Given, equation of normal $m x-y+c=0$
Parabola $y^2=16 x$
Comparing with general equation of parabola i.e.,
$y^2=4 a x \Rightarrow a=4$
Let the coordinates of point $P$ on parabola be $\left(4 t^2, 8 t\right)$
$\because$ Slope of tangent at point $P=\frac{1}{t}$ $\ldots$ (i)
Also, slope of tangent $=-\frac{1}{\text { Slope of normal }}=-\frac{1}{m}$ $\ldots$ (ii)
From Eqs. (i) and (ii), $\frac{1}{t}=-\frac{1}{m}$ or $t=-m$
$\therefore$ Coordinate of point is $P\left(4 m^2,-8 m\right)$.
Focus of parabola is $(a, 0)$ or $(4,0)$.
Given, focal distance $=40$
Using distance formula,
$\sqrt{\left(4 m^2-4\right)^2+(-8 m-0)^2}=40$
or $\quad \sqrt{\left(m^2-1\right)^2+(-2 m)^2}=10$
or $\sqrt{\left(m^2-1\right)^2+4 m^2}=10$ or $\sqrt{\left(m^2+1\right)^2}=10$
or $m^2+1=10$ or $m^2=9$ or $m= \pm 3$
Therefore, coordinate of point $P$
$\left(4 \times( \pm 3)^2,-8( \pm 3)\right) \quad \equiv(36, \pm 24)$
$\because \text { Point } P \text { lies on line } m x-y+c=0$
then it must satisfy equation of line
$( \pm 3)(36)-( \pm 48)+c=0$
$\pm(3 \times 36+48)+c=0$
or $\quad c= \pm 132$
or $\quad|c|=132$