The slope of the line touching both the parabolas $$y^2 = 4x$$ and $$x^2 = -32y$$ is:

Let us take an arbitrary straight line with slope $$m$$ and $$y$$-intercept $$c$$. In point-slope form this line is written as

$$y = m x + c.$$

We want this single line to touch both parabolas one after the other, so we shall impose the condition of tangency separately for each curve and finally make the two conditions compatible.

First parabola  $$y^{2}=4x$$

Substituting $$y = m x + c$$ in $$y^{2}=4x$$ we obtain

$$(m x + c)^{2}=4x.$$

This expands to

$$m^{2}x^{2}+2mcx+c^{2}-4x=0.$$

The above is a quadratic in $$x$$. For the line to be a tangent, both roots must coincide, i.e. the discriminant must vanish. Recall the quadratic discriminant formula $$\Delta = b^{2}-4ac.$$ Here $$a=m^{2},\; b=2mc-4,\; c=c^{2}.$$ So

$$\Delta=(2mc-4)^{2}-4(m^{2})(c^{2}).$$

Evaluating,

$$$\begin{aligned} \Delta &= (2mc-4)^{2}-4m^{2}c^{2}\\ &= 4m^{2}c^{2}-16mc+16-4m^{2}c^{2}\\ &= -16mc+16. \end{aligned}$$$

Setting $$\Delta=0$$ gives

$$-16mc+16=0 \;\Longrightarrow\; mc=1 \;\Longrightarrow\; c=\frac1m.$$

Second parabola  $$x^{2}=-32y$$

Again substitute the same line $$y = m x + c$$ into $$x^{2}=-32y$$:

$$x^{2} = -32(m x + c).$$

This rearranges to a quadratic in $$x$$ as

$$x^{2}+32m x+32c=0.$$

Here $$a=1,\; b=32m,\; c=32c.$$ Applying the discriminant condition $$\Delta = b^{2}-4ac = 0$$ we have

$$(32m)^{2}-4(1)(32c)=0.$$

Simplifying step by step,

$$$\begin{aligned} 1024m^{2}-128c &=0\\ \Rightarrow\; 32m^{2}-4c&=0\\ \Rightarrow\; c&=8m^{2}. \end{aligned}$$$

Common tangency condition

The same line must satisfy both relations for $$c$$. Therefore we equate

$$\frac1m = 8m^{2}.$$

Multiplying both sides by $$m$$ gives

$$1 = 8m^{3}.$$

Hence

$$m^{3}=\frac18 \;\Longrightarrow\; m = \sqrt[3]{\frac18}= \frac12.$$

Thus the slope of the line which touches both parabolas is

$$m = \frac12.$$

Hence, the correct answer is Option C.