Let E: x^2 a^2+ y^2 ~b^2=1, agt b and H: x^2 ~A^2- y^2 ~B^2=1. Let the distance between the foci of E and the foci of H be 2 √3. If a-A=2, and the…

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