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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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
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
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