The slope of the line touching both the parabolas $${y^2} = 4x$$  and $${x^2} = - 32y$$   is-

Solution :
Let tangent to $${y^2} = 4x$$   be $$y = mx + \frac{1}{m}$$
Since this is also tangent to $${x^2} = - 32y$$
$$\therefore {x^2} = - 32\left( {mx + \frac{1}{m}} \right)\,\,\, \Rightarrow {x^2} + 32mx + \frac{{32}}{m} = 0$$
Now, $$D=0$$
$$\eqalign{ & {\left( {32} \right)^2} - 4\left( {\frac{{32}}{m}} \right) = 0 \cr & \Rightarrow {m^3} = \frac{4}{{32}} \cr & \Rightarrow {m^3} = \frac{1}{8} \cr & \Rightarrow m = \frac{1}{2} \cr} $$