Find the area enclosed between the parabola y2=4ax and the line y = mx.
Given, the curve (represents a parabola with vertex (0, 0)).
y2=4ax ...(i)
and equation of line y=mx ...(ii)
From Eqs. (i) and (ii), we get (mx)2=4ax
⇒x(m2x−4a)=0⇒x=0 or x=4am2
When x=0,y=m×0=0,
When x=4am2,y=m×4am2=4am
∴ The points of intersection of both the curves are (0, 0) and (4am2,4am)
∴ The points of intersection of both the curves are (0, 0) and (4am2,4am)
∴ Required area
=∫4am20(√4ax−mx)dx
(∵y2=4ax,∴y=√4ax in the first quadrant.)
=2√a[x3232]4am20−[mx22]4am20=43√a(4am2)32−323.a2m3−8a2m3=8a23m3