Find the equation (s) of the common tangent(s) to the parabola `y^(2) - 4x - 2y + 5 =0` and `y^(2) = -4x`.

Explanation

The equation of tangent to `y^(2) = -4x` is
`y = mx-(1)/(m)" "…(i)`
The parabola `y^(2) - 4x - 2y + 5 = 0` can be written as
`y^(2) - 2y + 1 = 4x -4`
`rArr (y-1)^(2) = 4(x-1)`
The equation tangent to it be
`y-1=m(x-1)+(1)/(m)`
`rArr y = mx - m +(1)/(m) + 1" "...(ii)`
For the equation of common tangent to the parabolas (i) & (ii) represent same line
`therefore -(1)/(m) = -m+(1)/(m) + 1`
`m -(2)/(m) - 1= 0`
`rArr m^(2) - m-2=-0`
` m=2, -1`
`therefore` The required common tangents are
`y = 2x-(1)/(2) and 4x - y -1 = 0` and
`y = -x +1`
i.e. `x + y = 1`