Tangents PQ and PR are drawn to the parabola `y^(2) = 20(x+5)` and `y^(2) = 60 (x+15)`, respectively such that `/_RPQ = (pi)/(2)`. Then the locus of point P is

Options

A. `x + 10 = 0`
B. `x + 30 = 0`
C. `x +40 = 0`
D. `x +20 = 0`

Answer

D

Explanation

Tangent to parabola `y^(2) = 20 (x+5)` having slope m is
`y = m(x+5) + (5)/(m)`
or `m^(2) (x+5) -my +5 =0` (1)
Tangent to parabola `y^(2) = 60(a+15)` having slope m' is
`y = m' (x+15) +(15)/(m')` (2)
Given `m xx m' =-1`
Thus. (2) reduces to
`y =- (1)/(m) (x+15) -15m`
or `15m^(2) + my + (x+15) =0` (3)
Eliminating n from (1) and (3), we get locus `x + 20 =0`.