.82)(4.5,3)
\psplot[linewidth=1.6pt]{0}{4.5}{2 x mul sqrt}
\psplot[linewidth=1.6pt]{0}{4.5}{2 x mul sqrt -1 mul}
    \end{pspicture}
  \end{center}
\caption{Construction of points of the parabola}\label{fig:par-const}
\end{figure}

If one imagines that the
circles in Figure~\ref{fig:par-const} are not all in the same plane,
but serve as parallel bases of cones bounded by the same conic
surface, then one may be able to see how the curve arises as a
section of that surface.  However, an alternative approach to the
conic sections was given by Pappus of Alexandria
\cite[p.~492--503]{MR13:419a}; it may have been due 
originally to Euclid, although his works on conic
sections are lost. 
We can take the alternative approach as follows.

\chapter{Equations}\label{sect:analysis}

\section{Focus and directrix}\label{subsect:focus}

A \defn{conic section} $\zeta$ is determined by the following data:
\begin{compactenum}[1)]
  \item
a line $d$, called the \defn{directrix} of $\zeta$;
\item
a point $F$ (not on $d$), called the \defn{focus} of $\zeta$;
\item
a positive real number (or distance) $e$, called the
\defn{eccentricity} of $\zeta$.
\end{compactenum}
Then $\zeta$ comprises the points $P$ (in the plane of $d$ and $F$)
such that
\begin{equation*}
  \size{PF}=e\cdot\size{Pd}.
\end{equation*}
Some examples are in Figure~\ref{fig:e}, with the same directrix and
focus, but various eccentricities.  The examples are drawn (by
computer) by means of~\eqref{eqn:r=l} below.  (See also
Figure~\ref{fig:polar}.)  

\begin{figure}[ht!]
\psset{unit=8mm}
  \begin{center}
    \begin{pspicture}(-6,-6)(6,6)
%\psgrid(-6,-6)(6,6)
\psset{plotpoints=200}
      \psclip{\psframe[linecolor=white](-6,-6)(6,6)}
      \psline[linewidth=1.6pt](-1.5,-6)(-1.5,6)
%\uput[l](-1.5,0){$d$}
      \psdot(1.5,0)
%\uput[ur](1.5,0){$F$}
% e = 32
\parametricplot{89}{271}{96 1 32 t cos mul sub div t cos mul 1.5 add 
                         96 1 32 t cos mul sub div t sin mul}
\parametricplot{-88}{88}{96 1 32 t cos mul sub div t cos mul 1.5 add 
                         96 1 32 t cos mul sub div t sin mul}
% e = 16
\parametricplot{87}{273}{48 1 16 t cos mul sub div t cos mul 1.5 add 
                         48 1 16 t cos mul sub div t sin mul}
\parametricplot{-86}{86}{48 1 16 t cos mul sub div t cos mul 1.5 add 
                         48 1 16 t cos mul sub div t sin mul}
% e = 8
\parametricplot{83}{277}{24 1 8 t cos mul sub div t cos mul 1.5 add 
                         24 1 8 t cos mul sub div t sin mul}
\parametricplot{-82}{82}{24 1 8 t cos mul sub div t cos mul 1.5 add 
                         24 1 8 t cos mul sub div t sin mul}
\uput[ur](-0.7,-6){$8$}
\uput[ul](-2.4,-6){$8$}
% e = 4
\parametricplot{76}{284}{12 1 4 t cos mul sub div t cos mul 1.5 add 
                         12 1 4 t cos mul sub div t sin mul}
\parametricplot{-75}{75}{12 1 4 t cos mul sub div t cos mul 1.5 add 
                         12 1 4 t cos mul sub div t sin mul}
\uput[ur](0.05,-6){$4$}
\uput[ul](-3.4,-6){$4$}
% e = 2
\parametricplot{61}{299}{6 1 2 t cos mul sub div t cos mul 1.5 add
                         6 1 2 t cos mul sub div t sin mul}
\parametricplot{-59}{59}{6 1 2 t cos mul sub div t cos mul 1.5 add
                         6 1 2 t cos mul sub div t sin mul}
\uput[ur](1.5,-6){$2$}
\uput[dr](-6,-5){$2$}
% e = 1
\parametricplot{1}{359}{3 1 t cos sub div t cos mul 1.5 add
                        3 1 t cos sub div t sin mul}
%\psplot{0}{6}{6 x mul sqrt}
%\psplot{0}{6}{6 x mul sqrt neg}
\uput[ul](6,-5.85){$1$}
% e = 0.5
\parametricplot{0}{360}{1.5 1 0.5 t cos mul sub div t cos mul 1.5 add
1.5 1 0.5 t cos mul sub div t sin mul}
\uput[r](4.5,0){$\displaystyle\frac12$}
% e = 0.25
\parametricplot{0}{360}{0.75 1 0.25 t cos mul sub div t cos mul 1.5 add 
                        0.75 1 0.25 t cos mul sub div t sin mul}
\uput[r](2.5,0){$\displaystyle\frac14$}
% e = 0.125
\parametricplot{0}{360}{0.375 1 0.125 t cos mul sub div t cos mul 1.5 add 
                        0.375 1 0.125 t cos mul sub div t sin mul}
% e = 0.0625
\parametricplot{0}{360}{0.1875 1 0.0625 t cos mul sub div t cos mul 1.5 add 
                        0.1875 1 0.0625 t cos mul sub div t sin mul}
% e = 0.03125
\parametricplot{0}{360}{0.09375 1 0.03125 t cos mul sub div t cos mul 1.5 add 
                        0.09375 1 0.03125 t cos mul sub div t sin mul}
\endpsclip
    \end{pspicture}
  \end{center}
\caption{Conic sections of different eccentricities}\label{fig:e}
\end{figure}

Suppose we assign a rectangular coordinate system to the plane of
$\zeta$ in which $F$ has the coordinates $(h,k)$, and $d$ is defined by 
\begin{equation*}
Ax+By+C=0
\end{equation*}
(where $A\neq0$ or $B\neq0$).  Then $\zeta$ is defined by
\begin{equation*}
  \sqrt{(x-h)^2+(y-k)^2}=e\cdot\frac{\size{Ax+By+C}}{\sqrt{A^2+B^2}},
\end{equation*}
hence also by
\begin{equation}\label{eqn:rect1}
  (x-h)^2+(y-k)^2=e^2\cdot\frac{(Ax+By+C)^2}{A^2+B^2}.
\end{equation}
This equation is not very useful for showing the shape of
$\zeta$.  By choosing the rectangular coordinate system appropriately,
we can ensure
\begin{equation*}
  (h,k)=(0,0), \quad B=0, \quad A=1,\quad C>0.
\end{equation*}
Then $C$ is the distance between the focus and the directrix,
and~\eqref{eqn:rect1} becomes
\begin{equation}\label{eqn:rect2}
  x^2+y^2=e^2(x+C)^2.
\end{equation}

\section{The polar equation}

Equation~\eqref{eqn:rect2} is nicer than~\eqref{eqn:rect1},
but is still not
the most useful rectangular equation for $\zeta$.
However,~\eqref{eqn:rect2} becomes more useful when converted to polar
form.  Recall the conversion-equations:
\begin{equation*}
\left\{
\begin{aligned}[c]
  x&=r\cos\theta,\\
  y&=r\sin\theta;
\end{aligned}
\right.
\qquad\qquad
\left\{
\begin{aligned}[c]
&r^2=x^2+y^2,\\
&\tan\theta=\frac yx.
\end{aligned}
\right.
\end{equation*}
So the polar form of~\eqref{eqn:rect2} is
\begin{equation*}
  r^2=e^2(r\cos\theta+C)^2,
\end{equation*}
which is equivalent to
\begin{equation}\label{eqn:pm}
  \pm r=e(r\cos\theta+C).
\end{equation}
The plus-or-minus sign here is needed, unless we know that $r$ always
has the sign of
$r\cos\theta+C$, or always has the opposite sign.  It does not.  

However, note well that the same point can have different polar
coordinates; in particular, the same point has polar
coordinates $(r,\theta)$ 
and $(-r,\theta+\mpi)$.  We shall use this fact frequently.
The equation
\begin{equation}\label{eqn:polar0}
  -r=e(r\cos\theta+C)
\end{equation}
is equivalent to
\begin{equation*}
  -r=e(-r\cos(\theta+\mpi)+C).
\end{equation*}
Hence, if $(s,\phi)$ satisfies~\eqref{eqn:polar0}, then
$(-s,\phi+\mpi)$ satisfies
\begin{equation}\label{eqn:polar1}
  r=e(r\cos\theta+C).
\end{equation}
So we can take either~\eqref{eqn:polar0} or~\eqref{eqn:polar1} as
the polar equation for $\zeta$.
We can also derive~\eqref{eqn:pm} directly from the original
definition of $\zeta$; see Figure~\ref{fig:polar}.

\begin{figure}[ht!]
\psset{unit=8mm}
  \begin{center}
      \begin{pspicture}(-3,-1.5)(5,5)
    \psline{->}(-3,0)(5,0)
    \psline{->}(-2,-1.5)(-2,5)
\psdots(0,0)(3,4)(3,0)(-2,4)
\psline(0,0)(3,4)
\psline[linestyle=dashed](3,4)(-2,4)
\psline[linestyle=dashed](3,4)(3,0)
\psarc(0,0){0.5}{0}{53.13}
\uput[ur](0.5,0.15){$\theta$}
\uput[d](-1,0){$C$}
\uput[dr](1.5,2){$r$}
\uput[ul](1.5,2){$\size{PF}$}
\uput[d](1.5,0){$r\cos\theta$}
\uput[ul](0,0){$O$}
\uput[u](3,4){$P$}
\uput[d](3,0){$Q$}
\uput[l](-2,-1){$d$}
\uput[d](0,0){$F$}
\uput[d](0.5,4){$\size{Pd}$}
\uput[u](0.5,4){$C+r\cos\theta$}
\uput[ur](0,-1.5){$r=\size{e(C+r\cos\theta)}$}
  \end{pspicture}
  \end{center}
\caption{Derivation of the polar equation of a conic
  section}\label{fig:polar} 
\end{figure}

We can rewrite~\eqref{eqn:polar1} as
\begin{gather}\label{eqn:z0}
  r=er\cos\theta+eC,\\\notag
r-er\cos\theta=eC,\\\notag
r(1-e\cos\theta)=eC.
\end{gather}
Since $eC\neq0$, the factor $1-e\cos\theta$ will never be $0$, so we
can divide by it, obtaining
\begin{equation}\label{eqn:z}
r=\frac{eC}{1-e\cos\theta}.
\end{equation}
If we rewrite~\eqref{eqn:polar0} the same way, we get
\begin{equation}\label{eqn:zm}
r=\frac{eC}{-1-e\cos\theta}.
\end{equation}
Again, either~\eqref{eqn:z} or~\eqref{eqn:zm} by itself defines
$\zeta$. 


The line through the focus and parallel to the directrix is defined by
$\theta=\mpi/2$. 
By~\eqref{eqn:z} (or from the original definition of $\zeta$), this
line meets $\zeta$ in two points, $L_0$ and $L_1$, whose coordinates
are $(eC,\mpi/2)$ and $(eC,-\mpi/2)$.  It will be convenient to denote
the distance $\size{L_0L_1}$ by $2\ell$: this means defining
\begin{equation*}
  \ell=eC.
\end{equation*}
Then~\eqref{eqn:z0},~\eqref{eqn:z} and~\eqref{eqn:zm} can be rewritten as
\begin{gather}\label{eqn:r=er}
  r=er\cos\theta+\ell,\\\label{eqn:r=l}
r=\frac{\ell}{1-e\cos\theta}\\\label{eqn:r=lm}
r=\frac{\ell}{-1-e\cos\theta}.
\end{gather}

\section{Lines through the focus}

By~\eqref{eqn:r=l}, \emph{each} line $\theta=\phi$ through the origin
%(which is now the focus) 
meets $\zeta$ in two
points, namely  
\begin{equation*}
  \left(\frac{\ell}{1-e\cos\phi},\phi\right)\quad\text{ and }\quad
  \left(\frac{\ell}{1+e\cos\phi},\phi+\mpi\right),
\end{equation*}
unless $e\cos\phi=\pm 1$.  There are three possibilities,
corresponding the three kinds of conic sections:

\begin{asparaenum}
  \item
If $0<e<1$, then $\size{e\cos\theta}$ is never $1$, so every line
through the origin meets $\zeta$ at two points, \emph{and} these
points are on opposite sides of the origin; $\zeta$ is an
\defn{ellipse.}  See Figure~\ref{fig:ellipse}.

\begin{figure}[ht!]
  \begin{center}
    \begin{pspicture}(-3.5,-2.7)(4,2)
%\psgrid(-3,-3)(3,3)
      \psline{->}(-3.5,0)(4,0)
\psline(-3,-2.8)(-3,2)
\parametricplot[linewidth=1.6pt]{0}{360}{1.5 1 0.5 t cos mul sub div t cos mul
                        1.5 1 0.5 t cos mul sub div t sin mul}
\psline[showpoints=true](0,-1.5)(0,1.5)
\psdots(0,0)(-1,0)(3,0)
\psline[showpoints=true](1,-1.73)(-0.6,1.03)
\psarc(0,0){0.25}{0}{120}
\uput[ur](0.1,0.2){$\theta$}
\uput[u](0,1.5){$L_0$}
\uput[dl](0,-1.5){$L_1$}
\uput[ul](-0.377,1){$\left(\displaystyle\frac{\ell}{1-e\cos\theta},\theta\right)$} 
\uput[dr](0.8,-1.73){$\left(\displaystyle\frac{\ell}{1+e\cos\theta},\theta+\mpi\right)$}
\uput[dl](-1,0){$V$}
\uput[dr](3,0){$V'$}
\uput[r](-3,-2){$d$}
\uput[dl](0,0){$F$}
    \end{pspicture}
  \end{center}
\caption{The ellipse}\label{fig:ellipse}
\end{figure}

\item
If $e=1$, then every line through the origin meets $\zeta$ at two
points, which are are on opposite sides of the origin, \emph{unless}
the line is $\theta=0$:  This line meets $\zeta$ only at
$(\ell/2,\pi)$, halfway between the focus and the directrix.
Now $\zeta$ is a \defn{parabola.}  See Figure~\ref{fig:parabola}.

\begin{figure}[ht!]
  \begin{center}
    \begin{pspicture}(-3.4,-2.6)(3,3)
%\psgrid(-4,-5)(4,5)
\psset{unit=0.65cm}
\psline{->}(-3.5,0)(3,0)
\psline(-3,-2.3)(-3,4.4)
\parametricplot[linewidth=1.6pt]{70}{280}{%
3 1 t cos sub div t cos mul 
3 1 t cos sub div t sin mul}
\psdots(-1.5,0)(0,0)(0,3)(0,-3)(0.63,3.575)(-0.444,-2.517)
\psline(0,-3)(0,3)
\psline(0.63,3.575)(-0.444,-2.517)
\psarc(0,0){0.25}{0}{80}
\uput[ur](0.2,0.2){$\theta$}
\uput[ul](0,3){$L_0$}
\uput[ur](0,-3){$L_1$}
\uput[dr](0.63,3.575){$\left(\displaystyle\frac{\ell}{1-e\cos\theta},\theta\right)$} 
\uput[dl](-0.344,-2.417){$\left(\displaystyle\frac{\ell}{1+e\cos\theta},\theta+\mpi\right)$}
\uput[dl](-1.5,0){$V$}
\uput[r](-3,3){$d$}
\uput[dr](0,0){$F$}
%\uput[u](-2.25,0){$\displaystyle\frac{\ell}2$}
%\uput[u](-0.75,0){$\displaystyle\frac{\ell}2$}
    \end{pspicture}
  \end{center}
\caption{The parabola}\label{fig:parabola}
\end{figure}

\item
Suppose $e>1$. then $\cos\alpha=1/e$ for some $\alpha$ such that
$0<\alpha<\mpi/2$.  If $\alpha<\phi<2\mpi-\alpha$, then the line
$\theta=\phi$
meets $\zeta$ at two points, on opposite sides of the origin, as in
the ellipse and parabola.  If
$-\alpha<\phi<\alpha$, then the line $\theta=\phi$ meets $\zeta$ at
two points, on the \emph{same} side of the origin.  Each of the lines
$\theta=\alpha$ and $\theta=-\alpha$ meets $\zeta$ once, at
$(\ell/2,\mpi+\alpha)$ or $(\ell/2,\mpi-\alpha)$.  Here $\zeta$
is an \defn{hyperbola.}  It is really \emph{two} curves:
\begin{itemize}
  \item
$\zeta_0$, given by~\eqref{eqn:r=l}, where
    $\alpha<\theta<2\mpi-\alpha$;
  \item
$\zeta_1$, given by~\eqref{eqn:r=l}, where
    $-\alpha<\theta<\alpha$; or by~\eqref{eqn:r=lm}, where
    $\mpi-\alpha<\theta<\pi+\alpha$. 
\end{itemize}
See Figure~\ref{fig:hyperbola}.

\begin{figure}[ht!]
  \begin{center}
    \begin{pspicture}(-5,-2.6)(2,3.9)
%\psgrid(-5,-3)(2,4)
\psline{->}(-5,0)(2,0)
\psline(-1.333,-1.5)(-1.333,2.8)
\parametricplot[linewidth=1.6pt]{70}{280}{%
2 1 t cos 1.5 mul sub div t cos mul 
2 1 t cos 1.5 mul sub div t sin mul}
\parametricplot[linewidth=1.6pt]{-25}{25}{%
2 1 t cos 1.5 mul sub div t cos mul 
2 1 t cos 1.5 mul sub div t sin mul}
\psdots(0,0)(0,2)(0,-2)(0.604,2.843)(-0.317,-1.491)(-4.589,1.67)(-0.78,0.284)(-4,0)(-0.8,0)(-0.667,-0.745)(-0.667,0.745)
\psline(0,-2)(0,2)
\psline(0.604,2.843)(-0.317,-1.491)
\psline(-4.589,1.67)(0,0)%(1.149,-0.419)
\psarc(0,0){0.4}{0}{78}
\uput[ur](0.3,0.1){$\phi$}
\psarc(0,0){0.25}{0}{160}
\uput[ul](0.05,0.2){$\psi$}
\psline[linestyle=dashed]{->}(-0.667,-0.745)(2,2.236)
\psline[linestyle=dashed]{->}(-0.667,0.745)(2,-2.236)
\uput[r](1,1.2){$e\cos\theta=1$}
\uput[ur](1,-1.2){$e\cos\theta=1$}
\uput[ul](0,2){$L_0$}
\uput[ur](0,-2){$L_1$}
\uput[ul](0.704,2.743){$\left(\displaystyle\frac{\ell}{1-e\cos\phi},\phi\right)$} 
\uput[dl](-0.217,-1.391){$\left(\displaystyle\frac{\ell}{1+e\cos\phi},\phi+\mpi\right)$}
\uput[ul](0.704,2.743){$\left(\displaystyle\frac{\ell}{1-e\cos\phi},\phi\right)$} 
\uput[ur](-4.589,1.67){$\left(\displaystyle\frac{\ell}{-1-e\cos\psi},\psi\right)$}
\uput[dl](-0.8,0){$V$}
\uput[dr](-4,0){$V'$}
\uput[l](-1.333,-1){$d$}
%\uput[dl](0,0.1){$F$}
\uput[dr](1,3.4){$\zeta_0$}
\uput[ul](-4.2,-1){$\zeta_1$}
    \end{pspicture}
  \end{center}
\caption{The hyperbola}\label{fig:hyperbola}
\end{figure}
\end{asparaenum}

\section{Distances}

The line through the focus $F$ perpendicular to the directrix $d$ is
the \defn{axis} of $\zeta$.  Then $\zeta$ is symmetric about its axis,
because of the original definition, or by~\eqref{eqn:r=l}.
A point of $\zeta$ that lies on the axis
is a \defn{vertex} of $\zeta$.  Again, there are three cases:
\begin{asparaenum}
  \item
Say $0<e<1$, so
$\zeta$ is an ellipse.  Then $\zeta$ has a vertex $V$, with
coordinates $(\ell/(1+e),\mpi)$, and a vertex $V'$, given by
$(\ell/(1-e),0)$.  Since
\begin{equation*}
  0<1-e\leq1-e\cos\theta\leq1+e,
\end{equation*}
we have
\begin{equation}\label{eqn:ellipse-bounds}
  \frac{\ell}{1+e}\leq\frac{\ell}{1-e\cos\theta}\leq\frac{\ell}{1-e}.
\end{equation}
By~\eqref{eqn:r=l} then,
$V$ is the point of $\zeta$ that is closest to the focus, and $V'$ is the
point furthest from $F$.  Also,
\begin{equation*}
  \size{VV'}=\frac{\ell}{1+e}+\frac{\ell}{1-e}=\frac{2\ell}{1-e^2}.
\end{equation*}
See Figure~\ref{fig:extreme:e}.
\begin{figure}[ht!]
\psset{unit=7mm}
  \begin{center}
    \begin{pspicture}(-3,-3)(3,3)
\psset{plotpoints=200}
      \psplot{-3}{3}{9 x x mul sub sqrt}
      \psplot{-3}{3}{9 x x mul sub sqrt neg}
      \psplot{1.125 sqrt neg}{1.125 sqrt}{1.125 x x mul sub sqrt}
      \psplot{1.125 sqrt neg}{1.125 sqrt}{1.125 x x mul sub sqrt neg}
      \parametricplot[linewidth=1.6pt]{0}{360}{288 sqrt 6 sub 7 9 32 sqrt sub t cos mul
      sub div t cos mul
288 sqrt 6 sub 7 9 32 sqrt sub t cos mul sub div t sin mul}
\psdots(-1.06,0)(0,0)(3,0)
\uput[l](-1.06,0){$V$}
\uput[r](0,0){$F$}
\uput[r](3,0){$V'$}
    \end{pspicture}
  \end{center}
\caption{Extreme points in the ellipse}\label{fig:extreme:e}
\end{figure}

\item
Say $e=1$, so $\zeta$ is a parabola.  Then it has a unique vertex, $V$, with
coordinates $(\ell/2,\mpi)$.  As in the case of the ellipse, so in the
parabola, $V$ is the point of $\zeta$ closest to the focus; but there
is no furthest point.  See Figure~\ref{fig:extreme:p}.
\begin{figure}[ht!]
\psset{unit=7mm}
  \begin{center}
    \begin{pspicture}(-0.5,-3)(2.12,3)
\psset{plotpoints=200}
      \psplot[linewidth=1.6pt]{0}{4.5 sqrt}{18 sqrt x mul sqrt}
      \psplot[linewidth=1.6pt]{0}{4.5 sqrt}{18 sqrt x mul sqrt neg}
      \psplot{0}{4.5 sqrt}{1.125 x 1.125 sqrt sub dup mul sub sqrt}
      \psplot{0}{4.5 sqrt}{1.125 x 1.125 sqrt sub dup mul sub sqrt
      neg}
      \psdots(0,0)(1.06,0)
      \uput[l](0,0){$V$}
      \uput[r](1.06,0){$F$}
    \end{pspicture}
  \end{center}
\caption{Extreme point in the parabola}\label{fig:extreme:p}
\end{figure}
\item
Say $e>1$, so
$\zeta$ is an hyperbola.  Then it has two vertices, $V$ and $V'$, with
coordinates $(\ell/(e+1),\mpi)$ and $(\ell/(e-1),\mpi)$ respectively.
As before, suppose $\cos\alpha=1/e$, where $0<\alpha<\mpi/2$.  If
$-\alpha<\theta<\alpha$, then
\begin{gather*}
  \frac1e<\cos\theta\leq 1,\\
1<e\cos\theta\leq e,\\
0<e\cos\theta-1\leq e-1,\\
0<\frac{\ell}{e-1}\leq\frac{\ell}{e\cos\theta-1};
\end{gather*}
so $V'$ is the point of $\zeta_1$ closest to the focus.
If $\alpha<\theta<2\mpi-\alpha$, then
\begin{gather*}
  -1\leq\cos\theta<\frac1e,\\
-e\leq e\cos\theta<1,\\
-1<-e\cos\theta\leq e,\\
0<1-e\cos\theta\leq e+1,\\
\frac{\ell}{e+1}\leq\frac{\ell}{1-e\cos\theta};
\end{gather*}
so $V$ is the point of $\zeta_0$ closest to the focus.  Finally,
\begin{equation*}
  \size{VV'}=\frac{\ell}{e-1}-\frac{\ell}{e+1}=\frac{2\ell}{e^2-1}.
\end{equation*}
See Figure~\ref{fig:extreme:h}.  
\begin{figure}[ht!]
\psset{unit=7mm}
\begin{center}
\begin{pspicture}(-5,-3.5)(3,3.5)
\psset{plotpoints=200}
      \psplot{-3}{3}{9 x x mul sub sqrt}
      \psplot{-3}{3}{9 x x mul sub sqrt neg}
      \psplot{1.125 sqrt neg}{1.125 sqrt}{1.125 x x mul sub sqrt}
      \psplot{1.125 sqrt neg}{1.125 sqrt}{1.125 x x mul sub sqrt neg}  
      \psclip{\psframe[linecolor=white](-5,-3.5)(3,3.5)}
      \parametricplot[linewidth=1.6pt]{62}{298}{6 288 sqrt add 7 9 32
      sqrt add t cos mul sub div t cos mul
                                                6 288 sqrt add 7 9 32 
      sqrt add t cos mul sub div t sin mul}
      \parametricplot[linewidth=1.6pt]{-61}{61}{6 288 sqrt add 7 9 32
      sqrt add t cos mul sub div t cos mul
                                                6 288 sqrt add 7 9 32
      sqrt add t cos mul sub div t sin mul} 
\endpsclip
\psdots(-3,0)(-1.06,0)(0,0)
\uput[l](-3,0){$V'$}
\uput[l](-1.06,0){$V$}
\uput[r](0,0){$F$}
\end{pspicture}
  \end{center}
  \caption{Extreme points in the hyperbola}\label{fig:extreme:h}
\end{figure}
\end{asparaenum}

In both the ellipse and the hyperbola
then, the distance between the
two vertices is $2\ell/\size{e^2-1}$; this may also be denoted by
$2a$, so that
\begin{equation}\label{eqn:a}
  a=\frac{\ell}{\size{e^2-1}}.
\end{equation}

\section{Areas}

Let $P$ be an arbitrary point with coordinates $(r,\theta)$ on
$\zeta$, and let the foot of the 
perpendicular from $P$ to the axis of $\zeta$ be $Q$ (as in
Figure~\ref{fig:polar}). 
Then $Q$ has coordinates $(r\cos\theta,0)$.  We consider the position
of $Q$ with respect to the vertices:
\begin{asparaenum}
  \item
If $0<e<1$, then by~\eqref{eqn:ellipse-bounds} and~\eqref{eqn:r=er}
\begin{gather*}
  \frac{\ell}{1+e}\leq r\leq\frac{\ell}{1-e},\\
  \frac{\ell}{1+e}\leq er\cos\theta+\ell\leq\frac{\ell}{1-e},\\
  -\frac{\ell e}{1+e}\leq er\cos\theta\leq\frac{\ell e}{1-e},\\
  -\frac{\ell}{1+e}\leq r\cos\theta\leq\frac{\ell}{1-e};
\end{gather*}
so $Q$ is between $V$ and $V'$, and
\begin{gather*}
  \size{VQ}=r\cos\theta+\frac{\ell}{1+e},\\
  \size{V'Q}=\frac{\ell}{1-e}-r\cos\theta.
\end{gather*}
\item
If $e=1$, then
\begin{gather*}
  \frac{\ell}2\leq r=r\cos\theta+\ell,\\
-\frac{\ell}2\leq r\cos\theta,\\
\size{VQ}=r\cos\theta+\frac{\ell}2.
\end{gather*}
\item
If $e>1$, then there are two cases.
\begin{inparaenum}[(a)]
  \item
If $P$ is on $\zeta_0$, then
\begin{gather*}
  \frac{\ell}{1+e}\leq r=er\cos\theta+\ell,\\
-\frac{\ell e}{1+e}\leq er\cos\theta,\\
-\frac{\ell}{1+e}\leq r\cos\theta,\\
\size{VQ}=r\cos\theta+\frac{\ell}{e+1},\\
\size{V'Q}=r\cos\theta+\frac{\ell}{e-1}.
\end{gather*}
\item
If $P$ is on $\zeta_1$, then
\begin{gather*}
  \frac{\ell}{e-1}\leq r=-(er\cos\theta+\ell),\\
\frac{\ell e}{e-1}\leq-er\cos\theta,\\
\frac{\ell}{e-1}\leq-r\cos\theta,\\
\size{VQ}=-\left(r\cos\theta+\frac{\ell}{e+1}\right),\\
\size{V'Q}=-\left(r\cos\theta+\frac{\ell}{e-1}\right).
\end{gather*}
\end{inparaenum}
In either case, $Q$ is \emph{not} between $V$ and $V'$.
\end{asparaenum}

Now we can compute:
\begin{align}\notag
  \size{PQ}^2
&=r^2\sin^2\theta\\\notag
&=r^2-r^2\cos^2\theta\\\notag
&=(er\cos\theta+\ell)^2-r^2\cos^2\theta\\\label{eqn:last}
&=(r[e+1]\cos\theta+\ell)(r[e-1]\cos\theta+\ell).
\end{align}
There are two cases.
\begin{inparaenum}[(a)]
  \item
If $e=1$, then this equation becomes
\begin{equation}\label{eqn:parabola-old}
  \size{PQ}^2=
(2r\cos\theta+\ell)\cdot\ell=2\ell\cdot\size{VQ}.
\end{equation}
\item
If $e\neq1$, then
\begin{align}\notag
  \size{PQ}^2
&=(e^2-1)\left(r\cos\theta+\frac{\ell}{e+1}\right)
  \left(r\cos\theta+\frac{\ell}{e-1}\right)\\\label{eqn:sym}
&=\size{e^2-1}\cdot\size{VQ}\cdot\size{V'Q}\\\notag
&=2\ell\cdot\frac{\size{V'Q}}{\size{VV'}}\cdot\size{VQ}.
\end{align}
\end{inparaenum}

Let $VR$ be drawn perpendicular to the axis of $\zeta$ so that
$\size{VR}=2\ell$.  This line segment is called the \defn{latus
  rectum} of $\zeta$.  This is the term commonly used in English,
although it is the \emph{Latin} translation of the 
original Greek found in Apollonius; however, the literal English
translation, `upright side,' is used in \cite{MR0053831}.
Then the square with side $PQ$
\begin{compactitem}
  \item
is the area of the rectangle with sides $VQ$ and $VR$, if $\zeta$ is a
parabola;
\item
falls short of this area, if $\zeta$ is an ellipse;
\item
exceeds this area, if $\zeta$ is an hyperbola.
\end{compactitem}
This is what is suggested by the Greek names of the curves.  See
Figures~\ref{fig:areas-pe} and \ref{fig:areas-h}. 
\begin{figure}[ht!]
\psset{unit=15mm}
\centering
%\mbox{}\hfill  
\begin{pspicture}(-0.5,-1.5)(5.5,1.5)
    \psframe[fillstyle=solid,fillcolor=lightgray](0,-0.45)(3.5,0)
    \psframe[fillstyle=solid,fillcolor=lightgray](3.5,0)(4.748,1.248)
\psset{plotpoints=200}
    \psplot{0}{5}{0.45 x mul sqrt}
    \psplot{0}{5}{0.45 x mul sqrt neg}
    \psline(0,0)(5,0)
    \uput[l](0,0){$V$}
    \uput[dr](3.5,0){$Q$}
    \uput[ul](3.5,1.248){$P$}
    \uput[dl](0,-0.45){$R$}
\psdots(0,0)(0,-0.45)
  \end{pspicture}

  \begin{pspicture}(-3,-1.62)(3,1.5)
%\psgrid
\psframe[fillstyle=solid,fillcolor=lightgray](-2.5,-1.62)(-2,0)
\psframe[fillstyle=solid,fillcolor=lightgray](-2,0)(-1.1,0.9)
    \psline(-2.5,0)(2.5,0)
\psset{plotpoints=200}
    \parametricplot{0}{360}{0.9 1 0.8 t cos mul sub div t cos mul 2 sub
                            0.9 1 0.8 t cos mul sub div t sin mul}
\psline(-2.5,0)(-2.5,-1.8)(2.5,0)
%\psline(-2.5,-1.62)(-2,-1.62)(-2,0.9)(-1.1,0.9)(-1.1,0)
\uput[l](-2.5,0){$V$}
\uput[r](2.5,0){$V'$}
\uput[dl](-2.5,-1.62){$R$}
\uput[dr](-2.0,0){$Q$}
\uput[ul](-2,0.9){$P$}
\psdots(-2.5,0)(2.5,0)(-2.5,-1.8)
  \end{pspicture}

  \caption{Parabola and ellipse as determined by equations of areas}%
\label{fig:areas-pe}
\end{figure}
\begin{figure}[ht!]
\centering
\psset{unit=15mm}
  \begin{pspicture}(-3,-1.7)(3,1.5)
    %\psgrid(-3,-2)(3,2)
\psframe[fillstyle=solid,fillcolor=lightgray](1.5,0)(2.232,0.732)
\psframe[fillstyle=solid,fillcolor=lightgray](1,-1.0725)(1.5,0)
\psset{plotpoints=200}
    \psline(-2.5,0)(2.5,0)
\psplot{-2.5}{-1}{x x mul 1 sub sqrt 0.655 mul}
\psplot{-2.5}{-1}{x x mul 1 sub sqrt 0.655 mul neg}
\psplot{1}{2.5}{x x mul 1 sub sqrt 0.655 mul}
\psplot{1}{2.5}{x x mul 1 sub sqrt 0.655 mul neg}
\psline(-1,0)(1,-0.858)(1.5,-1.0725)
\psline(1,0)(1,-0.858)
\uput[ur](-1,0){$V'$}
\uput[ul](1,0){$V$}
\uput[ul](1.5,0.732){$P$}
\uput[dr](1.5,0){$Q$}
\uput[dl](1,-0.858){$R$}
\psdots(-1,0)(1,0)(1,-0.858)
  \end{pspicture}

  \begin{pspicture}(-3,-1.5)(3,1.5)
    %\psgrid(-3,-2)(3,2)
\psset{plotpoints=200}
\psframe[fillstyle=solid,fillcolor=lightgray](-1.5,0)(-2.232,0.732)
\psframe[fillstyle=solid,fillcolor=lightgray](1,0)(-1.5,0.2145)
    \psline(-2.5,0)(2.5,0)
\psplot{-2.5}{-1}{x x mul 1 sub sqrt 0.655 mul}
\psplot{-2.5}{-1}{x x mul 1 sub sqrt 0.655 mul neg}
\psplot{1}{2.5}{x x mul 1 sub sqrt 0.655 mul}
\psplot{1}{2.5}{x x mul 1 sub sqrt 0.655 mul neg}
\psline(-1.5,0.2145)(-1,0)(1,-0.858)
\psline(1,0)(1,-0.858)
\uput[dr](-1,0){$V'$}
\uput[dl](1,0){$V$}
\uput[ur](-1.5,0.732){$P$}
\uput[d](-1.5,0){$Q$}
\uput[dl](1,-0.858){$R$}
\psdots(-1,0)(1,0)(1,-0.858)
  \end{pspicture}

  \caption{Hyperbola as determined by equations of areas}\label{fig:areas-h}
\end{figure}

\section{The rectangular equations}

For the parabola, choose a rectangular coordinate system in which $V$ is
the origin and the $X$-axis is the axis of $\zeta$.
Then~\eqref{eqn:parabola-old} becomes
\begin{equation*}
  y^2=2\ell x.
\end{equation*}
This is the standard rectangular equation for a parabola.  The focus
is at $(\ell/2,0)$, and the directrix is given by $x+\ell/2=0$.

For the ellipse and the hyperbola, let the origin of a rectangular
coordinate system be the midpoint $O$ of $VV'$: this is the \defn{center}
of the conic section.  Let the $X$-axis contain the vertices.  Then
the vertices will
have coordinates $(\pm a,0)$.  By~\eqref{eqn:sym}, the curve is
symmetric about the new $Y$-axis.  In particular, the curve has, not
just one focus, but two foci; hence it has, not just one directrix,
but two directrices,
one for each focus.  The curve is now given by 
\begin{equation}\label{eqn:abs}
  y^2=\size{e^2-1}\cdot\size{x-a}\cdot\size{x+a}
  =\size{e^2-1}\cdot\size{x^2-a^2}.  
\end{equation}
Moreover, by the previous subsection, in the ellipse, $e^2-1$ and
$x^2-a^2$ are both negative; in
the hyperbola, positive.  Hence~\eqref{eqn:abs} can be written
\begin{gather*}
  y^2=(e^2-1)(x^2-a^2),\\
\frac{y^2}{a^2(e^2-1)}=\frac{x^2}{a^2}-1,\\
\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1.
\end{gather*}
Recalling~\eqref{eqn:a}, we can write 
\begin{equation}\label{eqn:al}
\frac{x^2}{a^2}\pm\frac{y^2}{a\ell}=1,  
\end{equation}
where the upper sign is for the ellipse, and the lower is for the
hyperbola.  We 
may let $b$ be the positive number such that
\begin{equation}\label{eqn:b}
  b^2=a\ell,
\end{equation}
so that~\eqref{eqn:al} becomes
\begin{equation}\label{eqn:ab}
\frac{x^2}{a^2}\pm\frac{y^2}{b^2}=1.  
\end{equation}
The $Y$-intercepts of the ellipse are $(0,\pm b)$; the
hyperbola has no $Y$-inter\-cepts. 
By~\eqref{eqn:a} and~\eqref{eqn:b},
\begin{equation*}
  e=\sqrt{1\mp\frac{b^2}{a^2}};
\end{equation*}
where again the upper sign is for the ellipse.  Also,
\begin{equation*}
  \size{FO}=a\mp\frac{\ell}{1+e}=a-\frac{a(1-e^2)}{1+e}=a-a(1-e)=ae;
\end{equation*}
so the foci are at
$(\pm ae,0)$.  Likewise, 
\begin{equation*}
  \size{dO}=ae\pm\frac{\ell}e=ae+\frac{a(1-e^2)}e=\frac ae;
\end{equation*}
so the directrices are given by $x\pm a/e=0$.

Finally, the hyperbola given by~\eqref{eqn:ab} does not meet the two
lines given by
\begin{equation*}
\frac{x^2}{a^2}-\frac{y^2}{b^2}=0.  
\end{equation*}
These lines---given also by $ay\pm bx=0$---are the \defn{asymptotes}
of the hyperbola.  Their slopes are $\pm b/a$.  In general, a line
through $O$ meets the hyperbola if and only if the slope of the line
is less than $b/a$ in absolute value.  Indeed, the equations
\begin{equation}
\begin{cases}\label{eqn:y=mx}
  y=mx,\\
\displaystyle\frac{x^2}{a^2}-\displaystyle\frac{y^2}{b^2}=1,  
\end{cases}
\end{equation}
if solved simultaneously, yield
\begin{gather}\notag
  \frac{x^2}{a^2}-\frac{m^2x^2}{b^2}=1,\\\label{eqn:xmx}
\frac{b^2}{a^2}-m^2=\frac{b^2}{x^2},\\\notag
\frac{b^2}{a^2}-m^2>0,\\\notag
m^2<\frac{b^2}{a^2},\\\notag
\size{m}<\frac ba;
\end{gather}
and if the last inequality holds, then there \emph{is} a simultaneous
solution, obtainable from~\eqref{eqn:xmx} and then~\eqref{eqn:y=mx}.

The two hyperbolas $x^2/a^2-y^2/b^2=\pm1$ have the same asymptotes.
Also, their foci are at the same distance from the center, namely
$\sqrt{a^2+b^2}$.  Such hyperbolas are \defn{conjugate.}
The ellipse $x^2/a^2+y^2/b^2=1$ is tangent to them at their
vertices.  See Figure~\ref{fig:eh}.  

\begin{figure}[ht!]
\psset{unit=10mm}
  \begin{center}
\rotatebox{90}{\begin{pspicture}(-6,-5.2)(6,5.2)
%\psgrid(-6,-6)(6,6)
\psclip{\psframe[linecolor=white](-6,-5.2)(6,5.2)}
\psset{plotpoints=200}
\psplot{-4}{4}{1 x x mul 16 div sub 8 mul sqrt}      
\psplot{-4}{4}{1 x x mul 16 div sub 8 mul sqrt neg}      
\psplot[linewidth=1.6pt]{-6}{-4}{x x mul 16 div 1 sub 8 mul sqrt}      
\psplot[linewidth=1.6pt]{4}{6}{x x mul 16 div 1 sub 8 mul sqrt neg}      
\psplot[linewidth=1.6pt]{4}{6}{x x mul 16 div 1 sub 8 mul sqrt}      
\psplot[linewidth=1.6pt]{-6}{-4}{x x mul 16 div 1 sub 8 mul sqrt neg}      
\psplot[linewidth=1.6pt]{-6}{6}{1 x x mul 16 div add 8 mul sqrt}      
\psplot[linewidth=1.6pt]{-6}{6}{1 x x mul 16 div add 8 mul sqrt neg}      
\psdots(-4,0)(4,0)(2.83,0)(-2.83,0)(4.899,0)(-4.899,0)(0,4.899)(0,-4.899)
\psline(-6,-4.24)(6,4.24)
\psline(-6,4.24)(6,-4.24)
\psline(5.66,-6)(5.66,6)
\psline(3.266,-6)(3.266,6)
\psline(-5.66,-6)(-5.66,6)
\psline(-3.266,-6)(-3.266,6)
\psline(-6,1.633)(6,1.633)
\psline(-6,-1.633)(6,-1.633)
\psline(0,-6)(0,6)
\psline(-6,0)(6,0)
\endpsclip
    \end{pspicture}}
  \end{center}
  \caption{Conjugate hyperbolas}\label{fig:eh}
\end{figure}

The segment joining the two vertices of an ellipse is the 
\defn{major axis} of the ellipse; the \defn{minor axis}
passes through the center, but is perpendicular to the major
axis.

A \defn{circle} can be described as an ellipse of eccentricity $0$.
Strictly, however, a circle 
is not a conic section by the definition given in
\S~\ref{subsect:focus}.  The circle does not have a directrix.
However, the circle is a kind of `limit' of the ellipses with the same
focus and latus rectum, as the directrix moves indefinitely far away
(which means the eccentricity tends to $0$).
See Figure~\ref{fig:limit}.

\begin{figure}[ht!]
\psset{unit=8mm}
  \begin{center}
    \begin{pspicture}(-6,-6)(6,6)
%\psgrid(-6,-6)(6,6)
\psclip{\psframe[linecolor=white](-6,-6)(6,6)}
\psdots(0,-3)(0,0)(0,3)
\psset{plotpoints=200}
      \parametricplot{-75}{75}{3 1 4 t cos mul sub div t cos mul
                               3 1 4 t cos mul sub div t sin mul}
      \parametricplot{76}{284}{3 1 4 t cos mul sub div t cos mul
                               3 1 4 t cos mul sub div t sin mul}
      \parametricplot{-59}{59}{3 1 2 t cos mul sub div t cos mul
                               3 1 2 t cos mul sub div t sin mul}
      \parametricplot{61}{299}{3 1 2 t cos mul sub div t cos mul
                               3 1 2 t cos mul sub div t sin mul}
      \parametricplot{1}{359}{3 1 t cos sub div t cos mul
                              3 1 t cos sub div t sin mul}
      \parametricplot{0}{360}{3 1 0.5 t cos mul sub div t cos mul
                              3 1 0.5 t cos mul sub div t sin mul}
      \parametricplot{0}{360}{3 1 0.25 t cos mul sub div t cos mul
                              3 1 0.25 t cos mul sub div t sin mul}
      \parametricplot{0}{360}{3 1 0.125 t cos mul sub div t cos mul
                              3 1 0.125 t cos mul sub div t sin mul}
      \parametricplot{0}{360}{3 1 0.0625 t cos mul sub div t cos mul
                              3 1 0.0625 t cos mul sub div t sin mul}
      \parametricplot{0}{360}{3 1 0.03125 t cos mul sub div t cos mul
                              3 1 0.03125 t cos mul sub div t sin mul}
      \parametricplot{0}{360}{3 1 0.015625 t cos mul sub div t cos mul
                              3 1 0.015625 t cos mul sub div t sin mul}
      \parametricplot[linewidth=1.6pt]{0}{360}{3 t cos mul
                              3 t sin mul}
\uput[dl](-5.6,6){$2$}
\uput[dl](-2.4,6){$4$}
\uput[dr](0.8,6){$4$}
\uput[dr](1.6,6){$2$}
\uput[dr](4.5,6){$1$}
\uput[l](6,0){$\displaystyle\frac12$}
\uput[r](4,0){$\displaystyle\frac14$}
\endpsclip
    \end{pspicture}
  \end{center}
\caption{The circle as a limit of conics}\label{fig:limit}
\end{figure}

%\addcontentsline{toc}{section}{\numberline{}Bibliography}
\bibliographystyle{plain}
%\bibliography{../references}

\begin{thebibliography}{10}

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\bibitem{MR0245397}
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\bibitem{MR13:419a}
Ivor Thomas, editor.
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\newblock With an English translation by the editor.

\bibitem{MR13:419b}
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\newblock With an English translation by the editor.

\end{thebibliography}

%\newpage
%\layout


\end{document}

