Let the eccentricity of an ellipse x ^ 2 a ^ 2 + y ^ 2 b ^ 2 = 1 is...

Given,

The eccentricity of an ellipse $\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$ is reciprocal to that of the hyperbola $2{x}^{2} - 2{y}^{2} = 1$,

Now eccentricity of rectangular hyperbola $2{x}^{2} - 2{y}^{2} = 1$ is ${e}_{H} = \sqrt{2}$

So, ${e}_{e} = \dfrac{1}{\sqrt{2}}$

Now, focus of hyperbola $= \left(\pm 1, 0\right)$

Now given that both curve intersect orthogonally, so ellipse and hyperbola are confocal

So, for ellipse $a{e}_{e} = 1 ⇒ a = \sqrt{2}$

Now length of latusrectum $L . R . = \dfrac{2{b}^{2}}{a} = 2a\left(1 - {{e}_{e}}^{2}\right)$

$= 2\sqrt{2} . \dfrac{1}{2} = \sqrt{2}$