QJEEM23Ma0a1e: JEE Main 2023 Mathematics - Class 11 JEEM23Ma0a1e Solution: Let the eccentricity of an ellipse x 2 a 2 + y 2 b 2 = 1 is…

Equation of hyperbola is $2 x^2-2 y^2=1$

$\Rightarrow \frac{x^2}{1 / 2}-\frac{y^2}{1 / 2}=1$

Here, $a=b$

$\therefore$ Eccentricity of rectangular hyperbola is $\sqrt{2}$

$\therefore$ Eccentricity of ellipse is $\frac{1}{\sqrt{2}}$

Since, ellipse intersects the hyperbola at right angles

$\therefore$ Ellipse and the hyperbola are confocal

$\therefore$ Foci of hyperbola $=(1,0)$

$\Rightarrow$ Foci of ellipse, $a e=1$

$$\begin{array}{ll}& \Rightarrow a\left(\frac{1}{\sqrt{2}}\right)=1\\ \\ & \Rightarrow a=\sqrt{2}\\ \\ & \therefore e^2=1-\frac{b^2}{a^2}\\ \\ & \Rightarrow \frac{1}{2}=1-\frac{b^2}{2}\\ \\ & \Rightarrow \frac{b^2}{2}=\frac{1}{2}\\ \\ & \Rightarrow b^2=1\end{array}$$

$\therefore$ Length of the latus rectum $=\frac{2 b^2}{a}=\frac{2}{\sqrt{2}}=\sqrt{2}$

$\therefore$ Square of length of the latus rectum $=2$