JEE Main 2025 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 3 April 2025, Shift 2, Maths Q12

Question 12 of 75 in this shift, Maths question 12 of 25, Section A.

Let

C

be the circle of minimum area enclosing the ellipse

E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

with eccentricity

\frac{1}{2}

and foci

(\pm 2, 0)

. Let

PQR

be a variable triangle, whose vertex

P

is on the circle

C

and the side

QR

of length

2a

is parallel to the major axis of

E

and contains the point of intersection of

E

with the negative

y

-axis. Then the maximum area of the triangle

PQR

is :

(1) $8\left(2+\sqrt{3}\right)$ Official answer
(2) $8\left(3+\sqrt{2}\right)$
(3) $6\left(2+\sqrt{3}\right)$
(4) $6\left(3+\sqrt{2}\right)$

Official answer

Option 1

NTA final key.

Chapter: Conic Sections
Topic: Ellipse Eccentricity
Idea tested: Center to Focus Distance

Same topic in other shifts

All Ellipse Eccentricity questions

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 42 of the 174 JEE Main shifts held from 2017 to 2026.