Let ( h , k ) lie on the circle C : x 2 + y 2 =4 and the point (2 h +1,3 k +2) lie on an ellipse with eccentricity e . Then the value of frac 5 e 2 is equal to

Mathematics · Ellipse

JEE Main 2026 — 24 January, Evening Shift — Question 22

Let $(\mathrm{h}, \mathrm{k})$ lie on the circle $\mathrm{C}: \mathrm{x}^{2}+\mathrm{y}^{2}=4$ and the point $(2 \mathrm{~h}+1,3 \mathrm{k}+2)$ lie on an ellipse with eccentricity $e$ . Then the value of $\frac{5}{e^{2}}$ is equal to

Answer: 9

Numerical answer — enter this value.

Step-by-step solution

Let $\mathrm{P} \equiv(2 \cos \theta, 2 \sin \theta)$

∴ coordinates of $\mathrm{Q}=(4 \cos \theta+1,6 \sin \theta+3)$

∴ locus of Q is $\left(\frac{\mathrm{x}-1}{4}\right)^{2}+\left(\frac{\mathrm{y}-3}{6}\right)^{2}=1$

$\therefore \mathrm{e}^{2}=1-\frac{16}{36}=\frac{5}{9}$

$\therefore \frac{5}{\mathrm{e}^{2}}=9$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Evening Shift
Subject: Mathematics
Chapter: Ellipse
Topic: Introduction to Ellipse