If $$y = mx + c$$ is the normal at a point on the parabola $$y^2 = 8x$$ whose focal distance is 8 units, then $$|c|$$ is equal to:

The parabola is $$y^{2}=8x$$. Writing this in the standard form $$y^{2}=4ax$$ we compare and obtain $$4a=8$$, so $$a=2$$. A convenient parametric representation of every point on this parabola is therefore $$x=at^{2}=2t^{2}$$ and $$y=2at=4t$$, where $$t$$ is any real parameter.

The focus of the parabola $$y^{2}=4ax$$ is always the point $$(a,0)$$. Here $$a=2$$, so the focus is $$F(2,0)$$. If we choose the point $$P(2t^{2},4t)$$ on the curve, the square of its focal distance is

$$FP^{2}=(2t^{2}-2)^{2}+(4t-0)^{2}.$$

The problem says that the focal distance is $$8$$ units, so

$$FP^{2}=8^{2}=64.$$

Substituting the coordinates of $$P$$,

$$\begin{aligned} (2t^{2}-2)^{2}+(4t)^{2}&=64.\\ \end{aligned}$$

First expand each term:

$$\begin{aligned} (2t^{2}-2)^{2}&=4(t^{2}-1)^{2},\\ (4t)^{2}&=16t^{2}. \end{aligned}$$

Hence

$$4(t^{2}-1)^{2}+16t^{2}=64.$$

Divide the entire equation by $$4$$ to simplify:

$$\begin{aligned} (t^{2}-1)^{2}+4t^{2}&=16.\\ \end{aligned}$$

Now expand $$ (t^{2}-1)^{2} $$:

$$ t^{4}-2t^{2}+1+4t^{2}=16.$$

Combine the like terms $$-2t^{2}+4t^{2}=2t^{2}$$ to get

$$ t^{4}+2t^{2}+1 = 16.$$

Observe that the left-hand side is a perfect square:

$$ (t^{2}+1)^{2}=16.$$

Taking the non-negative square root (since $$ (t^{2}+1) $$ is always positive),

$$ t^{2}+1=4 \quad\Longrightarrow\quad t^{2}=3.$$

Thus the admissible values of the parameter are $$t=\sqrt{3}$$ or $$t=-\sqrt{3}$$.

For the tangent at $$P(2t^{2},4t)$$, differentiate the parametric equations. We have $$dx/dt=4t$$ and $$dy/dt=4$$, so the slope of the tangent is

$$m_{\text{tan}}=\frac{dy/dt}{dx/dt}=\frac{4}{4t}=\frac{1}{t}.$$

The slope of the normal, being the negative reciprocal, is

$$m=-t.$$

The normal at $$P(x_{1},y_{1})$$ is written in point-slope form as

$$y-y_{1}=m(x-x_{1}).$$

To express it as $$y=mx+c$$ we rearrange to obtain $$c=y_{1}-mx_{1}.$$ We now compute $$c$$ for each value of $$t$$.

Case 1: $$t=\sqrt{3}$$.

The point on the curve is $$ \begin{aligned} x_{1}&=2t^{2}=2(\sqrt{3})^{2}=6,\\ y_{1}&=4t=4\sqrt{3}. \end{aligned} $$

The slope of the normal is $$m=-t=-\sqrt{3}.$$ Therefore

$$ \begin{aligned} c&=y_{1}-mx_{1}\\ &=4\sqrt{3}-\bigl(-\sqrt{3}\bigr)(6)\\ &=4\sqrt{3}+6\sqrt{3}\\ &=10\sqrt{3}. \end{aligned} $$

Case 2: $$t=-\sqrt{3}$$.

The point is $$ \begin{aligned} x_{1}&=2t^{2}=2(-\sqrt{3})^{2}=6,\\ y_{1}&=4t=4(-\sqrt{3})=-4\sqrt{3}. \end{aligned} $$

The slope of the normal is $$m=-t=\sqrt{3}.$$ Hence

$$ \begin{aligned} c&=y_{1}-mx_{1}\\ &=-4\sqrt{3}-\bigl(\sqrt{3}\bigr)(6)\\ &=-4\sqrt{3}-6\sqrt{3}\\ &=-10\sqrt{3}. \end{aligned} $$

In both cases the magnitude of the intercept is

$$|c|=10\sqrt{3}.$$

Hence, the correct answer is Option B.