Let P be a point on the ellipse frac{x^2}{9}+ | vedclass

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(D) Let $P = (3 \cos 	heta, 2 \sin 	heta)$ be a point on the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$.Since the line through $P$ is parallel to t

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$\frac{\sqrt{13}}{7}$

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(D) Let $P = (3 \cos 	heta, 2 \sin 	heta)$ be a point on the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$.Since the line through $P$ is parallel to the $y$-axis,its equation is $x = 3 \cos 	heta$.This line meets the circle $x^2 + y^2 = 9$ at $Q = (3 \cos 	heta, 3 \sin 	heta)$ (since $P$ and $Q$ are on the same side of the $x$-axis,we take the positive $y$-coordinate).Let $R = (h, k)$ be a point on $PQ$ such that $PR:RQ = 4:3$.Using the section formula,$h = 3 \cos 	heta$ and $k = \frac{4(3 \sin 	heta) + 3(2 \sin 	heta)}{4+3} = \frac{12 \sin 	heta + 6 \sin 	heta}{7} = \frac{18}{7} \sin 	heta$.Thus,$\cos 	heta = \frac{h}{3}$ and $\sin 	heta = \frac{7k}{18}$.Using $\cos^2 	heta + \sin^2 	heta = 1$,we get $\frac{h^2}{9} + \frac{49k^2}{324} = 1$.The locus is $\frac{x^2}{9} + \frac{y^2}{(18/7)^2} = 1$.Here $a^2 = 9$ and $b^2 = \frac{324}{49}$. Since $a^2 > b^2$,the eccentricity $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{324/49}{9}} = \sqrt{1 - \frac{36}{49}} = \sqrt{\frac{13}{49}} = \frac{\sqrt{13}}{7}$.

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