Ellipse Question 1
Question 1 - 2024 (01 Feb Shift 2)
Let $\mathrm{P}$ be a point on the ellipse $\frac{\mathrm{x}^{2}}{9}+\frac{\mathrm{y}^{2}}{4}=1$. Let the line passing through $\mathrm{P}$ and parallel to $\mathrm{y}$-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 $P Q$ such that $P R: R Q=4: 3$ as $P$ moves on the ellipse, is :
(1) $\frac{11}{19}$
(2) $\frac{13}{21}$
(3) $\frac{\sqrt{139}}{23}$
(4) $\frac{\sqrt{13}}{7}$
Show Answer
Answer (4)
Solution
$\mathrm{P}(3 \cos \theta, 2 \sin \theta)$
$\mathrm{Q}(3 \cos \theta, 3 \sin \theta)$
$\mathrm{h}=3 \cos \theta$
$\mathrm{k}=\frac{18}{7} \sin \theta$
$\therefore$ locus $=\frac{\mathrm{x}^{2}}{9}+\frac{49 \mathrm{y}^{2}}{324}=1$
$\mathrm{e}=\sqrt{1-\frac{324}{49 \times 9}}=\frac{\sqrt{117}}{21}=\frac{\sqrt{13}}{7}$
