Let $P$ be a point on the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$. Let the line passing through $P$ and parallel to $y$-axis meet the circle $x^2+y^2=9$ at point $Q$ such that $P$ and $Q$ are on the same side of the $x$-axis. Then the eccentricity of the locus of the point $R$ on $PQ$ such that $PR:RQ=4:3$ as $P$ moves on the ellipse, is:
Step-by-Step Solution
Key Concept: Parametrize $P=(3\cos\theta,2\sin\theta)$ on the ellipse. On the vertical line $x=3\cos\theta$, $Q=(3\cos\theta,3\sin\theta)$ on the circle. Use section formula with ratio $4:3$ to find $R=(h,k)$, then eliminate $\theta$ to get the locus.
$P=(3\cos\theta,2\sin\theta)$, $Q=(3\cos\theta,3\sin\theta)$. $R$ divides $PQ$ in ratio $4:3$: $h=3\cos\theta$, $k=\frac{18}{7}\sin\theta$. Locus: $\frac{x^2}{9}+\frac{49y^2}{324}=1$. $e=\sqrt{1-\frac{324}{441}}=\frac{\sqrt{117}}{21}=\frac{\sqrt{13}}{7}$.
Correct Answer: 4