the line lx + my +n=0 is a normal to the parabola y^2 = 4ax if

the equation of the given parabola is $y^2=4ax$
the equation of the normal to the given parabola is $y+tx=2at+a{t}^3$
i.e. $tx+y-(2at+a{t}^3)=0$ ........(1)
 the equation (1) and  the given equation of the normal $lx+my+n=0 ...........\left(2\right)$ represent the same line:
therefore 
$\frac{t}{l}=\frac{1}{m}=\frac{-(2at+a{t}^3)}{n}$
$$\begin{gathered} \frac{t}{l}=\frac{1}{m}\Rightarrow t=\frac{l}{m} ........\left(3\right) \\ and \frac{1}{m}=\frac{-(2at+a{t}^3)}{n} \\ -\frac{n}{m}=2at+a{t}^3 \\ -\frac{n}{m}=2a.\left(\frac{l}{m}\right)+a.{\left(\frac{l}{m}\right)}^3 [ substituting the value of t from eq(3\left)\right] \\ -\frac{n}{m}=\frac{2a.l}{m}+\frac{a.l^3}{m^3} \\ \Rightarrow -n.m^2=2a.l{m}^2+a.l^3 \\ \Rightarrow a.l^3+2al{m}^2+n{m}^2=0 \end{gathered}$$

hope this helps you