Given, parabola, $y^{2}=4 a x$
$\begin{array}{lr}\Rightarrow & \text { 2y } \frac{d y}{d x}=4 a \\ \Rightarrow \quad \frac{d y}{d x}=\frac{2 a}{y}\end{array}$
which is the slope of tangent.
Given, $l x+m y+n=0$
is an equation of tangent of the parabola $y^{2}=4 a x$
$\therefore$ Slope of tangent $=-\frac{l}{m}$
From Eqs. (i) and (ii)
$$
\begin{aligned} \frac{2 a}{y} &=-\frac{l}{m} \Rightarrow y=\frac{-2 a m}{l} \\ \Rightarrow \quad & y^{2}=4 a x \\ \Rightarrow \quad \frac{4 a^{2} m^{2}}{l^{2}} &=4 a x \\ \Rightarrow & x=\frac{a m^{2}}{l^{2}} \end{aligned}
$$
On putting the values of $x$ and $y$ in the following equation
$$
\begin{aligned}
b x+m y+n &=0 \\
l\left(\frac{a m^{2}}{l^{2}}\right)+m\left(\frac{-2 a m}{l}\right)+n &=0 \\
\frac{a m^{2}}{l}-\frac{2 a m^{2}}{l}+n &=0 \\
\Rightarrow \quad \quad \frac{a m^{2}}{l}=n \Rightarrow a m^{2} &=n l
\end{aligned}
$$
which is the required relation.