Question

Let \( P(2,4), Q(18,-12) \) be the points on the parabola \( y^{2}=8 x \). The equation of straight line having slope \( \frac{1}{2} \) and passing through the point of intersection of the tangents to the parabola drawn at the points \( P \) and \( Q \) is 1) \( 2 x-y=1 \) 2) \( 2 x-y=2 \) 3) \( x-2 y=1 \) 4) \( x-2 y=2 \)

Answer 1

P - (2,4) and Q - (18,-12)

Parabola- y^2 = 8x

differentiating - 2ydy/dx = 8 => dy/dx = 4/y

Dy/dx at point P = 1(slope of tangent at P)

Dy/dx at point Q = -1/3(slope of tangent at Q)

eq.of tangent at P - (y-4) = 1(x-2)

                                     x-y+2=0 ----------T1

eq.of tangent at Q - (y+12) = -1/3(x-18)

                                      x+3y+18=0 --------T2

Intersection point of T1 and T2 A - (-4 , -6)

eq.of line passing through the intersection point of both tangents (T1 and T2) is

(y+4) = 1/2(x+6)

x-2y = 2  Ans.

So,  (4) is the correct option.