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P is a point on the parabola whose ordinate equals its abscissa. A normal is drawn to the parabola at P to meet it again at Q . If S is the focus of the parabola, then the product of the slopes of S P and S Q is

Correct answer

-1

Step-by-step solution

Let, P a t 2 , 2 a t be a point on the parabola y 2 = 4 a x Since, the ordinate equals its abscissa. ⇒ a t 2 = 2 a t ⇒ t = 2 P ≡ 4 a , 4 a Equation of the normal at P 4 a , 4 a is y + 2 x = 2 a 2 + a 2 3 ⇒ y + 2 x = 12 a … i y 2 = 4 a x ⇒ y 2 = 2 a 12 a - y … [From i ] ⇒ y 2 + 2 a y - 24 a 2 = 0 ⇒ y - 4 a y + 6 a = 0 ⇒ y = 4 a or y = - 6 a ⇒ S ≡ a , 0 , P ≡ 4 a , 4 a , Q ≡ 9 a , - 6 a Slope of SP = 4 3 and slope of SQ = - 6 8 ⇒ Required product = 4 3 × - 6 8 = - 1

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