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{\bf Question}

If $c$ is a fixed positive constant, then $$\frac{x^2}{\lambda^2}
+ \frac{y^2}{\lambda^2-c^2} = 1 \hspace{.5in} c^2 < \lambda^2$$
defines a family of ellipses, any member of which is characterised
by a particular value of $\lambda$.  Show that every member of the
family $$\frac{x^2}{\mu^2} - \frac{y^2}{c^2-\mu^2} = 1
\hspace{.5in} \mu^2 < c^2$$ intersects any member of the first
family at right angles.

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{\bf Answer}

$\ds \frac{x^2}{a^2} +\frac{y^2}{b^2} = 1 {\rm \ \ gives\ \ }
\frac{dy}{dx} = -\frac{x}{y} \cdot \frac{b^2}{a^2}$

$\ds \frac{x^2}{c^2} -\frac{y^2}{d^2} = 1 {\rm \ \ gives\ \ }
\frac{dy}{dx} = \frac{x}{y} \cdot \frac{d^2}{c^2}$

So the product of slopes where a curve of each system intersects
is $$-\frac{x^2}{y^2} \cdot \frac{\lambda^2-c^2}{\lambda^2} \cdot
\frac{c^2-\mu^2}{\mu^2}$$ where they
intersect$$-\frac{x^2}{\lambda^2} + \frac{y^2}{\lambda^2 - c^2} =
1 = \frac{x^2}{\mu^2} - \frac{y^2}{c^2 - mu^2}$$ So $\ds
\frac{x^2}{y^2} \frac{(\lambda^2 - c^2)(c^2 -
\mu^2)}{\lambda^2\mu^2} = 1$

Thus the systems of curves are mutually orthogonal.

The foci are all at $(\pm c, 0)$

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