If y=m x+4 is a tangent to both the parabolas

$y=m x+4$...(i)

Tangent of $y^{2}=4 x$ is

$\Rightarrow \quad y=m x+\frac{1}{m}$...(ii)

$\left[\because\right.$ Equation of tangent of $y^{2}=4 a x$ is $\left.y=m x+\frac{a}{m}\right]$

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

$x^{2}=2 b y$, so solve both equations.

$x^{2}=2 b\left(\frac{x+16}{4}\right)$

$\Rightarrow \quad 2 x^{2}-b x-16 b=0$

$\Rightarrow \quad D=0 \quad$ [For tangent]

\Rightarrow b^{2}-4 \times 2 \times(-16 b)=0

$\Rightarrow b^{2}+32 \times 4 b=0$

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