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Let $P$ be a point on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line through $P$ parallel to the $Y$ -axis mects the circle $x^{2}+y^{2}=9$ at $Q$ where $P, Q$ are on the same side of the $X$ -axis. If $R$ is a point on $P Q$ such that $\frac{P R}{R Q}=\frac{1}{2},$ then the locus of $R$ is
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Verified Answer
The correct answer is:
$\frac{x^{2}}{9}+\frac{9 y^{2}}{49}=1$
Since, point $P$ on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
$\therefore$
$P(3 \cos \theta, 2 \sin \theta)$
Now, equation of line parallel of $Y$ -axis is
$$
x=3 \cos \theta
$$
and above line meets circle at $Q$
$\therefore$
$Q(3 \cos \theta, 3 \sin \theta)$
Given, $\frac{P R}{R Q}=\frac{1}{2}$
$\therefore h=\frac{3 \cos \theta+6 \cos \theta}{3}, k=\frac{3 \sin \theta+4 \sin \theta}{3}$
$\Rightarrow \quad h=3 \cos \theta, \quad k=\frac{7}{3} \sin \theta$
$\Rightarrow \quad \cos \theta=h / 3, \sin \theta=\frac{3 k}{7}$
Now, $\cos ^{2} \theta+\sin ^{2} \theta=h^{2} / 9+\frac{9 k^{2}}{49}=1$
Hence, locus of a point is $\frac{x^{2}}{9}+\frac{9 y^{2}}{49}=1$
$\therefore$
$P(3 \cos \theta, 2 \sin \theta)$
Now, equation of line parallel of $Y$ -axis is
$$
x=3 \cos \theta
$$
and above line meets circle at $Q$
$\therefore$
$Q(3 \cos \theta, 3 \sin \theta)$
Given, $\frac{P R}{R Q}=\frac{1}{2}$
$\therefore h=\frac{3 \cos \theta+6 \cos \theta}{3}, k=\frac{3 \sin \theta+4 \sin \theta}{3}$
$\Rightarrow \quad h=3 \cos \theta, \quad k=\frac{7}{3} \sin \theta$
$\Rightarrow \quad \cos \theta=h / 3, \sin \theta=\frac{3 k}{7}$
Now, $\cos ^{2} \theta+\sin ^{2} \theta=h^{2} / 9+\frac{9 k^{2}}{49}=1$
Hence, locus of a point is $\frac{x^{2}}{9}+\frac{9 y^{2}}{49}=1$
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