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(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let the point $P$ on the ellipse be $(a \cos 	heta, b \sin 	heta)$. The

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an ellipse

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(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let the point $P$ on the ellipse be $(a \cos 	heta, b \sin 	heta)$. The foci of the ellipse are $F_1(ae, 0)$ and $F_2(-ae, 0)$. The centroid $(h, k)$ of $	riangle P F_1 F_2$ is given by: $h = \frac{a \cos 	heta + ae - ae}{3} = \frac{a \cos 	heta}{3}$ $k = \frac{b \sin 	heta + 0 + 0}{3} = \frac{b \sin 	heta}{3}$ Thus,$\cos 	heta = \frac{3h}{a}$ and $\sin 	heta = \frac{3k}{b}$. Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we get: $\left(\frac{3h}{a}ight)^2 + \left(\frac{3k}{b}ight)^2 = 1$ $\frac{9h^2}{a^2} + \frac{9k^2}{b^2} = 1$ Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^2}{(a/3)^2} + \frac{y^2}{(b/3)^2} = 1$,which is an ellipse.

Let $P$ be an arbitrary point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $a > b > 0$. Suppose $F_1$ and $F_2$ are the foci of the ellipse. The locus of the centroid of the $\triangle P F_1 F_2$ as $P$ moves on the ellipse is

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