Consider the parabola with vertex 1 2 , 3 4 and the directrix y = 1 2 . Let P be the point where the parabola meets the line x = - 1 2 . If the normal to the parabola at P intersects the parabola again at the point Q . then ( PQ ) 2 is equal to :

Step-by-step solution

Equation of the parabola is x - 1 2 2 = y - 3 4       . . . 1 Put x = - 1 2 in equation 1 , we get y = 7 4 Hence, point P is P - 1 2 , 7 4 Now, differentiate equation 1 on both sides, we get y ' = 2 x - 1 2 At point P - 1 2 , 7 4 , the slope of tangent is - 2 . Therefore, the slope of normal is 1 2 Equation of normal at P - 1 2 ,   7 4 is x = 2   y - 4       . . . 2 On solving equations 1   &   2 , we get the intersection point Q 2 ,   3 ∴ ( PQ )