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Let the eccentricity of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is reciprocal to that of the hyperbola $2 x^2-2 y^2=1$. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is $................$.
$4$
$6$
$2$
$8$
Solution
$e _{ H }=\sqrt{2}$
$e _{ E }=\frac{1}{\sqrt{2}}$
Since the curves intersect each other orthogonally The ellipse and the hyperbola are confocal
$H: \frac{x^2}{1 / 2}-\frac{y^2}{1 / 2}=1$
$\Rightarrow \text { foci }=(1,0)$
For ellipse a.e $E _{ E }=1$
$\Rightarrow a =\sqrt{2}$
$\left(e_{ E }\right)^2=\frac{1}{2} \Rightarrow 1-\frac{ b ^2}{ a ^2}=\frac{1}{2} \Rightarrow \frac{ b ^2}{ a ^2}=\frac{1}{2}$
$\Rightarrow b ^2=1$
Length of L.R. $=\frac{2 b^2}{a}=\frac{2}{\sqrt{2}}=\sqrt{2}$
