1.

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

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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