Let the ellipse E_1: x^2/(a^2)+y^2/(b^2)=1, a > b and E_2: x^2/(A^2)+y^2/(B^2)=1, A < B have same eccentricity 1/(√ 3). Let the product of their…

Let the ellipse $\displaystyle \mathrm{E}_1: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a}>\mathrm{b}$ and $\displaystyle \mathrm{E}_2: \frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1, \mathrm{~A}<\mathrm{B}$ have same eccentricity $\displaystyle \frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\displaystyle \frac{32}{\sqrt{3}}$, and the distance between the foci of $\displaystyle \mathrm{E}_1$ be 4. If $\displaystyle \mathrm{E}_1$ and $\displaystyle \mathrm{E}_2$ meet at A, B, C and D, then the area of the quadrilateral ABCD equals :