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 $T$ is a point on $PQ$ such that $\frac{PR}{PQ}=\frac{1}{2}$, 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. $\frac{{9x}^2}{49}+\frac{y^2}{49}=1$

Answer 1

The correct option is C $\frac{x^2}{9}+\frac{y^2}{49}=1$

$P(3\cos \theta ,2\sin \theta );Q(3\cos \theta ,3\sin \theta )$
$h=\frac{3\cos \theta +6\cos \theta }{3},k=\frac{3\sin \theta +4\sin \theta }{3}$
$h=3\cos \theta ,k=\frac{7}{3}\sin \theta$
$\therefore {\sin }^2\theta +{\cos }^2\theta =1$
$\frac{h^2}{9}+\frac{{9k}^2}{49}=1$
locus is $\frac{x^2}{9}+\frac{y^2}{49}=1$