Solution:
Area enclosed between the parabola, $y^{2}=4 a x$, and the line, $y=m x$, is represented by the shaded area $\mathrm{OABO}$ as

Points of intersection of both the curves are $(0,0)$ and $\left(\frac{4 a}{m^{2}}, \frac{4 a}{m}\right)$.
Draw $A C$ perpendicular to $x$-axis.
$\therefore$ Area $\mathrm{OABO}=$ Area $\mathrm{OCABO}-$ Area $(\triangle \mathrm{OCA})$
$=\int_{0}^{4 a} x^{2} 2 \sqrt{a x} d x-\int^{4 a} m^{2} m x d x$
$=2 \sqrt{a}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{\frac{4 a}{m^{2}}}-m\left[\frac{x^{2}}{2}\right]_{0}^{\frac{4 a}{m^{2}}}$
$=\frac{4}{3} \sqrt{a}\left(\frac{4 a}{m^{2}}\right)^{\frac{3}{2}}-\frac{m}{2}\left[\left(\frac{4 a}{m^{2}}\right)^{2}\right]$
$=\frac{32 a^{2}}{3 m^{3}}-\frac{m}{2}\left(\frac{16 a^{2}}{m^{4}}\right)$
$=\frac{32 a^{2}}{3 m^{3}}-\frac{8 a^{2}}{m^{3}}$
$=\frac{8 a^{2}}{3 m^{3}}$ sq. units