The equation of the common tangent to the curves $${y^2} = 8x$$  and $$xy=-1$$  is-

Solution :
The given curves are
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,{y^2} = 8x.....(1) \cr & {\text{and }}xy = - 1.....(2) \cr} $$
If $$m$$ is the slope of tangent to (1), then equation of tangent is
$$y = mx + \frac{2}{m}$$
If this tangent is also a tangent to (2), then putting value of $$y$$ in curve (2)
$$\eqalign{ & x = \left( {mx + \frac{2}{m}} \right) = - 1 \cr & \Rightarrow m{x^2} + \frac{2}{m}x + 1 = 0 \cr & \Rightarrow {m^2}{x^2} + 2x + m = 0 \cr} $$
We should get repeated roots for the equation (condition of tangency)
$$\eqalign{ & \Rightarrow D = 0 \cr & \therefore {\left( 2 \right)^2} - 4{m^2}.m = 0 \cr & \Rightarrow {m^3} = 1\,\,\,\,\, \Rightarrow m = 1 \cr} $$
Hence required tangent is $$y=x +2$$