Mathematics Question from JEE Main 2025 · 29 January, Shift 1

Let the ellipse $E_1 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a > b$ and $E_2 : \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, $A < B$, have the same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$ and the distance between the foci of $E_1$ be $4$. If $E_1$ and $E_2$ meet at A, B, C, and D, then the area of the quadrilateral ABCD equals: