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

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(C) For the hyperbola $2x^2 - 2y^2 = 1$,we can write it as $\frac{x^2}{1/2} - \frac{y^2}{1/2} = 1$. Here $a^2 = 1/2$ and $b^2 = 1/2$.Eccentricity $e_H

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$2$

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(C) For the hyperbola $2x^2 - 2y^2 = 1$,we can write it as $\frac{x^2}{1/2} - \frac{y^2}{1/2} = 1$. Here $a^2 = 1/2$ and $b^2 = 1/2$.Eccentricity $e_H = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2}$.The eccentricity of the ellipse $e_E$ is the reciprocal of $e_H$,so $e_E = \frac{1}{\sqrt{2}}$.Since the ellipse and hyperbola intersect orthogonally,they are confocal.The foci of the hyperbola are $(\pm ae, 0) = (\pm \sqrt{1/2 	imes 2}, 0) = (\pm 1, 0)$.For the ellipse,$ae_E = 1$. Since $e_E = \frac{1}{\sqrt{2}}$,we have $a 	imes \frac{1}{\sqrt{2}} = 1$,so $a = \sqrt{2}$.Using $e_E^2 = 1 - \frac{b^2}{a^2}$,we get $\frac{1}{2} = 1 - \frac{b^2}{2}$,which implies $\frac{b^2}{2} = \frac{1}{2}$,so $b^2 = 1$.The length of the latus-rectum $L = \frac{2b^2}{a} = \frac{2(1)}{\sqrt{2}} = \sqrt{2}$.The square of the length of the latus-rectum is $(\sqrt{2})^2 = 2$.

Column $I$	Column $II$
$(A)$ Circle	$(p)$ The locus of the point $(h, k)$ for which the line $h x+k y=1$ touches the circle $x^2+y^2=4$
$(B)$ Parabola	$(q)$ Points $z$ in the complex plane satisfying $\|z+2\|-\|z-2\|= \pm 3$
$(C)$ Ellipse	$(r)$ Points of the conic have parametric representation $x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{2 t}{1+t^2}$
$(D)$ Hyperbola	$(s)$ The eccentricity of the conic lies in the interval $1 \leq x < \infty$
	$(t)$ Points $z$ in the complex plane satisfying $\operatorname{Re}(z+1)^2=\|z\|^2+1$

Let the eccentricity of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the reciprocal of the eccentricity of the hyperbola $2x^2 - 2y^2 = 1$. If the ellipse intersects the hyperbola at right angles,then the square of the length of the latus-rectum of the ellipse is $................$.

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