Let x =9 be a directrix of an ellipse E , whose centre is at the origin and eccentricity is 1/3 . Let P (α, 0), α 0 , be a focus of E and AB be a chord passing through P . Then the locus of the mid point of AB is :

Mathematics · Ellipse

JEE Main 2026 — 6 April, Evening Shift — Question 31

Let $\mathrm{x}=9$ be a directrix of an ellipse E , whose centre is at the origin and eccentricity is $\frac{1}{3}$ . Let $\mathrm{P}(\alpha, 0), \alpha>0$ , be a focus of E and AB be a chord passing through P . Then the locus of the mid point of AB is :

Option A:
$9y²=8x(1-x)$
Correct
Option B:
$3y²=4x(1-x)$
Option C:
$9y²=8x(x-1)$
Option D:
$3y²=4x(x-1)$

Answer: A

Step-by-step solution

$\frac{\mathrm{a}}{\mathrm{e}}=9 \Rightarrow \mathrm{a}=3$ $\mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)=9\left(1-\frac{1}{9}\right)=8$ $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1$ $S(\alpha, 0) \equiv S(1,0)$ $\mathrm{T}=\mathrm{S}_{1} \Rightarrow \frac{\mathrm{hx}}{9}+\frac{\mathrm{ky}}{8}=\frac{\mathrm{h}^{2}}{9}+\frac{\mathrm{k}^{2}}{8}$ $(1,0) \Rightarrow \frac{\mathrm{h}}{9}+0=\frac{\mathrm{h}^{2}}{9}+\frac{\mathrm{k}^{2}}{8}$ $\mathrm{h}=\mathrm{h}^{2}+\frac{9}{8} \mathrm{k}^{2}$ $9 y^{2}=8 x(1-x)$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Evening Shift
Subject: Mathematics
Chapter: Ellipse
Topic: Introduction to Ellipse