Let $\mathrm{E}:\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\mathrm{a}>\mathrm{b}\text{ and }\mathrm{H}:\frac{{\mathrm{x}}^2}{{\mathrm{A}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{B}}^2}=1$ Let the distance between the foci of E and the foci of H be $2\sqrt{3}$.If a – A = 2, and the ratio of the eccentricities of E and H is $\frac{1}{3}$, then the sum of the lengths of their latus rectums is equal to :

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