The condition for the line \(y=m x+c\) to be \(a\) normal to the parabola \(\mathrm{y}^{2}=4 \mathrm{ax}\) is

Given that, equation of parabola \(y^{2}=4 a x\), let the parametric coordinate is \(\left(\mathrm{am}^{2}, 2 \mathrm{am}\right)\).
\[
\begin{array}{ll}
\Rightarrow & 2 y \frac{d y}{d x}=4 a \\
\Rightarrow & \frac{d y}{d x}=\frac{2 a}{y}
\end{array}
\]
Slope of normal \(=\left(\frac{-y}{2 a}\right)\)
At
\[
\left(\mathrm{am}^{2}, 2 \mathrm{am}\right)=\frac{-2 \mathrm{am}}{2 \mathrm{a}}=-\mathrm{m}
\]
Now, the equation of normal to the parabola is
\[
\begin{gathered}
(\mathrm{y}-2 \mathrm{am})=(-\mathrm{m})\left(\mathrm{x}-\mathrm{am}^{2}\right) \\
\mathrm{y}-2 \mathrm{am}=-\mathrm{mx}+a \mathrm{~m}^{3} \\
\mathrm{mx}+\mathrm{y}-\left(2 \mathrm{am}+a \mathrm{~m}^{3}\right)=0 \quad \text{...(i)}
\end{gathered}
\]
Also, given the line
\[
y=m x+c \text { or } m x-y+c=0 \quad \text{...(ii)}
\]
is normal to parabola, then
On comparing \(\mathrm{c}=-2 \mathrm{am}-\mathrm{am}^{3}\)