The area (in square units) of the quadrilater | vedclass

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(D) For the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$,$a^2=9$ and $b^2=5$. The eccentricity $e = \sqrt{1-\frac{5}{9}} = \frac{2}{3}$. The foci are $(\pm

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$8$

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(D) For the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$,$a^2=9$ and $b^2=5$. The eccentricity $e = \sqrt{1-\frac{5}{9}} = \frac{2}{3}$. The foci are $(\pm ae, 0) = (\pm 2, 0)$.For the ellipse $\frac{x^2}{5}+\frac{y^2}{9}=1$,$a^2=9$ and $b^2=5$ (with major axis along the $y$-axis). The eccentricity $e = \sqrt{1-\frac{5}{9}} = \frac{2}{3}$. The foci are $(0, \pm ae) = (0, \pm 2)$.The vertices of the quadrilateral are $(2, 0), (0, 2), (-2, 0),$ and $(0, -2)$.This quadrilateral is a rhombus with diagonals of length $d_1 = 4$ and $d_2 = 4$.The area of the quadrilateral is $\frac{1}{2} 	imes d_1 	imes d_2 = \frac{1}{2} 	imes 4 	imes 4 = 8$ square units.

The area (in square units) of the quadrilateral formed by joining the foci of the two ellipses $\frac{x^2}{9}+\frac{y^2}{5}=1$ and $\frac{x^2}{5}+\frac{y^2}{9}=1$ is

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