Question

Find the area enclosed between the parabola $y^2=4ax$ and the line $y=mx$.

Answer 1

The area enclosed between the parabola, $y^2=4ax$ and the line, $y=mx$ is represented by the shaded area $OABO$ as
The points of intersection of both the curves are $(0,0)$ and $(\frac{4a}{m^2},\frac{4a}{m})$.
We draw $AC$ perpendicular to $x$-axis.

$\therefore AreaOABO=AreaOCABO-Area\left(\Delta OCA\right)$

$={\int }_0^{\frac{4a}{m^2}}2\sqrt{ax}dx-{\int }_0^{\frac{4a}{m^2}}mxdx$

$=2\sqrt{a}{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^{\frac{4a}{m^2}}-m{\left[\frac{x^2}{2}\right]}_0^{\frac{4a}{m^2}}$

$=\frac{4}{3}\sqrt{a}{\left(\frac{4a}{m^2}\right)}^{\frac{3}{2}}-\frac{m}{2}\left[{\left(\frac{4a}{m^2}\right)}^2\right]$

$=\frac{32a^2}{3m^3}-\frac{m}{2}\left(\frac{16a^2}{m^4}\right)$

$=\frac{32a^2}{3m^3}-\frac{8a^2}{m^3}$

$=\frac{8a^2}{3m^3}$ sq. units