JEE Main 2022 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 25 July 2022, Shift 2, Maths Q14

Question 14 of 90 in this shift, Maths question 14 of 30, Section A.

Let 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. Then the length of the latus rectum of the hyperbola is :

(1) $\frac{32}{9}$
(2) $\frac{18}{5}$
(3) $\frac{27}{4}$
(4) $\frac{27}{10}$ 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
9 Apr 2024, Shift 2 · Q15Let 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…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.