Let P be a point on the parabola y 2 = 4 a x , where a > 0 . The normal to the parabola at P meets the x -axis at a point Q . The area of the triangle P F Q , where F is the focus of the parabola, is 120 . If the slope m of the normal and a are both positive integers, then the pair a ,   m is
A. 2 ,   3
Given, Parabola y 2 = 4 a x , Now plotting the normal to parabola at point P a m 2 , - 2 a m , we get So, equation of normal at P a m 2 ,   - 2 a m is y = m x - 2 a m - a m 3 Hence, point Q 2 a + a m 2 , 0 And focus of parabola will be F a , 0 Now finding the area of ∆ P F Q we get, Area of ∆ P F Q = 1 2 a + a m 2 × 2 a m = 120 ⇒ a 2 m 1 + m 2 = 120 So, possible pair from option will be, a ,   m ≡ 2 ,   3 which satisfies above equation
Related: Mathematics — Parabola · All PYQ Banks