If y = mx + 4 is a tangent to both the parabolas, y ^ 2 = 4x and x...

Equation of tangent to ${y}^{2} = 4ax$ in slope Form is $y = mx + \dfrac{a}{m}$.

Comparing with given equation of tangent and parabola we get $\dfrac{1}{m} = 4 ⇒ m = \dfrac{1}{4}$.

Hence, line $y = \dfrac{x}{4} + 4$ is tangent to ${x}^{2} = 2by$ also.

Solving both, we get ${x}^{2} = 2b\left(\dfrac{x}{4} + 4\right) ⇒ 2{x}^{2} - bx - 16b = 0 . . ( 1 )$.

Since the line is a tangent, so \left(1\right)$must have equal roots.</p><p>$⇒ {\left(- b\right)}^{2} - 4 \cdot 2 \cdot \left(- 16b\right) = 0$</p><p>$⇒ {b}^{2} + 128b = 0 ⇒ b = 0, - 128$</p><p>But$b$can't be zero.</p><p>So,$b = - 128