If $m x-y+c=0$ is a normal at a point $P$ on the parabola $y^2=16 x$ and the focal distance of $P$ is 40 units, then $|c|=$

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 $\quad|c|=132$