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

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Question

(A) Let the point $P$ on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ be $(3\cos	heta, 2\sin	heta)$.The line through $P$ parallel to the $Y$-axis

Vedclass · Accepted Answer

$\frac{x^{2}}{9}+\frac{9y^{2}}{49}=1$

Vedclass · Answer

(A) Let the point $P$ on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ be $(3\cos	heta, 2\sin	heta)$.The line through $P$ parallel to the $Y$-axis is $x = 3\cos	heta$.This line meets the circle $x^{2}+y^{2}=9$ at $Q$. Substituting $x = 3\cos	heta$ into the circle equation:$(3\cos	heta)^{2} + y^{2} = 9$ $\Rightarrow 9\cos^{2}	heta + y^{2} = 9$ $\Rightarrow y^{2} = 9(1-\cos^{2}	heta) = 9\sin^{2}	heta$.Since $P$ and $Q$ are on the same side of the $X$-axis,$y$ must have the same sign as $2\sin	heta$. Thus,$Q = (3\cos	heta, 3\sin	heta)$.Let $R(h, k)$ be a point on $PQ$ such that $\frac{PR}{RQ} = \frac{1}{2}$. Using the section formula:$h = \frac{1(3\cos	heta) + 2(3\cos	heta)}{1+2} = 3\cos	heta$$k = \frac{1(3\sin	heta) + 2(2\sin	heta)}{1+2} = \frac{7\sin	heta}{3}$From these,$\cos	heta = \frac{h}{3}$ and $\sin	heta = \frac{3k}{7}$.Using the identity $\cos^{2}	heta + \sin^{2}	heta = 1$,we get:$(\frac{h}{3})^{2} + (\frac{3k}{7})^{2} = 1 \Rightarrow \frac{h^{2}}{9} + \frac{9k^{2}}{49} = 1$.Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{x^{2}}{9} + \frac{9y^{2}}{49} = 1$.

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