\documentclass[a4paper,12pt]{amsart}
\title{Math 304 examination, Tuesday, May 18, 2010}
\author{David Pierce}
%\email{dpierce@metu.edu.tr}
%\address{Mathematics Dept., Middle East Technical University, Ankara
%06531, Turkey}
%\urladdr{metu.edu.tr/~dpierce/Courses/303/}
%\date{Tuesday, March 30, 2010}
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\begin{document}
\maketitle
\thispagestyle{empty}
\begin{problem}\mbox{}
The ellipse $AEB$ is
determined as follows.
Triangle $ABC$ is given, the angle at $A$ being right. If a point $D$
is chosen at random on $AB$, and $DE$ is erected at right angles to
$AB$, then $E$ lies on the ellipse if (and only if) the square on $DE$
is equal to the rectangle $ADFG$ (which is formed by letting $ED$,
extended as necessary, meet $BC$ at $F$). Let also the circle $AHB$
with diameter $AB$ be given.
Find $h$ (in terms of the given straight lines) such that $h$ is
to $AB$ as the ellipse is to the circle.
Prove that your answer is correct, using Newton's lemmas as needed.
\end{problem}
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\newpage
\begin{problem}
We have used without proof Propositions I.33 and 49 of the
\emph{Conics} of Apollonius. This problem is an opportunity to prove
those propositions, using the techniques of Descartes and Newton as
appropriate.
A straight line $\ell$ (not shown), a curved line
$ABE$, and a straight line
$AC$ are given such that, whenever a point $B$ is chosen at random
on $ABE$, and straight line $BC$ is dropped
perpendicular to $AC$, then the square on $BC$ is equal to the
rectangle bounded by $\ell$ and $AC$. So $ABE$ is a parabola with
axis $AC$.
Let $B$ now be fixed; so we may write $BC=a$ and $AC=b$. Extend $CA$ to $D$
so that $AD=AC$. Draw straight line $DBK$, and let $c=BD$.
Let a point $E$ be chosen at random on the parabola $ABE$. Draw
straight lines $BF$ parallel to $AC$, and $EF$ parallel to $BD$.
\begin{enumerate}
\item
Show that the parabola
$ABE$ must indeed lie all on one side of $DBK$.
\item
Show that the square on $EF$ varies as $BF$, and find $m$ (in
terms of $a$, $b$, and $c$ only) such that $m\times BF$ is equal to the
square on $EF$. For your computations, let $x=EF$ and $y=BF$.
\item
Explain why $BD$ is tangent to the parabola at $B$.
\end{enumerate}
\end{problem}
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