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Let $\mathrm{P}$ be an arbitrary point on the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}-1, \mathrm{a}>\mathrm{b}>0 .$ Suppose $\mathrm{F}_{1}$ and $\mathrm{F}_{2}$ are the foci of the ellipse. The locus of the centroid of the triangle $\mathrm{PF}_{1} \mathrm{~F}_{2}$ as $\mathrm{P}$ moves on the ellipse is-
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Verified Answer
The correct answer is:
an ellipse
$$
\begin{array}{l}
\mathrm{P} \rightarrow \mathrm{a} \cos \theta, \mathrm{b} \sin \theta \\
\mathrm{G} \rightarrow\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{3}, \frac{\Sigma \mathrm{y}_{\mathrm{i}}}{3}\right) \\
\mathrm{F}_{1} \rightarrow(\mathrm{ae}, 0) \quad \mathrm{F}_{2} \rightarrow(-\mathrm{ae}, 0) \\
\mathrm{h}=\frac{\operatorname{acos} \theta+a \mathrm{e}-\mathrm{ae}}{3} \quad ; \quad \cos \theta=\frac{3 \mathrm{~h}}{\mathrm{a}}
\end{array}
$$
$$
\begin{array}{l}
k=\frac{b \sin \theta}{3} ; \quad \sin \theta=\frac{3 k}{b} \\
\cos ^{2} \theta+\sin ^{2} \theta=1 \\
\frac{9 h^{2}}{a^{2}}+\frac{9 k^{2}}{b^{2}}=1 \Rightarrow \frac{x^{2}}{\left(a^{2} / 9\right)}+\frac{y^{2}}{\left(b^{2} / 9\right)}=1 \\
\text { (Ellipse) }
\end{array}
$$
\begin{array}{l}
\mathrm{P} \rightarrow \mathrm{a} \cos \theta, \mathrm{b} \sin \theta \\
\mathrm{G} \rightarrow\left(\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{3}, \frac{\Sigma \mathrm{y}_{\mathrm{i}}}{3}\right) \\
\mathrm{F}_{1} \rightarrow(\mathrm{ae}, 0) \quad \mathrm{F}_{2} \rightarrow(-\mathrm{ae}, 0) \\
\mathrm{h}=\frac{\operatorname{acos} \theta+a \mathrm{e}-\mathrm{ae}}{3} \quad ; \quad \cos \theta=\frac{3 \mathrm{~h}}{\mathrm{a}}
\end{array}
$$
$$
\begin{array}{l}
k=\frac{b \sin \theta}{3} ; \quad \sin \theta=\frac{3 k}{b} \\
\cos ^{2} \theta+\sin ^{2} \theta=1 \\
\frac{9 h^{2}}{a^{2}}+\frac{9 k^{2}}{b^{2}}=1 \Rightarrow \frac{x^{2}}{\left(a^{2} / 9\right)}+\frac{y^{2}}{\left(b^{2} / 9\right)}=1 \\
\text { (Ellipse) }
\end{array}
$$
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