Let the eccentricity of an ellipse frac{x^2}{ | vedclass

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Question

$e _{ H }=\sqrt{2}$

$e _{ E }=\frac{1}{\sqrt{2}}$

Since the curves intersect each other orthogonally The ellipse and the hyperbola are confocal

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$e _{ H }=\sqrt{2}$

$e _{ E }=\frac{1}{\sqrt{2}}$

Since the curves intersect each other orthogonally The ellipse and the hyperbola are confocal

$H: \frac{x^2}{1 / 2}-\frac{y^2}{1 / 2}=1$

$\Rightarrow 	ext { foci }=(1,0)$

For ellipse a.e $E _{ E }=1$

$\Rightarrow a =\sqrt{2}$

$\left(e_{ E }ight)^2=\frac{1}{2} \Rightarrow 1-\frac{ b ^2}{ a ^2}=\frac{1}{2} \Rightarrow \frac{ b ^2}{ a ^2}=\frac{1}{2}$

$\Rightarrow b ^2=1$

Length of L.R. $=\frac{2 b^2}{a}=\frac{2}{\sqrt{2}}=\sqrt{2}$

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 $................$.

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