Let x = 9 be a directrix of an ellipse E, whose centre is at the origin and eccentricity is 1 3 . Let P( , 0), 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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Let x = 9 be a directrix of an ellipse E, whose centre is at the origin and eccentricity is 1 3 . Let P( , 0), > 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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Correct answer: A. 9y^2 = 8x(1-x)

Given the directrix of the ellipse is x = a e = 9 and eccentricity e = 1 3 . a 1/3 = 9 a = 3 The value of b^2 is given by b^2 = a^2(1 - e^2) = 9 (1 - 1 9 ) = 8. The equation of the ellipse is x^2 9 + y^2 8 = 1. The focus P( , 0) for > 0 is at (ae, 0) = (3 1 3 , 0 ) = (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 : hx 9 + ky 8 = h^2 9 + k^2 8 Since the chord passes through the focus P(1, 0), substituting x = 1 and y = 0 gives: h 9 = h^2 9 + k^2 8 k^2 8 = h 9 - 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)

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