Question

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 meets 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 $PQ$ such that $\frac{PR}{RQ}=\frac{1}{2}$ , then the locus of $R$ is

• A. $\frac{x^2}{9}+\frac{9y^2}{49}=1$
• B. $\frac{x^2}{49}+\frac{y^2}{9}=1$
• C. $\frac{x^2}{9}+\frac{y^2}{49}=1$
• D. $\frac{9x^2}{49}+\frac{y^2}{9}=1$

Answer 1

Answer: (A)

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, $\frac{PR}{RQ}=\frac{1}{2}$

$\therefore h=\frac{3\cos \theta +6\cos \theta }{3},k=\frac{3\sin \theta +4\sin \theta }{3}$
$\Rightarrow h=3\cos \theta ,k=\frac{7}{3}\sin \theta$
$\Rightarrow \cos \theta =h/3,\sin \theta =\frac{3k}{7}$
Now, ${\cos }^2\theta +{\sin }^2\theta =h^2/9+\frac{9k^2}{49}=1$
Hence, locus of a point is $\frac{x^2}{9}+\frac{9y^2}{49}=1$