JEE Main 2025 — Hyperbola Question with Solution

$\begin{aligned} & \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \text { foci are }(\mathrm{ae}, 0) \text { and }(-\mathrm{ae}, 0) \\ & \frac{\mathrm{x}^2}{\mathrm{~A}^2}+\frac{\mathrm{y}^2}{\mathrm{~B}^2}=1 \text { foci are }\left(\mathrm{Ae}^{\prime}, 0\right) \text { and }\left(-\mathrm{Ae}^{\prime}, 0\right) \\ & \Rightarrow 2 \mathrm{ae}=2 \sqrt{3} \Rightarrow \mathrm{ae}=\sqrt{3} \\ & \text { and } 2 \mathrm{Ae}^{\prime}=2 \sqrt{3} \Rightarrow \mathrm{Ae}^{\prime}=\sqrt{3} \\ & \Rightarrow \mathrm{ae}=\mathrm{Ae}^{\prime} \Rightarrow \frac{\mathrm{e}}{\mathrm{e}^{\prime}}=\frac{\mathrm{A}}{\mathrm{a}} \\ & \Rightarrow \frac{1}{3}=\frac{\mathrm{A}}{\mathrm{a}} \Rightarrow \mathrm{a}=3 \mathrm{~A} \end{aligned}$ Now $\mathrm{a}-\mathrm{A}=2\Rightarrow \mathrm{a}-\frac{\mathrm{a}}{3}-2\Rightarrow \mathrm{a}=3$ and $\mathrm{A}=1$ $\mathrm{Ae}=\sqrt{3}\Rightarrow \mathrm{e}=\frac{1}{\sqrt{3}}\text{ and }{\mathrm{e}}^{\prime }=\sqrt{3}$ $\begin{aligned} & \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right) \\ & \mathrm{b}^2=6 \end{aligned}$ and ${\mathrm{B}}^2={\mathrm{A}}^2\left({\left({\mathrm{e}}^{\prime }\right)}^2-1\right)=(2)\Rightarrow {\mathrm{B}}^2=2$ $\text{ sum of }\mathrm{LR}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}+\frac{2{\mathrm{B}}^2}{\mathrm{A}}=8$