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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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Verified Answer
The correct answer is:
$\frac{x^2}{100}+\frac{y^2}{4}=1$
Let the equation of required ellipse is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad(a>b)$
So, according to given information
$2 a e=8 \sqrt{6}$ $\ldots$ (i)
and $(2 a)(2 b)=80$ $\ldots$ (ii)
$\because \quad e=\sqrt{1-\frac{b^2}{a^2}}$,
So, $a^2\left(\frac{a^2-b^2}{a^2}\right)=96$ [from Eq. (i)]
$\Rightarrow \quad a^2-b^2=96$ $\ldots$ (iii)
and $\quad a b=20 \ldots$ (iv) [from Eq. (ii)]
$\therefore \quad\left(a^2+b^2\right)^2=\left(a^2-b^2\right)^2+(2 a b)^2$
$=(96)^2+(40)^2=10816 \Rightarrow a^2+b^2=104 \quad \ldots(\mathrm{v})$
From Eqs. (iii) and (v), we get
$a^2=100 \text { and } b^2=4$
Therefore, equation of required ellipse is
$\frac{x^2}{100}+\frac{y^2}{4}=1$
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \quad(a>b)$
So, according to given information
$2 a e=8 \sqrt{6}$ $\ldots$ (i)
and $(2 a)(2 b)=80$ $\ldots$ (ii)
$\because \quad e=\sqrt{1-\frac{b^2}{a^2}}$,
So, $a^2\left(\frac{a^2-b^2}{a^2}\right)=96$ [from Eq. (i)]
$\Rightarrow \quad a^2-b^2=96$ $\ldots$ (iii)
and $\quad a b=20 \ldots$ (iv) [from Eq. (ii)]
$\therefore \quad\left(a^2+b^2\right)^2=\left(a^2-b^2\right)^2+(2 a b)^2$
$=(96)^2+(40)^2=10816 \Rightarrow a^2+b^2=104 \quad \ldots(\mathrm{v})$
From Eqs. (iii) and (v), we get
$a^2=100 \text { and } b^2=4$
Therefore, equation of required ellipse is
$\frac{x^2}{100}+\frac{y^2}{4}=1$
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