If y = mx + 4 is a tangent to both the parabolas $y^2=4x$ and  $x^2=2by$, then b is equal to

The given line equation is 

y = mx+4                       ….(i)

 The equation of the tangent to the parabola $y^2=4x$ is   $y=mx+\frac{1}{m}$                        ….(ii)

from (i) and (ii)

 $4=\frac{1}{m}\Rightarrow m=\frac{1}{4}$

So line $y=\frac{1}{4}x+4$ is also tangent to parabola $x^2=2by$ , so solve

$x^2=2b\left(\frac{x+16}{4}\right)$, it must have only one solution, it implies that discriminant must be zero. 

$\Rightarrow 2x^2-bx-16b=0\Rightarrow D=0\Rightarrow b^2-4\times 2\times \left(-16b\right)=0$    

 $\Rightarrow b^2+32\times 4b=0$    

$b=-128,b=0$(not possible)