Let the ellipse, \({\mathrm{E}}_{1}:\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=…

Let the ellipse, \({\mathrm{E}}_{1}:\frac{{\mathrm{x}}^{2}}{{\mathrm{a}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{b}}^{2}}=1,\mathrm{a}>\mathrm{b}\) and \({\mathrm{E}}_{2}:\frac{{\mathrm{x}}^{2}}{{\mathrm{A}}^{2}}+\frac{{\mathrm{y}}^{2}}{{\mathrm{B}}^{2}}=1,\mathrm{A}<\mathrm{B}\) have 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:

[JEE Main 2025, 29 Jan (Shift 1)]