JEE Main 2026 — Ellipse Question with Solution

Given the directrix of the ellipse is $x=\frac{a}{e}=9$ and eccentricity $e=\frac{1}{3}$.

$\frac{a}{1/3}=9\Rightarrow a=3$

The value of $b^2$ is given by $b^2=a^2(1-e^2)=9\left(1-\frac{1}{9}\right)=8$.

The equation of the ellipse is $\frac{x^2}{9}+\frac{y^2}{8}=1$.

The focus $P(\alpha ,0)$ for $\alpha >0$ is at $(ae,0)=\left(3\times \frac{1}{3},0\right)=(1,0)$.

Let the midpoint of the chord $AB$ be $(h,k)$. The equation of the chord in terms of its midpoint is given by $T=S_1$ :

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

Since the chord passes through the focus $P(1,0)$, substituting $x=1$ and $y=0$ gives:

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

$\frac{k^2}{8}=\frac{h}{9}-\frac{h^2}{9}$

$9k^2=8h(1-h)$

Replacing $(h,k)$ with $(x,y)$, the locus of the midpoint is:

$9y^2=8x(1-x)$

Answer: $9y^2=8x(1-x)$