JEE Main 2025 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 7 April 2025, Shift 2, Maths Q12

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

Let the length of a latus rectum of an ellipse

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

be 10. If its eccentricity is the minimum value of the function

f(t)=t^2+t+\frac{11}{12}

t\in\mathbf{R}

, then

a^2+b^2

is equal to :

(1)115
(2)120
(3)125
(4)126Official answer

Official answer

Option 4

NTA final key.

Chapter: Conic Sections
Topic: Ellipse Eccentricity
Idea tested: Length of Latus Rectum

Same idea in other shifts

Asked 5× in all

4 Apr 2026, Shift 1 · Q22Consider the parabola $P: y^2=4kx$ and the ellipse $E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ . Let the line segment joining the points of…MediumNumerical value
9 Apr 2024, Shift 1 · Q12Let $f(x)=x^2+9$ , $g(x)=\frac{x}{x-9}$ and $a=f\circ g(10)$ , $b=g\circ f(3)$ . If $e$ and $l$ denote the eccentricity and the length of the…EasySingle correct
31 Jan 2024, Shift 1 · Q28Let the foci and length of the latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\ a>b$ be $(\pm5,0)$ and $\sqrt{50}$ ,…MediumNumerical value
31 Jan 2024, Shift 2 · Q14Let $P$ be a parabola with vertex $(2,3)$ and directrix $2x+y=6$ . Let an ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\ a>b$ , of…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 42 of the 174 JEE Main shifts held from 2017 to 2026.