Let P be a point on ellipse frac{x^2}{9}+frac{y^2}{4}=1 . Let line passing through P and parallel to

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10-2. Parabola, Ellipse, Hyperbola

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Let $P$ be a point on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$. 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 $P Q$ such that $P R: R Q=4: 3$ as $P$ moves on the ellipse, is :

A

$\frac{11}{19}$

B

$\frac{13}{21}$

C

$\frac{\sqrt{139}}{23}$

D

$\frac{\sqrt{13}}{7}$

(JEE MAIN-2024)

Solution

$ \mathrm{h}=3 \cos \theta $

$ \mathrm{k}=\frac{18}{7} \sin \theta $

$ \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}$

Standard 11

Mathematics

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