Let the foci of the ellipse frac{x^{2}}{9}+y^ | vedclass

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(D) Given the ellipse equation is $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$. Comparing with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we have $a^{2}=9$ a

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$x^{2}+y^{2}=8$

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(D) Given the ellipse equation is $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$. Comparing with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we have $a^{2}=9$ and $b^{2}=1$. The eccentricity $e$ is given by $e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$. The foci are $(\pm ae, 0) = (\pm 3 	imes \frac{2\sqrt{2}}{3}, 0) = (\pm 2\sqrt{2}, 0)$. Let $P(h, k)$ be a point such that the foci $F_{1}(2\sqrt{2}, 0)$ and $F_{2}(-2\sqrt{2}, 0)$ subtend a right angle at $P$. Thus, the product of the slopes of $PF_{1}$ and $PF_{2}$ is $-1$. $\frac{k-0}{h-2\sqrt{2}} 	imes \frac{k-0}{h+2\sqrt{2}} = -1$. $\frac{k^{2}}{h^{2}-(2\sqrt{2})^{2}} = -1$. $k^{2} = -(h^{2}-8)$. $h^{2}+k^{2} = 8$. Replacing $(h, k)$ with $(x, y)$, the locus is $x^{2}+y^{2}=8$.

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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