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

We are given that the straight line $$y = mx + 4$$ touches (is tangent to) both parabolas $$y^2 = 4x$$ and $$x^2 = 2by$$. We shall first impose the condition of tangency with the first parabola to determine the slope $$m$$, and then use this slope to find the unknown parameter $$b$$ for the second parabola.

For the parabola $$y^2 = 4x$$, we can rewrite it as $$x = \dfrac{y^2}{4}$$. Substituting this value of $$x$$ into the line $$y = mx + 4$$ gives

$$y \;=\; m\left(\dfrac{y^2}{4}\right) + 4.$$

Multiplying every term by $$4$$ to clear the denominator, we obtain

$$4y \;=\; my^2 + 16.$$

Rearranging all terms to one side produces a quadratic in $$y$$:

$$my^2 - 4y + 16 \;=\; 0.$$

For a line to be tangent to a parabola, this quadratic must possess exactly one real solution. The quadratic formula tells us that a quadratic $$Ay^2 + By + C = 0$$ has a repeated root when its discriminant $$\Delta = B^2 - 4AC$$ is zero. Here

$$A = m,\quad B = -4,\quad C = 16.$$

So we set

$$\Delta = (-4)^2 - 4(m)(16) = 0.$$

Simplifying, we have

$$16 - 64m = 0 \;\;\Longrightarrow\;\; 64m = 16 \;\;\Longrightarrow\;\; m = \dfrac{16}{64} = \dfrac14.$$

Thus the common tangent must be

$$y = \dfrac14\,x + 4.$$

Now we require this same line to be tangent to the second parabola $$x^2 = 2by$$. Substituting $$y = \dfrac14\,x + 4$$ into that parabola yields

$$x^2 \;=\; 2b\left(\dfrac14\,x + 4\right).$$

Distributing on the right side, we get

$$x^2 \;=\; \frac{2b}{4}\,x + 8b \;=\; \frac{b}{2}\,x + 8b.$$

Collecting all terms to one side gives a quadratic in $$x$$:

$$x^2 - \frac{b}{2}\,x - 8b = 0.$$

Again, tangency requires this quadratic to have exactly one real solution, so its discriminant must vanish. For a general quadratic $$ax^2 + bx + c = 0$$, the discriminant condition is $$b^2 - 4ac = 0$$. Here

$$a = 1,\quad b = -\dfrac{b}{2},\quad c = -8b.$$

Setting the discriminant to zero:

$$\left(-\dfrac{b}{2}\right)^2 - 4(1)(-8b) = 0.$$

Simplifying step by step,

$$\dfrac{b^2}{4} + 32b = 0.$$

Multiplying by $$4$$ to clear the denominator, we get

$$b^2 + 128b = 0.$$

Factoring,

$$b(b + 128) = 0.$$

This gives two possible values: $$b = 0$$ or $$b = -128$$. The case $$b = 0$$ would make the parabola $$x^2 = 2by$$ degenerate (it would collapse to the line $$x = 0$$), so we discard it. Therefore, the only admissible value is

$$b = -128.$$

Hence, the correct answer is Option C.