Question

Find the area enclosed between the parabola y2=4ax and the line y=mx

Answer 1

The area enclosed between the parabola,y2=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/m2,4a/m).

We draw AC perpendicular to x-axis.

∴Area OABO=Area OCABO-Area(ΔOCA)

=

\int_{0}^{\frac{4a}{m^3}} 2\sqrt{ax} \,dx$$$$-\int_{0}^{\frac{4a}{m^2}} mx \,dx

=2√a[$\frac{x^{\frac{3}{2}}}{\frac{3}{2}}$]4a/m20-m[$\frac{x^2}{2}$]4a/m2 0

=$\frac 43$√a$(\frac{4a}{m^2})^{3/2}$-$\frac m2$[($\frac{4a}{m^2}$)2]

=$\frac{32a^2}{3m^3}$-$\frac m2$($\frac {16a^2}{m^4}$)

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

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