If the line \(b x+m y+n=0\) is tangent to the parabola \(y^{2}=4 a x\), then

Given, parabola, \(y^{2}=4 a x\)
\(\Rightarrow \text { 2y } \frac{d y}{d x}=4 a\)
\(\Rightarrow \frac{d y}{d x}=\frac{2 a}{y}\)
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)
\(\frac{2 a}{y}=-\frac{l}{m} \Rightarrow y=\frac{-2 a m}{l}\)
\(\Rightarrow y^{2}=4 a x\)
\(\Rightarrow \frac{4 a^{2} m^{2}}{l^{2}} =4 a x\)
\(\Rightarrow x=\frac{a m^{2}}{l^{2}}\)
On putting the values of \(x\) and \(y\) in the following equation
\(b x+m y+n =0\)
\(l\left(\frac{a m^{2}}{l^{2}}\right)+m\left(\frac{-2 a m}{l}\right)+n =0\)
\(\frac{a m^{2}}{l}-\frac{2 a m^{2}}{l}+n=0\)
\(\Rightarrow \frac{a m^{2}}{l}=n \Rightarrow a m^{2} =n l\)
which is the required relation.