If $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ is the normal at a point on the parabola ${\mathrm{y}}^2=8\mathrm{x}$ whose focal distance is 8 units then $\left|c\right|=$

${\mathrm{y}}^2=8\mathrm{x}$

             $$\begin{gathered} \Rightarrow 4\mathrm{a}=8 \\ \Rightarrow \overline{)\mathrm{a}=2} \\ diff w.r.to x on both sides \\ \frac{dy}{dx}=\frac{8}{2y}=\frac{4}{y} \end{gathered}$$

$$\begin{gathered} Given focal dis\tan ce \left|{\mathrm{x}}_1+\mathrm{a}\right|=8 \\ \left|{\mathrm{x}}_1+2\right|=8 \\ \Rightarrow {\mathrm{x}}_1=\pm 6 \\ sub x=6 then \\ {\mathrm{y}}^2=6\times 8 \\ \Rightarrow \mathrm{y}=\pm 4\sqrt{3} \end{gathered}$$

points are $\left(6, 4\sqrt{3}\right) \& \left(6, -4\sqrt{3}\right)$ 

slope of normal at $\left(6, 4\sqrt{3}\right)$ is $\mathrm{m}=\pm \sqrt{3}$

$$\begin{gathered} \mathrm{So},\mathrm{normal} \mathrm{becomes} \mathrm{y}=\pm \sqrt{3}\mathrm{x}+\mathrm{c} \\ \mathrm{since} \mathrm{it} \mathrm{passes} \mathrm{through} \mathrm{the} \mathrm{point} (6,\pm 4\sqrt{3}) \mathrm{then} \\ \pm 4\sqrt{3}=\pm \sqrt{3}\left(6\right)+c \\ \Rightarrow c=\pm 10\sqrt{3} \\ \Rightarrow \left|\mathrm{c}\right|=10\sqrt{3} \end{gathered}$$