Let the ellipse \(\mathrm{E}_1: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a} \gt…
Let the ellipse \(\mathrm{E}_1: \frac{x^2}{\mathrm{a}^2}+\frac{y^2}{\mathrm{~b}^2}=1, \mathrm{a} \gt \mathrm{b}\) and \(\mathrm{E}_2: \frac{x^2}{\mathrm{~A}^2}+\frac{y^2}{\mathrm{~B}^2}=1, \mathrm{~A} \lt \mathrm{B}\) 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 \(A B C D\) equals :
\(\frac{12 \sqrt{6}}{5}\)
\(6 \sqrt{6}\)
\(\frac{18 \sqrt{6}}{5}\)
\(\frac{24 \sqrt{6}}{5}\)
Solution
$\begin{aligned} & 2 a e=4 \\ & \Rightarrow \quad a=2 \sqrt{3} \\ & \Rightarrow \quad 1-\frac{b^2}{12}=\frac{1}{3} \Rightarrow b^2=8 \\ & \\ & \frac{2 b^2}{a} \times \frac{2 A^2}{B}=\frac{32}{\sqrt{3}} \\ & \Rightarrow \frac{2 \times 8}{2 \sqrt{3}} \times \frac{2 A^2}{B}=\frac{32}{\sqrt{3}} \\ & \Rightarrow \frac{A^2}{B}=2 \Rightarrow A^2=2 B \\ & 1-\frac{A^2}{B}=\frac{1}{3} \\ & \Rightarrow B=3 \Rightarrow A^2=6\end{aligned}$ $E_1: \frac{x^2}{12}+\frac{y^2}{8}=1$ ....(i) $E_1: \frac{x^2}{6}+\frac{y^2}{9}=1$...(ii) On solving (i) & (ii) $\begin{aligned} &(x, y)=\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right), \left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right)\end{aligned}$ Four points are vertices of rectangle area $=\frac{24 \sqrt{6}}{5}$