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


{\bf Question}

The lines $\ds \frac{x}{a}-\frac{y}{b} = 0 \, \frac{x}{a} +
\frac{y}{b} = 0$ are the asymptotes of the hyperbola $\ds
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.$  If the asymptotes are at
right angles the hyperbola is called a rectangular hyperbola. Find
a condition on $a,b$ for this to be so.  Find the eccentricity of
a rectangular.  Find the equation of a rectangular hyperbola
referred to its asmptotes as axes.

\vspace{.25in}

{\bf Answer}

$\ds \frac{x}{a} - \frac{y}{b} = 0$ - slope $\ds \frac{b}{a}$

$\ds \frac{x}{a} + \frac{y}{b} = 0$ - slope $\ds -\frac{b}{a}$

So they are orthogonal iff $\frac{b^2}{a^2}=1 \hspace{.2in} {\rm
i.e.} b = \pm a$

So the two lines become $x=y$ $x=-y$ and the hyperbola becomes
$\ds \frac{x^2}{a^2} - \frac{y^2}{a^2} = 1$

So $a^2(1-e^2) = a^2 \hspace{.2in} 1 - e^2 = -1 \hspace{.2in} e =
\sqrt 2$

The X-Y axis are obtained by rotation of $45^\circ$

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\put(1,0){\vector(0,1){2}}

\put(0,1){\vector(1,0){2}}

\put(2,0){\vector(-1,1){2}}

\put(2.3,1){\makebox(0,0){x}}

\put(1,2.3){\makebox(0,0){y}}

\put(2,2.3){\makebox(0,0){X}}

\put(0,2.3){\makebox(0,0){Y}}


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\begin{eqnarray*} x & = & X\frac{1}{\sqrt2} - Y\frac{1}{\sqrt2} \\ y &
= & X \frac{1}{\sqrt2} + Y\frac{1}{\sqrt 2} \\ \\ \frac{x^2}{a^2}
- \frac{y^2}{a^2} & = & \frac{1}{2a^2} \left( (X-Y)^2 - (Y + X)^2
\right) = 1 \\ {\rm So \ \ } XY & = & -\frac{a^2}{2}
\end{eqnarray*}


\end{document}
