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{\bf Question}

An ellipsoid is generated by by rotating an ellipse about its
major axis.  The inside surface of the ellipsoid id silvered to
produce a mirror.  Show that a ray of light emanating from one
focus will be reflected to the other focus.

\vspace{.25in}

{\bf Answer}

{\bf Either}

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${}$

The equation to $\ds \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at
$(x_0, y_0)$ i s$$\frac{x_0}{a^2} + \frac{yy_0}{b^2} = 1$$ so
$y=0$ gives $\ds x = \frac{a^2}{x_0}$ which is the  x coordinate
of T.

So $\ds ST = \frac{a^2}{x_0} - ae \hspace{.2in} S'T =
\frac{a^2}{x_0} + ae$

The distance of P from the directrix $l_1$ is $\ds \frac{a}{e} -
x_0$

So $SP = e\left(\frac{a}{e} - x_0\right) = a-ex_0. \hspace{.3in}
s'P = a+ex_0$

So $\ds \frac{ST}{s'T} = \frac{SP}{S'P}$

Thus $PT$ is an external bisector of $SPS'$ hence by the angles
bisector theorem (converse) the angles are equal as marked.

Hence the reflection property.

${}$

{\bf or}

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${}$

Let P be any point on the ellipse with coordinates $(x_0,y_0)$

\begin{eqnarray*} {\rm The\ gradient\ of\ PT\ is\ }
-\frac{b^2x_0}{a^2y_0} & = & m_1 \\ {\rm The\ gradient\ of\ S'P\
is\ } \frac{y_0}{x_0+ae} & = & m_2' \\ {\rm The\ gradient\ of\ SP\
is\ } \frac{y_0}{x_0-ae} & = & m_2 \end{eqnarray*}
\newpage
\begin{eqnarray*} \tan K\hat{P}T & = & \frac{m_1 -
m_2'}{1+m_1m_2'} \\ & = & \frac{\ds \frac{y_0}{x_0+ae} +
\frac{b^2x_0}{a^2y_0}}{1 - \ds \frac{y_0}{x_0+ae} \cdot
\frac{b^2}{a^2y_0}} \\ & = & \frac{a_2y_0^2 + b^2x_0^2 +
b^2aex_0}{(a^2-b^2)x_0y_0 + a^3y_0e} \\ & = & \frac{a^2b^2 +
b^2aex_0}{a^3y_0e + a^2e^2x_0y_0} \\ & = & \frac{b^2}{aey_0}
\end{eqnarray*}

To obtain $\ds \frac{m_1 - m_2}{1 + m_1m_2}$ replace $a$ by $-a$
in above, so that $\ds \tan H \hat{P} T = -\frac{b}{ay_oe}$

So $K\hat{P}T = \pi - H \hat{P} T$

Therefore $L\hat{P}S' = S \hat{P} T$ QED

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