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

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