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Q. If $e_1$ is the eccentricity of the ellipse $\frac {x^2}{16}+\frac{y^2}{25}=1$ and $e_2$ is the eccentricity of the hyperbola passing through the foci of the ellipse and $e_1e_2=1$ then equation of the hyperbola is

IIT JEEIIT JEE 2006Conic Sections

Solution:

The eccentricity of $\frac {x^2}{16}+\frac {y^2}{25}=1$ is
$e_1=(\sqrt{1-\frac{16}{25}})=\frac{3}{5}$
$\therefore e_2=\frac{5}{3}$ $[\because e_1e_2=1]$
$\Rightarrow $ Foci of ellipse $(0,\pm 3)$
$\Rightarrow $ Equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=-1.$