JEE Main 2026 · MathsNumerical answerNumerical valueMediumMulti-step

JEE Main 4 April 2026, Shift 1, Maths Q22

Question 22 of 75 in this shift, Maths question 22 of 25, Section B.

Consider 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 intersection of P and E, be their latus rectums. If the eccentricity of E is

e

, then

e^2+2\sqrt2

is equal to ______.

Official answer

NTA final key.

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

Same idea in other shifts

Asked 5× in all

7 Apr 2025, Shift 2 · Q12Let 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…MediumSingle correct
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.