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.