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 ___________.

Equation of hyperbola is $2 x^2-2 y^2=1$ <br/><br/>$\Rightarrow \frac{x^2}{1 / 2}-\frac{y^2}{1 / 2}=1$ <br/><br/>Here, $a=b$ <br/><br/>$\therefore$ Eccentricity of rectangular hyperbola is $\sqrt{2}$ <br/><br/>$\therefore$ Eccentricity of ellipse is $\frac{1}{\sqrt{2}}$ <br/><br/>Since, ellipse intersects the hyperbola at right angles <br/><br/>$\therefore$ Ellipse and the hyperbola are confocal <br/><br/>$\therefore$ Foci of hyperbola $=(1,0)$ <br/><br/>$\Rightarrow$ Foci of ellipse, $a e=1$ <br/><br/>$$ \begin{array}{ll} &\Rightarrow a\left(\frac{1}{\sqrt{2}}\right)=1 \\\\ &\Rightarrow a=\sqrt{2} \\\\ &\therefore e^2=1-\frac{b^2}{a^2} \\\\ &\Rightarrow \frac{1}{2}=1-\frac{b^2}{2} \\\\ &\Rightarrow \frac{b^2}{2}=\frac{1}{2} \\\\ &\Rightarrow b^2=1 \end{array} $$ <br/><br/>$\therefore$ Length of the latus rectum $=\frac{2 b^2}{a}=\frac{2}{\sqrt{2}}=\sqrt{2}$ <br/><br/>$\therefore$ Square of length of the latus rectum $=2$