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 $(\pm ae,0)$ 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]
$\Rightarrow h^2-8=-k^2$
$\Rightarrow x^2+y^2=8$