Let the ellipses E_1:x^2a^2+y^2b^2=1 (ab) and E_2:x^2A^2+y^2B^2=1 (A

Question

Let the ellipses $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\ (a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1\ (A<B)$ have the same eccentricity $\dfrac{1}{\sqrt3}$. Let the product of their lengths of latus rectums be $\dfrac{32}{\sqrt3}$, and the distance between the foci of $E_1$ be $4$. If $E_1$ and $E_2$ meet at $A,B,C,D$, then the area of quadrilateral $ABCD$ equals:

Accepted Answer

$2ae=4$ with $e=\dfrac{1}{\sqrt3}\Rightarrow a=2\sqrt3$; $b^2=a^2(1-e^2)=12\times\dfrac23=8$.

$L_1=\dfrac{2b^2}{a}=\dfrac{16}{2\sqrt3}=\dfrac{8}{\sqrt3}$. Since $L_1L_2=\dfrac{32}{\sqrt3}$, $L_2=4=\dfrac{2A^2}{B}\Rightarrow A^2=2B$.

For $E_2$: $e^2=1-\dfrac{A^2}{B^2}=\dfrac13 \Rightarrow A^2=\dfrac{2B^2}{3}$. Combined with $A^2=2B$: $2B=\dfrac{2B^2}{3}\Rightarrow B=3,\ A^2=6$.

$E_1:\dfrac{x^2}{12}+\dfrac{y^2}{8}=1$, $E_2:\dfrac{x^2}{6}+\dfrac{y^2}{9}=1$. Solving simultaneously: $y^2=\dfrac{36}{5},\ x^2=\dfrac65$, giving the rectangle vertices $\left(\pm\dfrac{\sqrt6}{\sqrt5},\pm\dfrac{6}{\sqrt5}\right)$.

$$\text{Area}=2\cdot\frac{\sqrt6}{\sqrt5}\times2\cdot\frac{6}{\sqrt5}=\frac{24\sqrt6}{5}.$$

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