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A point P of the parabola y2 = 4ax lies on the line y = x. The normal chord PQ, normal at P, subtends an angle at the focus of the parabola which equal(A) 60º(B) 45º(C) 30º(D) 90º |
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Answer» Correct option (D) 90º Explanation : Let P be (at2, 2at) so that t = 2 because it lies on the line y = x. Suppose normal at P meets the curve at Q = (at'2 , 2at'). So t'2 = -t - 2/t = -2 - 1 (∴ t -= 2) Therefore, Q = (9a, - 6a) and P = (4a, 4a). Now Slop of SP x Slop of SQ = (4a/4a - a)(6a/a - 9a) = 4/3(-6/8) = -1 Hence,∠PSQ = 90º |
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