Find the area of the region enclosed by the parabola `y^2=4a x\ a n d` the line `y=m xdot`

Explanation

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 `((4a)/(m^2) , (4a)/m )`
we draw AC perpendicular to x-axis.
Area `OABO`=Area (`OCABO`)−Area (`triangleOCA`)
=` int_0^((4a)/(m^2)) 2sqrt(ax)dx - int_0^((4a)/(m^2)) mx dx`
`= 2sqrt(a) [ x^(3/2)/(3/2) ]_0^((4a)/(m^2)) - m[x^2/2]_0^((4a)/(m^2))`
`= 4/3 sqrt(a) ( 4a/m^2)^(3/2) - m/2[( (4a)/(m^2))^2]`
`= (32a^2)/(3m^3) - m/2 ( (16a^2)/m^4)`
`= (32a^2)/(3m^3) - (8a^2)/m^3`
`= (8a^2)/(3m^3)` sq. units