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(D) Given eccentricity $e = \frac{3}{5}$ and distance between foci $2ae = 6$.$2a(\frac{3}{5}) = 6 \Rightarrow a = 5$.Using the relation $b^2 = a^2(1 -

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$40$

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(D) Given eccentricity $e = \frac{3}{5}$ and distance between foci $2ae = 6$.$2a(\frac{3}{5}) = 6 \Rightarrow a = 5$.Using the relation $b^2 = a^2(1 - e^2)$:$b^2 = 25(1 - \frac{9}{25}) = 25(\frac{16}{25}) = 16 \Rightarrow b = 4$.The vertices of the ellipse are $(\pm a, 0)$ and $(0, \pm b)$,which are $(\pm 5, 0)$ and $(0, \pm 4)$.The quadrilateral formed by these vertices is a rhombus with diagonals of length $2a = 10$ and $2b = 8$.Area of the quadrilateral $= \frac{1}{2} 	imes d_1 	imes d_2 = \frac{1}{2} 	imes 10 	imes 8 = 40$ sq. units.

Consider an ellipse,whose centre is at the origin and its major axis is along the $x-$ axis. If its eccentricity is $\frac{3}{5}$ and the distance between its foci is $6$,then the area (in sq. units) of the quadrilateral inscribed in the ellipse,with the vertices as the vertices of the ellipse,is

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