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\begin{document}
\title{Conic sections\\
  with and without algebra}
%\subtitle{Antalya Algebra Days XXI}
\author{David Pierce}
\date{May 2019}
\publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\url{mat.msgsu.edu.tr/~dpierce/}\\
\url{polytropy.com}}
\maketitle

\addchap{Preface}

This document concerns a contributed talk,
delivered Wednesday, May 15, 2019, 16:10--16:30,
the first day of Antalya Algebra Days XXI,
Nesin Mathematics Village,
\c Sirince, Sel\c cuk, Izmir, Turkey.
\begin{compactdesc}
  \item[Abstract] is as submitted,
    then edited before publication in the conference booklet
    (an error remained, as noted).
  \item[Notes on abstract] spell out the algebra,
    not covered in the talk itself
    (except with a reference to the abstract).
\item[Notes of talk] are an approximation,
from written notes and memory,
of what I actually said in the talk,
mostly out loud only.
On the chalkboards of the Ni\c sanyan Library, I drew the diagrams,
but mostly wrote only the symbolic equations and proportions.
Since Tuna Alt\i nel had been imprisoned on the previous Saturday,
and Ay\c se Berkman had discussed the case when opening the meeting,
I alluded to the case
(and to the blocking of \textsl{Wikipedia} in Turkey).
\item[Notes for talk]
  were prepared and printed
before travelling from Istanbul to \c Sirince
(via Ankara, that previous Saturday).
I knew that I would not have time to write out all of the computations.
Only later did I streamline the written talk with the idea denoted by $\propto$.
\item[Additional remarks]
were originally considered for inclusion in the talk,
only there would not be enough time.
\end{compactdesc}

\tableofcontents
\listoffigures

\chapter{Abstract}
This is about how history may reveal a forgotten vision of mathematics.
Mathematics is universal, but ways of understanding it are not.
As has been argued in general terms \cite{FU},
the Ancients did not secretly use algebra.
This should not be understood pejoratively.
People without keyboards or even ready supplies of paper
may develop ways of understanding that we never feel the need for.

\begin{sloppypar}
  The square on the ordinate varies as the abscissa, in a parabola;
  in the ellipse and hyperbola,
  as the rectangle bounded by the two abscissas
  \cite{MR1660991}.
  In Cartesian terms \cite{Descartes-Geometry},
  the proportions are expressed
  by algebraic equations:
  \begin{align*}
    y^2&=\ell x,&
    y^2&=\frac{\ell x}{2d}(2d-x).
  \end{align*}
  With laborious algebraic manipulations,
  John Wallis
  \cite{Wallis}
  re-established that the curves defined above,
  if they contain $(a,b)$,
  and if $c=a+d$,%%%%%
  \footnote{Should be $c=d-a$.}
    are fixed
  under the respective affine transformations
  \begin{align*}
    &\left\{
    \begin{gathered}
      x'=x-\frac{2by}{\ell}+a\\
      y'=-y+b
    \end{gathered}
    \right\},
    &&\left\{
    \begin{gathered}
      x'=\frac{cx}d-\frac{2by}{\ell}+a\\
      y'=-\frac{bx}d-\frac{cy}d+b
    \end{gathered}
    \right\}.
  \end{align*}
  It has been asserted \cite{Rosenfeld}
  that Apollonius somehow understood this;
  but all modern proofs that I have found,
  as by de Witt, Euler, and Hugh Hamilton,
  lack the clarity of Apollonius's non-algebraic proof,
  which uses areas in a way
  that does not reduce to manipulations of lengths
  in the Cartesian fashion.
  The key is that the equations for the conics
  can be written as equations of a parallelogram
  with a triangle and a trapezoid respectively,
  all lying on one plane
  (which need only be an affine plane).
\end{sloppypar}

\chapter{Notes on abstract}

We confirm some features of the given affine transformations
  \begin{align}\label{eqn:par-tran}
      x'&=x-\frac{2by}{\ell}+a,
      &y'&=-y+b,\\\label{eqn:cen-tran}
      x'&=\frac{cx}d-\frac{2by}{\ell}+a,
      &y'&=-\frac{bx}d-\frac{cy}d+b,
  \end{align}
  of the parabola and central conic respectively.

\section{Characterization}

That \eqref{eqn:par-tran}
interchanges
$(0,0)$ and $(a,b)$
is easy to see.
To see that \eqref{eqn:cen-tran} does too,
we need that,
since the central conic is given by
\begin{equation}\label{eqn:cen}
      y^2=\frac{\ell x}{2d}(2d-x)
\end{equation}
and contains $(a,b)$, in particular
\begin{equation}\label{eqn:2b^2/l}
  \frac{2b^2}{\ell}=\frac ad(2d-a)=\frac ad(c+d),
\end{equation}
where
\begin{equation}\label{eqn:c=d-a}
  c=d-a
\end{equation}
(this corrects the abstract).
Moreover, \eqref{eqn:cen-tran} fixes $(d,0)$ in the central conic,
but \eqref{eqn:par-tran}
interchanges any $(x,0)$ and $(a+x,b)$.

\section{Derivation}

Taking $(0,0)$ to $(a,b)$, \eqref{eqn:cen-tran} must have the form
\begin{align*}
  x'&=Ax+By+a,&y'&=Cx+Dy+b.
\end{align*}
Since the transformation fixes $(d,0)$,
from \eqref{eqn:c=d-a} we have
\begin{align}\label{eqn:AC}
  A&=\frac cd,&C=-\frac bd.
\end{align}
Since \eqref{eqn:cen-tran} will take $(a,b)$ to $(0,0)$,
from \eqref{eqn:2b^2/l} we have
\begin{gather}\label{eqn:B}
  0=\frac{ca}{bd}+B+\frac ab
  =B+\frac a{bd}(c+d)
  =B+\frac{2b}{\ell},\\\label{eqn:D}
  0=-\frac ad+D+1
  =D+\frac cd,
\end{gather}
which yields \eqref{eqn:cen-tran}.
Alternatively, we obtain \eqref{eqn:AC} and \eqref{eqn:D},
then derive $B$ from
$AD-BC=-1$
instead of \eqref{eqn:B}.

\section{Confirmation}

Under the change of coordinates given by
\begin{equation*}
  x=d-t,
\end{equation*}
the transformation \eqref{eqn:cen-tran} becomes
\begin{align}\label{eqn:t'y'}
      t'&=\frac{ct}d+\frac{2by}{\ell},&
    y'&=\frac{bt}d-\frac{cy}d,
\end{align}
and \eqref{eqn:cen} becomes
\begin{gather}\notag
  y^2=\frac{\ell}{2d}(d^2-t^2),\\\label{eqn:cen-con}
  \frac{t^2}{d^2}+\frac{2y^2}{\ell d}=1
\end{gather}
($d$ is positive in the ellipse, negative in the hyperbola).
One easily checks that $(t',y')$
as in \eqref{eqn:t'y'} satisfies \eqref{eqn:cen-con},
since $(c,b)$ does in particular, so that
\begin{equation*}
  \frac{c^2}{d^2}+\frac{2b^2}{\ell d}=1.
\end{equation*}

\chapter{Notes of talk}

This talk is both mathematics and history
concerning Apollonius of Perga.
Such work has practical application
as an example of the principle that
being good at one thing (such as mathematics and history)
does not necessarily make you good at another thing
(such as history or mathematics):
\begin{compactitem}
\item
  Being good at getting elected president of a country
  does not make you good at choosing the rectors of your country's universities.
\item
  Being good at locking people up
  does not make you good at knowing who should be locked up.
\end{compactitem}
The flier for the Maryam Mirzakhani poster exhibition%%%%%
\footnote{Hung in the arcade outside the Library.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  says,
``Mathematics is universal.''
I say the same in my abstract, with the qualification that
\emph{understanding} is not universal.
For example,
in the first talk this morning, by Rostislav Grigorchuk,
we heard of four papers on the arXiv,
two asserting that a certain group was amenable,
the other two that it was not.
The authors on one side cannot go beat up those on other side with sticks;
they cannot denounce them to the police to have them locked up.
All authors have to consider the possibility that they are wrong.%%%%%
\footnote{Apparently they \emph{were} all wrong,
  and the status of the group in question, one of the Thompson groups,
  is yet undecided.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
In this sense, the world needs more mathematics.

By a theorem in Euclid,
when two chords of a circle cut one another,
the rectangles formed by the parts are equal to one another.
Algebraically, from \figref{fig:cir-cho},
\begin{figure}
  \subfloat[]{\label{fig:cir-cho}
    \begin{pspicture}(-2,-2)(2,2)
      \pscircle(0,0){2}
      \SpecialCoor
      \pnode(2;100){A}
      \pnode(2;265){B}
      \pnode(2;190){C}
      \pnode(2;30){D}
      \psset{PointSymbol=none,PointName=none}
      \pstInterLL ABCDO
      \ncline AO\naput{$a$}
      \ncline OB\naput{$b$}
      \ncline CO\nbput{$x$}
      \ncline OD\nbput{$y$}
    \end{pspicture}}
  \hfill
    \subfloat[]{\label{fig:diag}
      \begin{pspicture}(6,4)
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={-135,-45,45,135}]
        (0.5,0.5)A(5.5,0.5)B(5.5,3.5)C(0.5,3.5)D
        \pstHomO[HomCoef=0.35,PosAngle=-90]AB[E]
        \pstTranslation[PosAngle=90] ADE[F]
        \pstInterLL[PosAngle=135] ACEFG
        \pstTranslation[PosAngle=180] EAG[H]
        \pstTranslation EBG[K]
        \pspolygon(A)(B)(C)(D)
        \psline(A)(C)
        \psline(E)(F)
        \psline(H)(K)
        \psset{linestyle=none}
        \ncline HD\nbput{$a$}
        \ncline DF\nbput{$b$}
        \ncline KB\nbput{$x$}
        \ncline BE\nbput{$y$}
    \end{pspicture}}
\caption{Intersecting chords of a circle}
  
\end{figure}
\begin{equation}\label{eqn:ab=xy}
  ab=xy.
\end{equation}
What does this mean?
For Euclid, in \figref{fig:diag},%%%%%
\footnote{In the talk, I did not label the points,
  except those along the diagonal.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{equation}\label{eqn:abxy}
  ab=xy\iff AGC\text{ is straight.}
\end{equation}
If we take this as a definition,
why is the equality so defined transitive?
For Euclid, the equivalence is a theorem:
if $AGC$ is indeed straight, then,
being congruent, the two large triangles $ABC$ and $CDA$ are equal,
as are the small ones
($AEG$ and $GHA$, and $GKC$ and $CFG$);
equals being subtracted from equals,
the remaining rectangles are equal.

In short, for Euclid,
transitivity of equality of areas is axiomatic,
and \eqref{eqn:abxy} is a theorem.

Alternatively, one may use axioms like Hilbert's,
concerning only equality of lengths;
then one has to prove transitivity of equality of areas
as defined by \eqref{eqn:abxy}.

One may rewrite \eqref{eqn:ab=xy} as
\begin{equation*}
\rat ax\prop\rat yb,
\end{equation*}
``the ratio of $a$ to $x$ is the same as the ratio of $y$ to $b$.''
Transitivity of sameness of ratio is a consequence of, or \emph{is,}
what \textsl{Wikipedia} calls the Intercept Theorem.%%%%%
\footnote{In \figref{fig:diag}, assuming only $HG\parallel DC$,
  we may define a quaternary relation of proportionality by
  $\rat{AH}{HD}\prop\rat{AG}{GC}$.
  If, like the Ancients,
  we call this a \emph{sameness} of ratios,
  this sameness should be transitive
  (like any relation called ``sameness'').
  Thus, since also $GE\parallel CB$
  (but without using that $ABCD$ is a parallelogram),
  it should follow that $HE\parallel DB$.
  This is (a case of) Desargues's Theorem,
  which however follows trivially from the Intercept Theorem,
  if ratios are independently defined
  (as in the Eudoxan definition given by Euclid,
  or in a vector space),
  so that sameness of ratio is automatically transitive.}
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
In one expression of this theorem,
if $\bm a$ and $\bm b$ are independent vectors as in \figref{fig:vec},
\begin{figure}
  \centering
  \begin{pspicture}(10,4)
    \psset{PointName=none,PointSymbol=none}
%    \psgrid
    \pstGeonode(0.5,2)O(9.5,0.5)D(8,3.5)C
    \pstHomO[HomCoef=0.55]OD[B]
    \pstHomO[HomCoef=0.6]OC[A]
    \uput[l](O){$\bm0$}
    \uput[u](C){$\lambda\bm a$}
    \uput[u](A){$\bm a$}
    \uput[d](D){$\mu\bm b$}
    \uput[d](B){$\bm b$}
    \ncline OD\ncline OC\ncline AB\ncline CD
  \end{pspicture}
  \caption{Intercept Theorem}
  \label{fig:vec}
\end{figure}
and $\lambda$ and $\mu$ are nonzero scalars,
then
\begin{equation*}
  \bm a-\bm b\parallel\lambda\bm a-\mu\bm b\iff\lambda=\mu.
\end{equation*}
The \textsl{Wikipedia} article uses a \emph{normed} vector space,
but the norm is needless, unless one wants a ratio of $\bm a$ to $\bm b$.
I might edit the article,
but the Turkish state blocks our access.
We have ways of \emph{reading} \textsl{Wikipedia,}
but they do not allow editing;
thus Turkey deprives the world of our knowledge and understanding.

Returning to the circle,
if, as in \figref{fig:circ-VW},
  \begin{figure}
    \subfloat[]{\label{fig:circ-VW}
    \psset{PointSymbol=none}
    \begin{pspicture}(0,-2.2)(5,2.3)
%          \psgrid
      \pstGeonode[PosAngle={180,0}](0.5,0){\Wv}(4.5,0){\Ver}
      \pstCircleAB{\Wv}{\Ver}
      \pstMiddleAB[PointName=none]{\Wv}{\Ver}{\mbat}
      \pstHomO[HomCoef=0.8,PosAngle=45]{\Wv}{\Ver}[{\Xp}]
      \pstBissectBAC[PointName=none,linestyle=none]{\Wv}{\Xp}{\Ver}Z
      \pstInterLC[PosAngleA=45,PosAngleB=-45]{\Xp}{Z}{\mbat}{\Wv}{\Pnto}{\Pnt}
      \ncline{\Wv}{\Ver}\ncline{\Pnt}{\Pnto}
    \end{pspicture}
      }
      \hfill
    \subfloat[]{\label{fig:circ-VWM}
    \psset{PointSymbol=none}
    \begin{pspicture}(0,-2.2)(5,2.3)
%          \psgrid
      \pstGeonode[PosAngle={180,0}](0.5,0){\Wv}(4.5,0){\Ver}
      \pstCircleAB{\Wv}{\Ver}
      \pstMiddleAB[PointName=none]{\Wv}{\Ver}{\mbat}
      \pstHomO[HomCoef=0.25,PosAngle=45]{\Wv}{\Ver}[{\Mp}]
      \pstHomO[HomCoef=0.8,PosAngle=45]{\Wv}{\Ver}[{\Xp}]
      \pstBissectBAC[PointName=none,linestyle=none]{\Wv}{\Mp}{\Ver}Z
      \pstInterLC[PosAngleA=120,PosAngleB=-120]{\Mp}Z{\mbat}{\Wv}{\Dpo}{\Dp}
      \pstTranslation[PointName=none]{\Mp}Z{\Xp}[{\hide}]
      \pstInterLC[PosAngleA=45,PosAngleB=-45]{\Xp}{\hide}{\mbat}{\Wv}{\Pnto}{\Pnt}
      \ncline{\Wv}{\Ver}\ncline{\Dp}{\Dpo}\ncline{\Pnt}{\Pnto}
    \end{pspicture}
      }
      \caption{Circle with diameter and orthogonal chords}
  \end{figure}
  a circle has diameter ${\Ver}{\Wv}$,
  and chord ${\Pnt}{\Pnto}$ is orthogonal to this, we have
  \begin{equation*}
    {\Xp}{\Pnt}^2={\Ver}{\Xp}\cdot{\Xp}{\Wv}.
  \end{equation*}
  This too assumes a norm, even to define the circle itself.
  However, without the norm, in an affine plane, we can express
  \begin{equation*}
    {\Xp}{\Pnt}^2\propto{\Ver}{\Xp}\cdot{\Xp}{\Wv},
  \end{equation*}
  which means,
  if we draw chord ${\Dp}{\Dpo}$ parallel to ${\Pnt}{\Pnto}$
  as in \figref{fig:circ-VWM},
  \begin{equation*}
          (\rat{\vec{\Xp}{\Pnt}}{\vec{\Mp}{\Dp}})^2
        \prop
    (\rat{\vec{\Ver}{\Xp}}{\vec{\Ver}{\Mp}})(\rat{\vec{\Xp}{\Wv}}{\vec{\Mp}{\Wv}}).
  \end{equation*}
  Apollonius shows that the same proportion holds
  for a section of a cone as in \figref{fig:section},
\begin{figure}
  \centering
  \psset{unit=40mm}
  \begin{pspicture}(-1,-1)(1,1.6)
%    \psgrid
    \psset{Alpha=-34,Beta=33,PointSymbol=none
    }
    \pstThreeDCircle(0,0,0)(1,0,0)(0,1,0)
    \pstThreeDPut(0.2,0.5,1.8){\pnode{\Apex}}\nput{90}{\Apex}{${\Apex}$}
    \pstThreeDPut(0,-1,0){\pnode{\Beat}}\nput{45}{\Beat}{${\Beat}$}
    \ncline{\Apex}{\Beat}
    \pstThreeDPut(0, 1,0){\pnode{\Com}}\nput{225}{\Com}{${\Com}$}
    \ncline{\Beat}{\Com}
    \pstThreeDPut[SphericalCoor](1,130,0){\pnode{\Dp}}\nput{-60}{\Dp}{${\Dp}$}
    \pstThreeDPut[SphericalCoor](1,50,0){\pnode{\Dpo}}\nput{-170}{\Dpo}{${\Dpo}$}
    \ncline{\Dp}{\Dpo}
    \pstInterLL[PosAngle=5]{\Beat}{\Com}{\Dp}{\Dpo}{\Mp}
    \pstHomO[PosAngle=70,PointName={\Ver},HomCoef=.4]{\Apex}{\Beat}[{\Ver}]
    \pstInterLL[PosAngle=-135]{\Ver}{\Mp}{\Apex}{\Com}{\Wv}
    \ncline{\Ver}{\Wv}\ncline{\Apex}{\Wv}
    \pstMiddleAB[PosAngle=-75,PointName=none] {\Ver}{\Wv}{\Koc}
    \pstTranslation[PosAngle=180,PointName=none] {\Mp}{\Dp}{\Koc}[{\hide}]
      \pstHomO[PosAngle=-30,PointName=none,
      HomCoef=\pstDistAB{\Koc}{\Ver}\space dup
      dup mul \pstDistAB{\Koc}{\Mp}\space dup mul sub sqrt div]{\Koc}{\hide}[{\Lk}]
      \pstHomO[PosAngle=150,PointName=none,HomCoef=-1]{\Koc}{\Lk}[{\Lok}]
%      \ncline{\Lk}{\Lok}
      \newcommand{\points}{19}
      % begin multido
      \multido{\i=1+1}{\points}{
        \pstHomO[PointName=none,
          PosAngle=225,
          HomCoef=\i\space \points\space 1 add div 360 mul cos]
        {\Koc}{\Lk}[Y_\i]
        \pstHomO[PointName=none,
          PosAngle=135,%PointSymbol=default,
          HomCoef=\i\space \points\space 1 add div 360 mul sin]
                {\Koc}{\Ver}[X_\i]
                \pstTranslation[%PointSymbol=*,
                  PointName=none,
          PosAngle=-75]{\Koc}{X_\i}{Y_\i}[Z_\i]
      } % end multido
      \pstGenericCurve[GenCurvFirst={\Lk},GenCurvLast={\Lk}]{Z_}1{\points}
%      \psset{PointSymbol=*,linestyle=dotted}
      \uput[-30](Z_1){$\Pnt$}
      \uput[150](Z_9){$\Pnto$}
      \ncline{Z_1}{Z_9}
%      \psdots(Z_1)(Z_9)
      \pstInterLL[PosAngle=30]{Z_1}{Z_9}{\Ver}{\Wv}{\Xp}
  \end{pspicture}
  \caption{Cone, sectioned}\label{fig:section}
  
\end{figure}
where ${\Beat}{\Com}$ is a diameter of the circular base,
${\Dp}{\Dpo}$ is orthogonal to this,
and the apex ${\Apex}$ is any point not in the plane of the base.
Considering the section in isolation,
as in \figref{fig:sect},
\begin{figure}
  \centering
  \begin{pspicture}(-4.6,-3.6)(6.2,3.1)
%    \psgrid
    \psset{PointSymbol=none}
    \parametricplot0{360}{t cos 4 mul t sin 3 mul}
    \pstGeonode[PosAngle={90,-60,75,0,0,-90},
      PointName={default,default,default,none,none,default}
    ]
    (0,0){\Koc}
    (! -12 cos 4 mul -12 sin 3 mul){\Ver}
    (!  35 cos 4 mul  35 sin 3 mul){\Dp}
    (! 0 47 cos)x
    (0,1)t
    (! -95 cos 4 mul -95 sin 3 mul){\Pnt}
    \pstTranslation[PointName=none] t{\Ver}x
    \pstInterLL[PosAngle=-105]{x'}x{\Koc}{\Ver}{\Mp}
    \pstTranslation[PointName=none] x{\Ver}t
    \pstHomO[HomCoef=0.9,PosAngle=-105]{\Koc}{\Ver}[{\Edp}]
    \pstTranslation[PointName=none]{\Mp}{\Dp}{\Ver,\Pnt}[u,v]
    \pstInterLL[PosAngle=75]{\Ver}u{\Koc}{\Dp}{\Endp}
    \pstInterLL[PosAngle=-60]{\Pnt}v{\Koc}{\Ver}{\Xp}
    \pstInterLL[PosAngle=75]{\Pnt}{\Xp}{\Koc}{\Dp}{\Ypn}
    \pstTranslation[PointName=none]{\Edp}{\Dp}{\Pnt}[w]
    \pstInterLL[PosAngle=75]{\Pnt}w{\Koc}{\Ver}{\Yp}
    \pstTranslation[PosAngle=150]{\Ver}{\Koc}{\Koc}[\Wv]
    \ncline{\Wv}{\Ver}
    \ncline{\Mp}{\Dp}
    \ncline{\Ver}{\Endp}
    \ncline{\Xpn}{\Endp}
    \ncline{\Dp}{\Edp}
    \ncline{\Pnt}{\Ypn}
    \ncline{\Pnt}{\Yp}
    \ncline{\Koc}{\Endp}
  \end{pspicture}
  
  \caption{Conic section, isolated}
\label{fig:sect}  
\end{figure}
if we draw ${\Dp}{\Edp}$ at an arbitrary angle,
and let ${\Pnt}{\Yp}$ be parallel,
then
\begin{equation*}
  {\Xp}{\Pnt}^2\propto{\Xp}{\Pnt}{\Yp},
\end{equation*}
while
\begin{equation*}
  {\Xp}{\Wv}
  ={\Xp}{\Koc}+{\Koc}{\Wv}
  ={\Xp}{\Koc}+{\Ver}{\Koc}
  \propto{\Xp}{\Ypn}+{\Ver}{\Endp},
\end{equation*}
so that
\begin{equation*}
  {\Ver}{\Xp}\cdot  {\Xp}{\Wv}
\propto{\Ver}{\Xp}{\Ypn}{\Endp}.
\end{equation*}
Thus
\begin{equation*}
  {\Xp}{\Pnt}{\Yp}
\propto{\Ver}{\Xp}{\Ypn}{\Endp}.
\end{equation*}
We turn this into an equation by making
  \begin{equation*}
    {\Mp}{\Vern}{\Edp}=
  {\Ver}{\Mp}{\Vern}{\Endp},
  \end{equation*}
  as in \figref{fig:sec},
\begin{figure}
  \centering
  \begin{pspicture}(-4.6,-3.6)(6.2,3.1)
%    \psgrid
    \psset{PointSymbol=none}
    \parametricplot0{360}{t cos 4 mul t sin 3 mul}
    \pstGeonode[PosAngle={90,-60,75,0,0,-90},
      PointName={default,default,default,none,none,default}]
    (0,0){\Koc}
    (! -12 cos 4 mul -12 sin 3 mul){\Ver}
    (!  35 cos 4 mul  35 sin 3 mul){\Dp}
    (! 0 47 cos)x
    (0,1)t
    (! -85 cos 4 mul -85 sin 3 mul){\Pnt}
    \pstTranslation[PointName=none] t{\Ver}x
    \pstInterLL[PosAngle=-105]{x'}x{\Koc}{\Ver}{\Mp}
    \pstTranslation[PointName=none] x{\Ver}t
    \pstInterLL[PosAngle=-60]{t'}t{\Koc}{\Ver}{\Edp}
    \pstTranslation[PointName=none]{\Mp}{\Dp}{\Ver,\Pnt}[u,v]
    \pstInterLL[PosAngle=75]{\Ver}u{\Koc}{\Dp}{\Endp}
    \pstInterLL[PosAngle=-60]{\Pnt}v{\Koc}{\Ver}{\Xp}
    \pstInterLL[PosAngle=90]{\Pnt}{\Xp}{\Koc}{\Dp}{\Ypn}
    \pstTranslation[PointName=none]{\Edp}{\Dp}{\Pnt}[w]
    \pstInterLL[PosAngle=75]{\Pnt}w{\Koc}{\Ver}{\Yp}
    \pstInterLL[PosAngle=-150]{\Pnt}{\Yp}{\Koc}{\Dp}{\Xpn}
    \pstTranslation[PosAngle=150]{\Ver}{\Koc}{\Koc}[\Wv]
    \pstInterLL[PosAngle=10]{\Dp}{\Edp}{\Ver}{\Endp}{\Fp}
    \ncline{\Wv}{\Edp}
    \ncline{\Mp}{\Dp}
    \ncline{\Ver}{\Endp}
    \ncline{\Xpn}{\Endp}
    \ncline{\Dp}{\Edp}
    \ncline{\Pnt}{\Ypn}
    \ncline{\Pnt}{\Yp}
  \end{pspicture}
  
  \caption{Conic section, final analysis}
\label{fig:sec}  
\end{figure}
by ensuring
\begin{equation*}
          \rat{\vec{\Koc}{\Mp}}{\vec{\Koc}{\Ver}}
        \prop\rat{\vec{\Koc}{\Ver}}{\vec{\Koc}{\Edp}}.
\end{equation*}
Now
\begin{equation*}
  {\Xp}{\Pnt}{\Yp}
={\Ver}{\Xp}{\Ypn}{\Endp}.
\end{equation*}
We obtain from this
\begin{equation*}
  {\Ypn}{\Pnt}{\Xpn}
  =
  {\Yp}{\Xpn}{\Endp}{\Ver}
\end{equation*}
by adding the quadrilateral
${\Xp}{\Yp}{\Xpn}{\Ypn}$.
Thus we show that the conic section is preserved by what,
in modern terms,
is the affine transformation fixing ${\Koc}$
and interchanging ${\Ver}$ and ${\Dp}$.
The transformation is expressed algebraically in the abstract.
One modern commentator says that such a transformation
is what Apollonius is using;
but he is not using a Cartesian algebra of lengths,
he is using areas.



\chapter{Notes for talk}

Suppose, as in \figref{fig:circle},
  \begin{figure}
    \subfloat[Circle]{\label{fig:circle}
    \psset{PointSymbol=none}
    \begin{pspicture}(0,-2.8)(6,2.5)
%          \psgrid
      \pstGeonode[PosAngle={180,0}](0.5,0){\Com}(5.5,0){\Beat}
      \pstCircleAB{\Com}{\Beat}
      \pstMiddleAB[PointName=none]{\Com}{\Beat}{\mbat}
      \pstHomO[HomCoef=0.25,PosAngle=45]{\Com}{\Beat}[{\Mp}]
      \pstHomO[HomCoef=0.8,PosAngle=45]{\Com}{\Beat}[{\Nmp}]
      \pstBissectBAC[PointName=none,linestyle=none]{\Com}{\Mp}{\Beat}Z
      \pstInterLC[PosAngleA=120,PosAngleB=-120]{\Mp}Z{\mbat}{\Com}{\Dpo}{\Dp}
      \pstTranslation[PointName=none]{\Mp}Z{\Nmp}[{\hide}]
      \pstInterLC[PosAngleA=45,PosAngleB=-45]{\Nmp}{\hide}{\mbat}{\Com}{\Jop}{\Jp}
      \ncline{\Com}{\Beat}\ncline{\Dp}{\Dpo}\ncline{\Jp}{\Jop}
    \end{pspicture}
      }
      \hfill
      \subfloat[Cone]{\label{fig:ax}
  \psset{unit=20mm}
  \begin{pspicture}(-1,-0.7)(1,2.1)
%    \psgrid
  \psset{Alpha=120}
\pstThreeDCircle(0,0,0)(1,0,0)(0,1,0)
\pstThreeDPut(-0.3,0.8,1.8){\pnode{\Apex}}\nput{90}{\Apex}{${\Apex}$}
\pstThreeDPut(0,-1,0){\pnode{\Com}}\nput{210}{\Com}{${\Com}$}
\ncline{\Apex}{\Com}
\pstThreeDPut(0, 1,0){\pnode{\Beat}}\nput{ 30}{\Beat}{${\Beat}$}
\ncline{\Apex}{\Beat}\ncline{\Com}{\Beat}
\pstThreeDPut[SphericalCoor](1,-135,0)
             {\pnode{\Dpo}}\nput{130}{\Dpo}{${\Dpo}$}
\pstThreeDPut[SphericalCoor](1,-40,0)
                          {\pnode{\Dp}}\nput{-90}{\Dp}{${\Dp}$}
\ncline{\Dp}{\Dpo}
\pstInterLL[PosAngle=60,PointSymbol=none]
           {\Com}{\Beat}{\Dp}{\Dpo}{\Mp}
  \end{pspicture}
      }
      \caption{Circle as base of cone}
\end{figure}
diameter ${\Beat}{\Com}$ of a circle
  bisects parallel chords ${\Dp}{\Dpo}$ and ${\Jp}{\Jop}$
  at ${\Mp}$ and ${\Nmp}$ respectively.
  From
  \begin{align*}
        {\Nmp}{\Jp}^2&={\Beat}{\Nmp}\cdot{\Nmp}{\Com},&
        {\Mp}{\Dp}^2&={\Beat}{\Mp}\cdot{\Mp}{\Com}
  \end{align*}
  follows
        \begin{equation}\label{eqn:circle}
          (\rat{\vec{\Nmp}{\Jp}}{\vec{\Mp}{\Dp}})^2
        \prop
    (\rat{\vec{\Beat}{\Nmp}}{\vec{\Beat}{\Mp}})(\rat{\vec{\Nmp}{\Com}}{\vec{\Mp}{\Com}}),
        \end{equation}
        an \textbf{affine} proportion
        (all ratios are of \emph{parallel} directed segments).
        Now let the circle be the base of a cone with apex $\Apex$,
        as in \figref{fig:ax}.
Let a plane containing ${\Mp}{\Dp}$,
but not ${\Apex}$,
cut ${\Apex}{\Beat}$
at ${\Ver}$
and ${\Apex}{\Com}$ at ${\Wv}$
(or perhaps not at all).
We obtain, as in \figref{fig:el},
\begin{figure}
  \centering
  \psset{unit=51mm}
  \begin{pspicture}(-1,-1.2)(1,1.6)
%    \psgrid
    \psset{Alpha=-34,Beta=33,PointSymbol=none
    }
    \pstThreeDCircle(0,0,0)(1,0,0)(0,1,0)
    \pstThreeDPut(-0.3,0.8,2.3){\pnode{\Apex}}\nput{90}{\Apex}{${\Apex}$}
    \pstThreeDPut(0,-1,0){\pnode{\Beat}}\nput{45}{\Beat}{${\Beat}$}
    \ncline{\Apex}{\Beat}
    \pstThreeDPut(0, 1,0){\pnode{\Com}}\nput{225}{\Com}{${\Com}$}
    \ncline{\Beat}{\Com}
    \pstThreeDPut[SphericalCoor](1,130,0){\pnode{\Dp}}\nput{-60}{\Dp}{${\Dp}$}
    \pstThreeDPut[SphericalCoor](1,50,0){\pnode{\Dpo}}\nput{-170}{\Dpo}{${\Dpo}$}
    \ncline{\Dp}{\Dpo}
    \pstInterLL[PosAngle=-60]{\Beat}{\Com}{\Dp}{\Dpo}{\Mp}
    \pstHomO[PosAngle=70,PointName={\Ver},HomCoef=.4]{\Apex}{\Beat}[{\Ver}]
    \pstInterLL[PosAngle=-110]{\Ver}{\Mp}{\Apex}{\Com}{\Wv}
    \ncline{\Ver}{\Wv}\ncline{\Apex}{\Wv}
    \pstMiddleAB[PosAngle=-75,PointName=none] {\Ver}{\Wv}{\Koc}
    \pstTranslation[PosAngle=180,PointName=none] {\Mp}{\Dp}{\Koc}[{\hide}]
      \pstHomO[PosAngle=-30,PointName=none,
      HomCoef=\pstDistAB{\Koc}{\Ver}\space dup
      dup mul \pstDistAB{\Koc}{\Mp}\space dup mul sub sqrt div]{\Koc}{\hide}[{\Lk}]
      \pstHomO[PosAngle=150,PointName=none,HomCoef=-1]{\Koc}{\Lk}[{\Lok}]
%      \ncline{\Lk}{\Lok}
      \newcommand{\points}{19}
      % begin multido
      \multido{\i=1+1}{\points}{
        \pstHomO[PointName=none,
          PosAngle=225,
          HomCoef=\i\space \points\space 1 add div 360 mul cos]
        {\Koc}{\Lk}[Y_\i]
        \pstHomO[PointName=none,
          PosAngle=135,%PointSymbol=default,
          HomCoef=\i\space \points\space 1 add div 360 mul sin]
                {\Koc}{\Ver}[X_\i]
                \pstTranslation[%PointSymbol=*,
                  PointName=none,
          PosAngle=-75]{\Koc}{X_\i}{Y_\i}[Z_\i]
      } % end multido
      \pstGenericCurve[GenCurvFirst={\Lk},GenCurvLast={\Lk}]{Z_}1{\points}
      \psset{PointSymbol=*,linestyle=dotted}
      \uput[30](Z_1){$\Pnt$}
      \uput[150](Z_9){$\Pnto$}
      \ncline{Z_1}{Z_9}
      \psdots(Z_1)(Z_9)
     \pstInterLL{Z_1}{Z_9}{\Ver}{\Wv}{\Xp} \pstTranslation[PointName=none,PointSymbol=none]{\Com}{\Beat}{\Xp}[{\hide}]
      \pstInterLL[PosAngle=180] {\Xp}{\hide}{\Apex}{\Beat}{\Qxr}
      \pstInterLL[PosAngle=150] {\Xp}{\hide}{\Apex}{\Com}{\Rxq}
      \pstInterLL[PosAngle=-90] {\Apex}{\Xp}{\Beat}{\Com}{\Nmp}
      \pstTranslation[PointName=none,PointSymbol=none] {\Mp}{\Dp}{\Nmp}[{\hide}]
      \pstInterLL[PosAngle=-90]{\Nmp}{\hide}{\Apex}{Z_1}{\Jp}
      \pstInterLL[PosAngle=150]{\Nmp}{\hide}{\Apex}{Z_9}{\Jop}
      \ncline{\Apex}{\Nmp}\ncline{\Apex}{\Jp}\ncline{\Apex}{\Jop}\ncline{\Jp}{\Jop}
      \ncline{\Qxr}{\Rxq}
  \end{pspicture}
  \caption{Ellipse in cone}\label{fig:el}
  
\end{figure}
a \textbf{conic section,}
with arbitrary point ${\Pnt}$.
Let
\begin{equation*}
  \Pnt\Pnto\parallel\Dp\Dpo.
\end{equation*}
Then
\begin{equation*}
  \Pnt\Xp=\Xp\Pnto,
\end{equation*}
so $\Ver\Mp$ is a \textbf{diameter}
(\gr{di'ametros}) of the section.
Let also
\begin{equation*}
  {\Qxr}{\Rxq}\parallel\Beat\Com.
\end{equation*}
From \eqref{eqn:circle} and Thales's Theorem,
without using ${\Wv}$,
we compute
\begin{equation}\label{eqn:gen}
  (\rat{\vec{\Xp}{\Pnt}}{\vec{\Mp}{\Dp}})^2
  \prop
  (\rat{\vec{\Ver}{\Xp}}{\vec{\Ver}{\Mp}})
  (\rat{\vec{\Rxq}{\Xp}}{\vec{\Com}{\Mp}}),
\end{equation}
since, in detail,
    \begin{align*}
      %&\phantom{{}\prop{}}
      (\rat{\vec{\Xp}{\Pnt}}{\vec{\Mp}{\Dp}})^2
      &\prop
      (\rat{\vec{\Xp}{\Pnt}}{\vec{\Nmp}{\Jp}})^2
      (\rat{\vec{\Nmp}{\Jp}}{\vec{\Mp}{\Dp}})^2\\
      &\prop
      (\rat{\vec{\Qxr}{\Xp}}{\vec{\Beat}{\Nmp}})
      (\rat{\vec{\Xp}{\Rxq}}{\vec{\Nmp}{\Com}})
      (\rat{\vec{\Beat}{\Nmp}}{\vec{\Beat}{\Mp}})
      (\rat{\vec{\Nmp}{\Com}}{\vec{\Mp}{\Com}})\\
      &\prop
      (\rat{\vec{\Qxr}{\Xp}}{\vec{\Beat}{\Mp}})
      (\rat{\vec{\Xp}{\Rxq}}{\vec{\Mp}{\Com}})\\
      &\prop
      (\rat{\vec{\Ver}{\Xp}}{\vec{\Ver}{\Mp}})
      (\rat{\vec{\Rxq}{\Xp}}{\vec{\Com}{\Mp}}).
\end{align*}
When ${\Wv}$ exists, we have now
\begin{equation}\label{eqn:cen-Ap}
  (\rat{\vec{\Xp}{\Pnt}}{\vec{\Mp}{\Dp}})^2
  \prop
  (\rat{\vec{\Ver}{\Xp}}{\vec{\Ver}{\Mp}})
  (\rat{\vec{\Xp}{\Wv}}{\vec{\Mp}{\Wv}}).
\end{equation}
Let the midpoint of
${\Ver}{\Wv}$ be ${\Koc}$.
We show every line,
as ${\Dp}{\Koc}$,
through ${\Koc}$ is a new diameter,
with respect to which
a proportion as in \eqref{eqn:cen-Ap} is satisfied.
In the plane of the section,
as in \figref{fig:ell},
\begin{figure}
  \centering
  \begin{pspicture}(-4.6,-3.6)(6.2,3.1)
%    \psgrid
    \psset{PointSymbol=none}
    \parametricplot0{360}{t cos 4 mul t sin 3 mul}
    \pstGeonode[PosAngle={90,-60,75,0,0,-90},
      PointName={default,default,default,none,none,default}]
    (0,0){\Koc}
    (! -12 cos 4 mul -12 sin 3 mul){\Ver}
    (!  35 cos 4 mul  35 sin 3 mul){\Dp}
    (! 0 47 cos)x
    (0,1)t
    (! -85 cos 4 mul -85 sin 3 mul){\Pnt}
    \pstTranslation[PointName=none] t{\Ver}x
    \pstInterLL[PosAngle=-105]{x'}x{\Koc}{\Ver}{\Mp}
    \pstTranslation[PointName=none] x{\Ver}t
    \pstInterLL[PosAngle=-60]{t'}t{\Koc}{\Ver}{\Edp}
    \pstTranslation[PointName=none]{\Mp}{\Dp}{\Ver,\Pnt}[u,v]
    \pstInterLL[PosAngle=75]{\Ver}u{\Koc}{\Dp}{\Endp}
    \pstInterLL[PosAngle=-60]{\Pnt}v{\Koc}{\Ver}{\Xp}
    \pstInterLL[PosAngle=90]{\Pnt}{\Xp}{\Koc}{\Dp}{\Ypn}
    \pstTranslation[PointName=none]{\Edp}{\Dp}{\Pnt}[w]
    \pstInterLL[PosAngle=75]{\Pnt}w{\Koc}{\Ver}{\Yp}
    \pstInterLL[PosAngle=-150]{\Pnt}{\Yp}{\Koc}{\Dp}{\Xpn}
    \pstTranslation[PosAngle=150]{\Ver}{\Koc}{\Koc}[\Wv]
    \pstInterLL[PosAngle=10]{\Dp}{\Edp}{\Ver}{\Endp}{\Fp}
    \ncline{\Wv}{\Edp}
    \ncline{\Mp}{\Dp}
    \ncline{\Ver}{\Endp}
    \ncline{\Xpn}{\Endp}
    \ncline{\Dp}{\Edp}
    \ncline{\Pnt}{\Ypn}
    \ncline{\Pnt}{\Yp}
  \end{pspicture}
  
  \caption{Ellipse, isolated}
\label{fig:ell}  
\end{figure}
we let point ${\Edp}$ satisfy
\begin{equation}\label{eqn:KM}
          \rat{\vec{\Koc}{\Mp}}{\vec{\Koc}{\Ver}}
        \prop\rat{\vec{\Koc}{\Ver}}{\vec{\Koc}{\Edp}},
\end{equation}
and then
${\Yp}$, and ${\Endp}$ satisfy
      \begin{align*}
    {\Edp}{\Dp}&\parallel{\Pnt}{\Yp},
    &{\Mp}{\Dp}&\parallel{\Ver}{\Endp}.
      \end{align*}
      By similarity of the triangles,
      \begin{equation}\label{eqn:XP}    
        (\rat{\vec{\Xp}{\Pnt}}{\vec{\Mp}{\Dp}})^2
        \prop
        \rat{{\Xp}{\Pnt}{\Yp}}{{\Mp}{\Vern}{\Edp}}.
      \end{equation}
      Also
      \begin{equation}
        \label{eqn:VX}
  (\rat{\vec{\Ver}{\Xp}}{\vec{\Ver}{\Mp}})
  (\rat{\vec{\Xp}{\Wv}}{\vec{\Mp}{\Wv}})
  \prop\rat{{\Xp}{\Ypn}{\Endp}{\Ver}}
      {{\Mp}{\Vern}{\Endp}{\Ver}},
      \end{equation}
      since
      \begin{align*}
        \rat{\vec{\Xp}{\Wv}}{\vec{\Mp}{\Wv}}
        &\prop\rat{(\vec{\Xp}{\Koc}+\vec{\Koc}{\Wv})}
            {(\vec{\Mp}{\Koc}+\vec{\Koc}{\Wv})}\\
        &\prop\rat{(\vec{\Xp}{\Koc}+\vec{\Ver}{\Koc})}
            {(\vec{\Mp}{\Koc}+\vec{\Ver}{\Koc})}\\
        &\prop\rat{(\vec{\Xp}{\Ypn}+\vec{\Ver}{\Endp})}
            {(\vec{\Mp}{\Ypn}+\vec{\Ver}{\Endp})}.
      \end{align*}
  Since also, by \eqref{eqn:KM},
  \begin{equation}\label{eqn:tri}
    {\Ver}{\Fp}{\Edp}={\Endp}{\Fp}{\Dp},
  \end{equation}
so that
  \begin{equation*}
    {\Mp}{\Vern}{\Edp}=
  {\Mp}{\Vern}{\Endp}{\Ver},
  \end{equation*}
  this with \eqref{eqn:cen-Ap}, \eqref{eqn:XP}, and \eqref{eqn:VX}
yield
\begin{equation}\label{eqn:cen-alt}
  {\Xp}{\Pnt}{\Yp}={\Xp}{\Ypn}{\Endp}{\Ver},
\end{equation}
an alternative formulation of \eqref{eqn:cen-Ap}.
Adding ${\Yp}{\Xpn}{\Ypn}{\Xp}$
to either side of \eqref{eqn:cen-alt},
we obtain
\begin{equation*}
  {\Ypn}{\Pnt}{\Xpn}
  =
  {\Yp}{\Xpn}{\Endp}{\Ver},
\end{equation*}
since, in detail,
\begin{gather*}
  {\Xp}{\Pnt}{\Yp}+{\Yp}{\Xpn}{\Ypn}{\Xp}
  ={\Xp}{\Pnt}{\Yp}{\Xpn}{\Ypn}
  ={\Pnt}{\Xpn}{\Ypn},\\
  {\Yp}{\Xpn}{\Ypn}{\Xp}+  {\Xp}{\Ypn}{\Endp}{\Ver}
  =  {\Yp}{\Xpn}{\Ypn}{\Endp}{\Ver}
  =  {\Yp}{\Xpn}{\Endp}{\Ver}.
\end{gather*}
From \eqref{eqn:tri} again,
\begin{equation*}
  {\Ver}{\Yp}{\Xpn}{\Endp}{\Ver}=
  {\Yp}{\Xpn}{\Dp}{\Edp},
\end{equation*}
and therefore
\begin{equation*}
    {\Ypn}{\Pnt}{\Xp}
  =
  {\Yp}{\Xpn}{\Dp}{\Edp},
\end{equation*}
which is \eqref{eqn:cen-alt} with respect to the new diameter.

\chapter{Additional remarks}

In English, as in Greek,
\begin{compactenum}[1)]
\item
  a \textbf{parable} (\gr{parabol'h}) illustrates a moral;
\item
  \textbf{hyperbole} (\gr{<uperbol'h}) exaggerates;
\item
  \textbf{ellipsis} (\gr{>'elleiyis}) leaves out.
\end{compactenum}
By one account,
that of Hilbert and Cohn-Vossen
\cite{MR0046650},
these terms are assigned respectively to certain curves
for the following reason.
As is proved by Pappus and was apparently known to Euclid,
each of the curves is the locus of $P$,
where
\begin{equation*}
  \frac{\abs{PF}}{\abs{Pd}}=e,
\end{equation*}
where
\begin{compactitem}
  \item
    $F$ is the \textbf{focus,}
  \item
    $d$ is the \textbf{directrix,} and
  \item
    $e$ is the \textbf{eccentricity,}
which respectively
\begin{compactenum}[(1)]
  \item
    equals,
  \item
    exceeds, and
  \item
    falls short of unity,
\end{compactenum}
\end{compactitem}
as in \figref{fig:ecc}.
  \begin{figure}
      \psset{unit=20mm}
  \centering
%  \mbox{}\vfill
  \begin{pspicture*}(-2.7,-1.1)(2.7,2.5)
    %\psgrid
    \psline(-10,-1)(10,-1)
    \parametricplot[linestyle=dashed]{-260}{80}{1 t sin 1 sub neg div dup t cos mul exch t sin mul}
    \parametricplot[plotpoints=200]{-225}{44}{2 sqrt dup t sin mul 1 sub neg div dup t cos mul exch t sin mul}
    \parametricplot[plotpoints=360]0{360}{0.5 sqrt dup t sin mul 1 sub neg div dup t cos mul exch t sin mul}
    \psdot(0,0)
%    \uput[dr](-0.7,2){$e=1/\surd2$}
%    \uput[dr](1,-0.3){$e=\surd2$}
  \end{pspicture*}

  \caption{Conics with eccentricities $\surd2$, $1$, and $1/\surd2$}
  \label{fig:ecc}
    
  \end{figure}
  In the Cartesian plane,
  the curve where
  \begin{compactitem}
    \item
      $F$ is $(0,0)$ and
      \item
    $d$ is $y=-1$
  \end{compactitem}
  has polar equation
  \begin{equation*}
    \frac r{1+r\sin\theta}=e
  \end{equation*}
  or
  \begin{equation*}
    r=\frac{e}{1-e\sin\theta}.
  \end{equation*}
  Hibert and Cohn-Vossen show how any curve so defined
  is a section of a \emph{right} circular cone.

Apollonius works with an arbitrary circular cone,
possibly oblique.
One will not get the focus-directrix property this way.
Apollonius introduces the terms
\begin{inparaenum}[(1)]
\item
  \emph{parabola,}
\item
  \emph{hyperbola,} and
\item
  \emph{ellipse}
\end{inparaenum}
for another reason,
namely that,
  for each point of the curve,
  the square on the \emph{ordinate} respectively
  \begin{compactenum}[(1)]
  \item
    equals,
  \item
    exceeds, and
  \item
    falls short of
  \end{compactenum}
  the rectangle bounded by the \emph{abscissa}
  and the \emph{latus rectum.}
In particular, in the \textbf{parabola} of \figref{fig:par},
\begin{figure}
  \centering
  \psset{unit=51mm}
  \begin{pspicture}(-1,-0.5)(1,2.3)
    %\psgrid
    %    \psset{Alpha=145,PointSymbol=none}
        \psset{Alpha=-28,PointSymbol=none}
    \pstThreeDCircle(0,0,0)(1,0,0)(0,1,0)
    %    \pstThreeDPut(-0.3,0.8,2){\pnode A}\nput{90}A{$A$}
        \pstThreeDPut(-0.3,0.8,3){\pnode{\Apex}}\nput{90}{\Apex}{${\Apex}$}
    \pstThreeDPut(0,-1,0){\pnode{\Beat}}\nput{-75}{\Beat}{${\Beat}$}
    \ncline{\Apex}{\Beat}
    \pstThreeDPut(0, 1,0){\pnode{\Com}}\nput{-120}{\Com}{${\Com}$}
    \ncline{\Apex}{\Com}\ncline{\Beat}{\Com}
    \pstThreeDPut[SphericalCoor](1,160,0){\pnode{\Dp}}\nput{-60}{\Dp}{${\Dp}$}
    \pstThreeDPut[SphericalCoor](1,20,0){\pnode{\Dpo}}\nput{175}{\Dpo}{${\Dpo}$}
    \ncline{\Dp}{\Dpo}
    \pstInterLL[PosAngle=-90]{\Beat}{\Com}{\Dp}{\Dpo}{\Mp}
    \pstTranslation[PointName=none]{\Com}{\Apex}{\Mp}[{\hide}]
    \pstInterLL[PosAngle=-135]{\Apex}{\Beat}{\Mp}{\hide}{\Ver}
    \ncline{\Mp}{\Ver}
    \pstTranslation[PointName=none] {\Mp}{\Dp}{\Ver}[{\hide}]
%    \psset{PointSymbol=*}
    % begin multido
    \newcommand{\points}{13}
    \multido{\i=1+1}{\points}{
      \pstHomO[PointName=none,
        HomCoef=2 \i\space mul \points\space 1 add div 1 sub]
          {\Ver}{\hide}[Y_\i]
      \pstHomO[PointName=none,
  HomCoef=\pstDistAB{\Ver}{Y_\i}\space \pstDistAB{\Ver}{\hide}\space div dup mul]
              {\Ver}{\Mp}[X_\i]
      \pstTranslation[PointName=none,
                  PosAngle=45]{\Ver}{X_\i}{Y_\i}[Z_\i]
      } % end multido
    \pstGenericCurve[GenCurvFirst={\Dpo},GenCurvLast={\Dp}]{Z_}1{\points}
    \uput[r](Z_12){${\Pnt}$}
    \uput[l](Z_2){${\Pnto}$}
    \uput[10](X_2){${\Xp}$}
    \pstInterLL[PosAngle=-80]{\Apex}{\Xp}{\Beat}{\Com}{\Nmp}
    \pstTranslation[PointName=none] {\Mp}{\Com}{X_2}[{\hide}]
    \pstInterLL[PosAngle=-135]{\hide}{X_2}{\Apex}{\Com}{\Rxq}
    \pstInterLL[PosAngle=180]{\hide}{X_2}{\Apex}{\Beat}{\Qxr}
    \pstTranslation[PointName=none] {\Mp}{\Dp}{\Nmp}[{\hide}]
    \pstInterLL[PosAngle=-45] {\Nmp}{\hide}{\Apex}{Z_12}{\Jp}
    \pstInterLL[PosAngle=-75] {\Nmp}{\hide}{\Apex}{Z_2}{\Jop}
    \psset{linestyle=dotted}
    \ncline{\Qxr}{\Rxq}\ncline{Z_12}{Z_2}
    \ncline{\Apex}{\Nmp}
    \ncline{\Jp}{\Jop}\ncline{\Apex}{\Jp}\ncline{\Apex}{\Jop}
    \psdots({\Jp})(Z_12)({\Qxr})(Z_2)({\Jop})({\Rxq})({\Nmp})(X_2)
    %\pscurve({\Rxq})(Z_12)(B')(Z_2)({\Rxq})
  \end{pspicture}
  \caption{Parabola}\label{fig:par}
  
\end{figure}
we have
\begin{equation*}
  {\Mp}{\Ver}\parallel{\Apex}{\Com},
\end{equation*}
and then \eqref{eqn:gen} becomes
\begin{equation*}%\label{eqn:parab-Ap}
  (\rat{\vec{\Xp}{\Pnt}}{\vec{\Mp}{\Dp}})^2
  \prop\rat{\vec{\Ver}{\Xp}}{\vec{\Ver}{\Mp}},
\end{equation*}
meaning the square of the ratio of the ordinates
is the ratio of the abscissas.
In the Euclidean plane, if
\begin{align*}
  \vec{\Xp}{\Pnt}&=y,
 &\vec{\Ver}{\Xp}&=x
\end{align*}
in Cartesian fashion,
 for some $\ell$,
 \begin{equation*}
   y^2=\ell x.
 \end{equation*}
 Here $\ell$ is the length of
 the \textbf{\itshape latus rectum}
 or upright side.
 Applied to this,
 the square on the ordinate ${\Xp}{\Pnt}$
 becomes a rectangle whose base is the abscissa ${\Ver}{\Xp}$.

 An \textbf{hyperbola} is shown in \figref{fig:hyp};
\begin{figure}% hyperbola
  \centering
  \psset{unit=49mm}
  \begin{pspicture}(-1,-0.6)(1,2.5)
%    \psgrid
    \psset{Alpha=120,PointSymbol=none
    }
    \pstThreeDCircle(0,0,0)(1,0,0)(0,1,0)
    \pstThreeDPut(-0.3,0.8,2){\pnode{\Apex}}\nput{0}{\Apex}{${\Apex}$}
    \pstThreeDPut(0,-1,0){\pnode{\Beat}}\nput{210}{\Beat}{${\Beat}$}
    \ncline{\Apex}{\Beat}
    \pstThreeDPut(0, 1,0){\pnode{\Com}}\nput{ 30}{\Com}{${\Com}$}
    \ncline{\Beat}{\Com}
   \pstThreeDPut[SphericalCoor](1,210,0){\pnode{\Dp}}\nput{135}{\Dp}{${\Dp}$}
   \pstThreeDPut[SphericalCoor](1,-30,0){\pnode{\Dpo}}\nput{-90}{\Dpo}{${\Dpo}$}
    \ncline{\Dp}{\Dpo}
    \pstInterLL[PosAngle=-90]{\Beat}{\Com}{\Dp}{\Dpo}{\Mp}
    \pstHomO[PosAngle=120,PointName={\Ver},HomCoef=.35]{\Apex}{\Beat}[{\Ver}]
    \pstInterLL[PosAngle=80]{\Ver}{\Mp}{\Apex}{\Com}{\Wv}
    \ncline{\Mp}{\Wv}\ncline{\Com}{\Wv}
    \pstMiddleAB[PosAngle=210] {\Ver}{\Wv}{\Koc}
    \pstTranslation[PosAngle=180,PointName=none] {\Mp}{\Dp}{\Koc}[{\hide}]
      \pstHomO[PosAngle=135,
      HomCoef=\pstDistAB{\Koc}{\Ver}\space dup
     dup mul \pstDistAB{\Koc}{\Mp}\space dup mul exch sub sqrt div]{\Koc}{\hide}[{\Lk}]
      \pstHomO[PosAngle=-45,HomCoef=-1]{\Koc}{\Lk}[{\Lok}]
      \ncline{\Lk}{\Lok}
      % begin multido
      \newcommand{\points}{19}
      \multido{\i=1+1}{\points}{
        \pstHomO[PointName=none,PosAngle=225,
          HomCoef=2 \i\space mul \points\space 1 add div 1 sub]
            {\Koc}{\hide}[Y_\i]
        \pstHomO[PointName=none,PosAngle=135,
          HomCoef=\pstDistAB{\Koc}{Y_\i}\space
          \pstDistAB{\Koc}{\Lk} div dup mul 1 add sqrt]
                {\Koc}{\Ver}[X_\i]
        \pstTranslation[PointName=none,PosAngle=45]
          {\Koc}{X_\i}{Y_\i}[Z_\i]
      } % end multido
      \pstGenericCurve[GenCurvFirst={\Dpo},GenCurvLast={\Dp}]{Z_}1{\points}
      \psset{PointSymbol=*,linestyle=dotted}
      \uput[ul](Z_16){${\Pnt}$}
      \uput[dr](Z_4){${\Pnto}$}
      \uput[100](X_4){${\Xp}$}
      \ncline{Z_16}{Z_4}
      \psdots(Z_16)(Z_4)(X_4)
      \pstTranslation[PointName=none,PointSymbol=none] {\Beat}{\Mp}{X_4}[{\hide}]
      \pstInterLL[PosAngle=135] {\Apex}{\Beat}{X_4}{\hide}{\Qxr}
      \pstInterLL[PosAngle=30] {\Apex}{\Com}{X_4}{\hide}{\Rxq}
      \ncline{\Qxr}{\Rxq}
      \pstInterLL[PosAngle=-100] {\Apex}{X_4}{\Beat}{\Com}{\Nmp}
      \pstTranslation[PointName=none,PointSymbol=none] {\Mp}{\Dp}{\Nmp}[{\hide}]
      \pstInterLL[PosAngle=135] {\Nmp}{\hide}{\Apex}{Z_16}{\Jp}
      \pstInterLL[PosAngle=-90] {\Nmp}{\hide}{\Apex}{Z_4}{\Jop}
      \ncline{\Apex}{\Nmp}\ncline{\Jp}{\Jop}\ncline{\Apex}{\Jp}\ncline{\Apex}{\Jop}
  \end{pspicture}
  \caption{Hyperbola}\label{fig:hyp}
  
\end{figure}
computations are as for the ellipse of \figref{fig:el},
so that
the square of the ratio of ordinates
is the product of the ratios of the abscissas.
Again in the Euclidean plane,
if now
\begin{equation*}
  \vec{\Ver}{\Wv}=2d,
\end{equation*}
then for some $\ell$,
\begin{equation*}
  y^2=\frac{\ell x}{2d}(2d-x)
  =\ell x-\frac{\ell x^2}{2d},
\end{equation*}
meaning the square on the ordinate
\begin{compactitem}
\item
  exceeds (when $d<0$) and
\item
  falls short of (otherwise)
\end{compactitem}
the rectangle bounded by abscissa and \emph{latus rectum}
by the rectangle on the abscissa
similar to that bounded by \emph{latus transversum}
and \emph{latus rectum.}


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\bibliography{../../../references}

\end{document}
