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


{\bf Question}

Find the equation of the ellipse with foci $(-1,-1), \, (2,3)$ and
major axis of length 10.  Find the centre of this ellipse and
calculate its eccentricity.

\vspace{.25in}

{\bf Answer}

\setlength{\unitlength}{.5in}
\begin{picture}(9,6)
\put(0,2){\vector(1,0){5}}

\put(5.3,2){\makebox(0,0){x}}

\put(2.3,0){\vector(0,1){5}}

\put(2.3,5.3){\makebox(0,0){y}}

\put(0.5,0.5){\vector(1,1){4.5}}

\put(5.3,5.3){\makebox(0,0){X}}

\put(5,1){\vector(-1,1){4}}

\put(1,5.3){\makebox(0,0){Y}}

\multiput(3,0)(0,0.2){25}{\circle*{0.05}}

\put(3,5){\vector(0,1){0}}

\put(3,5.3){\makebox(0,0){$\eta$}}

\multiput(0,3)(0.2,0){25}{\circle*{0.05}}

\put(5,3){\vector(1,0){0}}

\put(5.3,3){\makebox(0,0){$\xi$}}

\put(1,1){\circle*{0.1}}

\put(1.3,1){\makebox(0,0){$s_1$}}

\put(4,4){\circle*{0.1}}

\put(4.3,4){\makebox(0,0){$s_2$}}

\put(3.4,3.2){\makebox(0,0){$\alpha$}}

\put(8,4){\makebox(0,0){ $\begin{array}{l} s_1 = (-1,-1) \\ s_2 =
(2,3) \\ s_1s_2 = 5 \\ {\rm Major\ axis} = 10 \\ {\rm so\ } e =
\frac{1}{2} \\ {\rm The\ centre\ is\ } (\frac{1}{2},1) \\ \lq \lq
a " = 5 \end{array}$}}

\end{picture}


Referred to the X-Y axis the equation is $$\frac{x^2}{25} +
\frac{Y^2}{25(1 - \frac{1}{4})} = 1 \hspace{.2in}\frac{X^2}{25} +
\frac{4y^2}{75} = 1 \Rightarrow 3X^2 +4Y2 = 75$$

X-Y axis are obtained from $\xi - \eta$ axis where $\sin \alpha =
\frac{4}{5} \, \, \, \cos \alpha = \frac{3}{5}$ \begin{eqnarray*}
X & = & \xi \cos \alpha + \eta \sin \alpha = \frac{3}{5} \xi +
\frac{4}{5} \eta \\ Y & = & \eta \cos \alpha - \xi \sin \alpha =
\frac{3}{5} \eta - \frac{4}{5} \xi \end{eqnarray*} which gives
$$91\xi^2 - 24\xi\eta + 84\eta^2 = 1875$$ Now $\ds \xi =
x-\frac{1}{2}$ and $\ds \eta = y-1$

So the equation is $$364x^2 - 96xy + 336y^2 - 268x - 624y -
7121=0$$






\end{document}