Q96 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Mathematics

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

  • A
    $\frac{x^2}{100}+\frac{y^2}{64}=1$
  • B
    $\frac{x^2}{100}+\frac{y^2}{16}=1$
  • C
    $\frac{x^2}{10}+\frac{y^2}{4}=1$
  • Correct answer
    D
    $\frac{x^2}{100}+\frac{y^2}{4}=1$

Solution

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

Hence, the correct answer is option: 'D'.

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