Locus of Midpoint of Chord of Ellipse E

To find the locus of the midpoint of the chord $AB$ of the ellipse $\text{E}$, we proceed step-by-step:

1. Find the parameters of the ellipse: The centre of the ellipse is at the origin $(0,0)$ and its directrix is given as the vertical line $x = 9$. Therefore, the standard equation of the ellipse is: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

   The directrix is given by $x = \frac{a}{e} = 9$. Given that the eccentricity $e = \frac{1}{3}$, we can solve for the semi-major axis $a$: $a = 9e = 9 \times \frac{1}{3} = 3 \implies a^2 = 9$

   The semi-minor axis $b$ is calculated using the relation $b^2 = a^2(1 - e^2)$: $b^2 = 9 \left(1 - \frac{1}{9}\right) = 8$

   Thus, the equation of the ellipse $\text{E}$ is: $\frac{x^2}{9} + \frac{y^2}{8} = 1$

2. Find the focus $P(\alpha, 0)$: The focus with $\alpha > 0$ is at $(ae, 0)$: $\alpha = ae = 3 \times \frac{1}{3} = 1$ So, the focus $P$ is $(1, 0)$.

3. Find the equation of the chord with midpoint $(h, k)$: The equation of a chord of the ellipse $\frac{x^2}{9} + \frac{y^2}{8} = 1$ whose midpoint is $(h, k)$ is given by $T = S_1$: $\frac{hx}{9} + \frac{ky}{8} - 1 = \frac{h^2}{9} + \frac{k^2}{8} - 1$ $\frac{hx}{9} + \frac{ky}{8} = \frac{h^2}{9} + \frac{k^2}{8}$

4. Apply the condition that the chord passes through $P(1, 0)$: Substituting $x = 1$ and $y = 0$ into the chord equation: $\frac{h(1)}{9} + \frac{k(0)}{8} = \frac{h^2}{9} + \frac{k^2}{8}$ $\frac{h}{9} = \frac{h^2}{9} + \frac{k^2}{8}$

5. Derive the locus: Rearranging the terms: $\frac{k^2}{8} = \frac{h}{9} - \frac{h^2}{9}$ $\frac{k^2}{8} = \frac{h(1 - h)}{9}$ $9k^2 = 8h(1 - h)$

   Replacing $(h, k)$ with $(x, y)$, the locus of the midpoint of the chord $AB$ is: $9y^2 = 8x(1 - x)$

This corresponds to Option A: $9y^2 = 8x(1 - x)$.