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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
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Verified Answer
The correct answer is:
$x^{2}+y^{2}=8$
$\frac{x^{2}}{9}+\frac{y^{2}}{1}=1 \Rightarrow e=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}$
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]
$\Rightarrow \quad h^{2}-8=-k^{2}$
$\Rightarrow \quad x^{2}+y^{2}=8$
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]
$\Rightarrow \quad h^{2}-8=-k^{2}$
$\Rightarrow \quad x^{2}+y^{2}=8$
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