Mathematics · Ellipse

JEE Main 2024 — 1 February, Shift 2 — Question 5

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 PQ such that $\mathrm{PR}: \mathrm{RQ}=4: 3$ as P moves on the ellipse, is :

Option A:
$\frac{11}{19}$
Option B:
$\frac{13}{21}$
Option C:
$\frac{\sqrt{139}}{23}$
Option D:
$\frac{\sqrt{13}}{7}$
Correct

Answer: D

Step-by-step 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 $=\frac{\mathrm{x}^{2}}{9}+\frac{49 \mathrm{y}^{2}}{324}=1$

$e=\sqrt{1-\frac{324}{49 \times 9}}=\frac{\sqrt{117}}{21}=\frac{\sqrt{13}}{7}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Mathematics
Chapter: Ellipse
Topic: Introduction to Ellipse