Let the eccentricity of an ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ is reciprocal to that of the hyperbola 2x² - 2y² =1. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is_______.

Given equation of the ellipse is
$\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 ......\left(1\right)$
$Also equation of hyperbola is 2x^2 - 2y^2 =1$
$\Rightarrow {\mathrm{x}}^2-{\mathrm{y}}^2=\frac{1}{2}$
Let e₁, e₂ be the eccentricities of eq’s (1) & (2)
Now e₂ = $\sqrt{2}$
Given
${\mathrm{e}}_1=\frac{1}{{\mathrm{e}}_2}=\frac{1}{\sqrt{2}}$
Since ellipse and hyperbola are intersect orthogonally both the curves has same focii. Focii of hyperbola are ($\pm$1, 0)
Now ae₁ = 1
$\Rightarrow \mathrm{a}\cdot \frac{1}{\sqrt{2}}=1\Rightarrow \mathrm{a}=\sqrt{2}$
WKT length of the latus rectum of ellipse = 2b²/a
$$\begin{gathered} \begin{array}{l}\Rightarrow (\text{ L.L.R }{)}^2={\left(\frac{2{\mathrm{b}}^2}{\mathrm{a}}\right)}^2\\ ={\left(\frac{2{\mathrm{a}}^2\left(1-{\mathrm{e}}_1^2\right)}{\mathrm{a}}\right)}^2\\ ={\left(2\mathrm{a}\left(1-{\mathrm{e}}_1^2\right)\right)}^2\\ ={\left(2\cdot \sqrt{2}\left(1-\frac{1}{2}\right)\right)}^2\end{array} \\ \begin{array}{l}={\left(2\sqrt{2}\cdot \frac{1}{2}\right)}^2=(\sqrt{2}{)}^2=2\\ \Rightarrow (\text{ L.L.R }{)}^2=2\end{array} \end{gathered}$$