JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 9)
Let $\mathrm{P}$ be a point on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. Let the line passing through $\mathrm{P}$ and parallel to $y$-axis meet the circle $x^2+y^2=9$ at point $\mathrm{Q}$ such that $\mathrm{P}$ and $\mathrm{Q}$ 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 :
$\frac{13}{21}$
$\frac{\sqrt{139}}{23}$
$\frac{\sqrt{13}}{7}$
$\frac{11}{19}$
Explanation
_1st_February_Evening_Shift_en_9_1.png)
_1st_February_Evening_Shift_en_9_2.png)
$\begin{aligned} & \mathrm{h}=3 \cos \theta \\\\ & \mathrm{k}=\frac{18}{7} \sin \theta\end{aligned}$
$\begin{aligned} & \therefore \text { locus }=\frac{\mathrm{x}^2}{9}+\frac{49 \mathrm{y}^2}{324}=1 \\\\ & \mathrm{e}=\sqrt{1-\frac{324}{49 \times 9}}=\frac{\sqrt{117}}{21}=\frac{\sqrt{13}}{7}\end{aligned}$
Comments (0)
