Let $x=9$ 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 :

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Ellipse

Let

x = 9

be a directrix of an ellipse

E

, whose centre is at the origin and eccentricity is

\frac{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 :

Answer: (A)

From

\frac{a}{e} = 9

with

e = \frac{1}{3}

a = 3

, and

b^{2} = a^{2} (1 - e^{2}) = 9 \cdot \frac{8}{9} = 8

, so the ellipse is

\frac{x ^{2}}{9} + \frac{y ^{2}}{8} = 1

with focus

P (a e, 0) = (1, 0)

. A chord with midpoint

(h, k)

satisfies

T = S_{1}

\frac{h x}{9} + \frac{k y}{8} = \frac{h ^{2}}{9} + \frac{k ^{2}}{8}

. Passing through

(1, 0)

gives

\frac{h}{9} = \frac{h ^{2}}{9} + \frac{k ^{2}}{8}

, i.e.

9 k^{2} = 8 h (1 - h)

. The locus of the midpoint is

9 y^{2} = 8 x (1 - x)

JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option