Mathematics - Hyperbola Question with Solution | TestHub

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

Now $\mathrm{a}-\mathrm{A}=2\Rightarrow \mathrm{a}-\frac{\mathrm{a}}{3}-2\Rightarrow \mathrm{a}=3$ and $\mathrm{A}=1$
$\mathrm{Ae}=\sqrt{3}\Rightarrow \mathrm{e}=\frac{1}{\sqrt{3}}\text{ and }{\mathrm{e}}^{\prime }=\sqrt{3}$
$\begin{array}{rl}& {\mathrm{b}}^2={\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right)\\ & {\mathrm{b}}^2=6\end{array}$
and ${\mathrm{B}}^2={\mathrm{A}}^2\left({\left({\mathrm{e}}^{\prime }\right)}^2-1\right)=(2)\Rightarrow {\mathrm{B}}^2=2$
$\text{ sum of }\mathrm{LR}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}+\frac{2{\mathrm{B}}^2}{\mathrm{A}}=8$