Given that, equation of parabola $y^{2}=4 a x$, let the parametric coordinate is $\left(a m^{2}, 2 a m\right)$.
$\Rightarrow 2 y \frac{d y}{d x}=4 a$
$\Rightarrow \frac{d y}{d x}=\frac{2 a}{y}$
Slope of normal $= \left(\frac{-y}{2 a}\right)$
At $\left(a m^{2}, 2 a m\right)=\frac{-2 a m}{2 a}=-m$
Now, the equation of normal to the parabola is
$(y-2 a m)=(-m)\left(x-a m^{2}\right)$
$y-2 a m=-mx +a m^{3}$
$mx +y-\left(2am +a m^{3}\right)=0$...(i)
Also, given the line
$y=mx +c$ or $m x-y +c=0$...(ii)
is normal to parabola, then
On comparing $c=-2 a m-a m^{3}$