\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

Find the polar equation of the ellipse $$\frac{x^2}{9} +
\frac{y^2}{4} = 1$$ with the origin at a focus.  Find the polar
equation with the origin at the other focus.

\vspace{.25in}

{\bf Answer}

$$\frac{x^2}{9} + \frac{y^2}{4} = 1\hspace{.3in} \lq \lq a = 3, \,
b=2 " $$  So $\ds 1 - e^2 = \frac{4}{9} \Rightarrow
e^2=\frac{5}{9} \Rightarrow e=\frac{\sqrt5}{3}$.  The Foci are at
$(\pm \sqrt 5, 0)$

When $x=\sqrt5$ we have $\ds \frac{5}{9}+\frac{y^2}{4} = 1
\Rightarrow y= \pm\frac{4}{3}$

So $\ds l = \frac{4}{3}$

Thus the polar equation is $$\frac{4}{3r} = 1 - \frac{\sqrt5}{3}
\cos \theta$$

Referred to the other focus the equation is $$\frac{4}{3r} = 1 +
\frac{\sqrt5}{3} \cos \theta$$

\end{document}