Equation of a common tangent to the parabola `y^(2)=4x` and the hyperbola xy=2 is

Options

A. x+2y+4=0
B. x+y+1=0
C. x-2y+4=0
D. 4x+2y+1=0

Answer

A

Explanation

Let the equation of tangent to `y^2=4x` be `y=mx+1/m`
Solving it with xy=2 we get : `x(mx+1/m)=2 rArr mx^2+x/m-2=0 rArr m^2x^2+x-2x=0`
For condition of tangency D =0 `rArr 1+8m^3 =0 rArr m^3=-1/8 rArr m=-1/2`
Put it in (1) we get `y=-1/2x-2 rArr x+2y+4=0`