JEE Main 2024 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 9 April 2024, Shift 2, Maths Q15

Question 15 of 90 in this shift, Maths question 15 of 30, Section A.

Let the foci of a hyperbola

H

coincide with the foci of the ellipse

E:\frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1

and the eccentricity of the hyperbola

H

be the reciprocal of the eccentricity of the ellipse

E

. If the length of the transverse axis of

H

\alpha

and the length of its conjugate axis is

\beta

, then

3\alpha^2+2\beta^2

is equal to

(1) $205$
(2) $242$
(3) $237$
(4) $225$ Official answer

Official answer

Option 4

NTA final key.

Chapter: Conic Sections
Topic: Ellipse Eccentricity
Idea tested: Coordinates of foci

Same idea in other shifts

Asked 5× in all

7 Apr 2025, Shift 2 · Q13Let $e_1$ and $e_2$ be the eccentricities of the ellipse $\frac{x^2}{b^2}+\frac{y^2}{25}=1$ and the hyperbola…MediumSingle correct
25 Jul 2022, Shift 2 · Q14Let the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{7}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{\alpha}=\frac{1}{25}$ coincide.…MediumSingle correct
25 Feb 2021, Shift 2 · Q72A hyperbola passes through the foci of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ and its transverse and conjugate axes coincide…MediumSingle correct
25 Jul 2021, Shift 2 · Q76If a tangent to the ellipse $x^2+4y^2=4$ meets the tangents at the extremities of its major axis at B and C, then the circle with BC as…HardSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.