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JEE Main 202624 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2026 · 24 January · Evening Shift · Mathematics · Ellipse

Let (h, k) lie on the circle C: x^ 2 +y^ 2 =4 and the point (2...

Let

(h, k)

lie on the circle

\mathrm{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

\_\_\_\_

.

Correct answer

0

Step-by-step solution

Since

(h, k)

lies on circle

x^2 + y^2 = 4

, we have

h = 2\cos\theta

and

k = 2\sin\theta

.
The transformed point is

(2h+1, 3k+2) = (4\cos\theta + 1, 6\sin\theta + 2)

.
Rearranging:

\frac{(x-1)^2}{16} + \frac{(y-2)^2}{36} = 1

This is an ellipse with

a^2 = 36

,

b^2 = 16

, so

e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{16}{36} = \frac{5}{9}

Therefore

\frac{5}{e^2} = \frac{5}{5/9} = 9

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