Let $$x = 9$$ be a directrix of an ellipse E, whose centre is at the origin and eccentricity is $$\dfrac{1}{3}$$. 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 :

For an ellipse centered at the origin, the equation of the directrix is $$x = \frac{a}{e}$$ and the focus is $$(ae, 0)$$.

$$\frac{a}{e} = 9 \implies \frac{a}{1/3} = 9 \implies a = 3$$

$$b^2 = 3^2\left(1 - \left(\frac{1}{3}\right)^2\right) = 9\left(1 - \frac{1}{9}\right) = 8$$

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

$$\alpha = ae = 3 \times \frac{1}{3} = 1 \implies P(1, 0)$$ (focus)

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

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

$$\frac{h(1)}{9} + \frac{k(0)}{8} = \frac{h^2}{9} + \frac{k^2}{8}$$

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

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

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