Let E: x^2/(a^2)+y^2/(b^2)=1, a > 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…

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