JEE Main 2025 — 22 January, Evening Shift — Question 16

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ foci are $(a e, 0)$ and $(-a e, 0)$

$\frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1$ foci are $\left(\mathrm{Ae}^{\prime}, 0\right)$ and $\left(-\mathrm{Ae}^{\prime}, 0\right)$

$\Rightarrow 2 \mathrm{ae}=2 \sqrt{3} \Rightarrow \mathrm{ae}=\sqrt{3}$

and $2 \mathrm{Ae}^{\prime}=2 \sqrt{3} \Rightarrow \mathrm{Ae}^{\prime}=\sqrt{3}$

$\Rightarrow \mathrm{ae}=\mathrm{Ae}^{\prime} \Rightarrow \frac{\mathrm{e}}{\mathrm{e}^{\prime}}=\frac{\mathrm{A}}{\mathrm{a}}$

$\Rightarrow \frac{1}{3}=\frac{\mathrm{A}}{\mathrm{a}} \Rightarrow \mathrm{a}=3 \mathrm{~A}$

Now $\mathrm{a}-\mathrm{A}=2 \Rightarrow \mathrm{a}-\frac{\mathrm{a}}{3}-2 \Rightarrow \mathrm{a}=3$ and $\mathrm{A}=1$

$\mathrm{ae}=\sqrt{3} \Rightarrow \mathrm{e}=\frac{1}{\sqrt{3}}$ and $\mathrm{e}^{\prime}=\sqrt{3}$

$\mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)$

$b^{2}=6$ and

$\mathrm{B}^{2}=\mathrm{A}^{2}\left(\left(\mathrm{e}^{\prime}\right)^{2}-1\right)=(2) \Rightarrow \mathrm{B}^{2}=2$

sum of $L R=\frac{2 b^{2}}{a}+\frac{2 \mathrm{~B}^{2}}{\mathrm{~A}}=8$