Question

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. $x^{2}+y^{2}=4$
• D. $x^{2}+y^{2}=8$

Answer 1

Answer: (D)

$\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$
$\Rightarrow e=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}$
Two foci are $(\pm a e, 0)$ i.e., $(\pm 2 \sqrt{2}, 0)$
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]
$\Rightarrow h^{2}-8=-k^{2}$
$\Rightarrow x^{2}+y^{2}=8$