JEE Main 2026 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 6 April 2026, Shift 2, Maths Q12

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

Let

x=9

be a directrix of an ellipse E, whose centre is at the origin and eccentricity is

\frac{1}{3}

. Let P

(\alpha,0)

\alpha>0

, be a focus of E and AB be a chord passing through P. Then the locus of the mid point of AB is :

(1) $9y^2=8x(1-x)$ Official answer
(2) $3y^2=4x(1-x)$
(3) $9y^2=8x(x-1)$
(4) $3y^2=4x(x-1)$

Official answer

Option 1

NTA final key.

Chapter: Conic Sections
Topic: General Conic Section
Idea tested: Chord Bisected at a Given Point (T=S1)

Same idea in other shifts

Asked 7× in all

2 Apr 2026, Shift 2 · Q11Let $O$ be the origin, and $P$ and $Q$ be two points on the rectangular hyperbola $xy = 12$ such that the midpoint of the line segment…MediumSingle correct
4 Apr 2026, Shift 1 · Q13Suppose that two chords, drawn from the point $(1,2)$ on the circle $x^2+y^2+x-3y=0$ are bisected by the $y$ -axis. If the other ends of…HardSingle correct
4 Apr 2024, Shift 2 · Q15Let PQ be a chord of the parabola $y^2=12x$ and the midpoint of PQ be at $(4,1)$ . Then, which of the following point lies on the line…MediumSingle correct
9 Apr 2024, Shift 2 · Q27Consider the circle $C: x^2+y^2=4$ and the parabola $P: y^2=8x$ . If the set of all values of $\alpha$ , for which three chords of the…HardNumerical value
27 Jan 2024, Shift 1 · Q16The length of the chord of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$ , whose mid point is $\left(1, \dfrac{2}{5}\right)$ , is equal to :HardSingle correct
26 Aug 2021, Shift 2 · Q73The locus of the mid points of the chords of the hyperbola $x^2-y^2=4$ , which touch the parabola $y^2=8x$ , is :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.