Question
Let $x = 9$ be a directrix of an ellipse $E$, whose centre is at the origin and eccentricity is $\frac{1}{3}$. 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 :
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Solution
Sol. [Key Concept: The directrix $\frac{a}{e} = 9$ with $e = \frac{1}{3}$ gives $a = 3,b^{2} = 8$; the mid-point $(h,k)$ of a focal chord satisfies $T = S_{1}$ about focus $S(\alpha,0)$.]
$${\frac{a}{e} = 9 \Rightarrow a = 3 }{b^{2} = a^{2}\left( 1 - e^{2} \right) = 9\left( 1 - \frac{1}{9} \right) = 8 }{\frac{x^{2}}{9} + \frac{y^{2}}{8} = 1 }{S(\alpha,0) \equiv S(1,0) }{T = S_{1} \Rightarrow \frac{hx}{9} + \frac{ky}{8} = \frac{h^{2}}{9} + \frac{k^{2}}{8} }{(1,0) \Rightarrow \frac{h}{9} + 0 = \frac{h^{2}}{9} + \frac{k^{2}}{8} }{h = h^{2} + \frac{9}{8}k^{2} }{9y^{2} = 8x(1 - x) }$$
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