Q60 / 160 · WBJEE 2012 · Mathematics

Let the foci of the ellipse $\frac{x^{2}}{9}+y^{2}=1$ subtend a right angle at a point $P$. Then, the locus of $P$ is

A
$x^{2}+y^{2}=1$
B
$x^{2}+y^{2}=2$
C
$x^{2}+y^{2}=4$
Correct answer
D
$x^{2}+y^{2}=8$

Solution

$\frac{x^{2}}{9}+\frac{y^{2}}{1}=1 \Rightarrow e=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}$ 0) ie, $(\pm 2 \sqrt{2}, 0)$ Two foci are $(\pm a e,$ Let $P(h, k)$ by any point on the ellipse $\therefore$ $\frac{k-0}{h-2 \sqrt{2}} \times \frac{k-0}{h+2 \sqrt{2}}=-1$ [From given condition] $\Rightarrow \quad h^{2}-8=-k^{2}$ $\Rightarrow \quad x^{2}+y^{2}=8$

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

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