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(D) Let the equation of the required ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad (a>b)$.According to the given information,the distance from

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$\frac{x^2}{100}+\frac{y^2}{4}=1$

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(D) Let the equation of the required ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad (a>b)$.According to the given information,the distance from each focus to the centre is $ae$,so the sum of distances is $ae + ae = 2ae = 8\sqrt{6} \Rightarrow ae = 4\sqrt{6}$.Squaring both sides,$a^2e^2 = 16 	imes 6 = 96$.Since $e^2 = 1 - \frac{b^2}{a^2}$,we have $a^2(1 - \frac{b^2}{a^2}) = 96$,which simplifies to $a^2 - b^2 = 96 \quad (i)$.The smallest rectangle in which the ellipse is inscribed has dimensions $2a 	imes 2b$. Thus,$4ab = 80$ $\Rightarrow ab = 20$ $\Rightarrow a^2b^2 = 400 \quad (ii)$.Using the identity $(a^2+b^2)^2 = (a^2-b^2)^2 + 4a^2b^2$,we get $(a^2+b^2)^2 = (96)^2 + 4(400) = 9216 + 1600 = 10816$.Taking the square root,$a^2+b^2 = \sqrt{10816} = 104 \quad (iii)$.Adding $(i)$ and $(iii)$,$2a^2 = 200 \Rightarrow a^2 = 100$.Subtracting $(i)$ from $(iii)$,$2b^2 = 8 \Rightarrow b^2 = 4$.Therefore,the equation of the ellipse is $\frac{x^2}{100}+\frac{y^2}{4}=1$.

If the sum of the distances from the foci to the centre $O(0,0)$ of an ellipse is $8 \sqrt{6}$ units and the area of the smallest rectangle in which that ellipse is inscribed is $80$ sq. units,then the equation of such an ellipse is

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