The slope of the line touching both the parabolas y2=4x  and x2=−32y is

Equation of the tangent at t2,2t to the parabola y2=4x is ty=x+t2                          (1)and the tangent at t−16t′,−8t2 to the parabola  x2=−32yt′x=y−8t′2                                                  (2)Since (1) and (2) represent the same line, comparing we get 1−t′=−t1=t2−8t′2⇒t=2Hence the required slope of the tangent is 1t=12.