Question

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

Answer 1

Given, the curve (represents a parabola with vertex (0, 0)).

$y^2=4ax$ ...(i)

and equation of line $y=mx$ ...(ii)

From Eqs. (i) and (ii), we get $(mx{)}^2=4ax$

$\Rightarrow x(m^{2}x-4a)=0\Rightarrow x=0orx=\frac{4a}{m^2}$

When $x=0,y=m\times 0=0,$

When $x=\frac{4a}{m^2},y=m\times \frac{4a}{m^2}=\frac{4a}{m}$

$\therefore$ The points of intersection of both the curves are (0, 0) and $(\frac{4a}{m^2},\frac{4a}{m})$

$\therefore$ The points of intersection of both the curves are (0, 0) and $(\frac{4a}{m^2},\frac{4a}{m})$

$\therefore$ Required area

$={\int }_0^{\frac{4a}{m^2}}(\sqrt{4ax}-mx)dx$
$(\because y^2=4ax,\therefore y=\sqrt{4ax}$ in the first quadrant.)
$=2\sqrt{a}[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}{]}_0^{\frac{4a}{m^2}}-[m\frac{x^2}{2}{]}_0^{\frac{4a}{m^2}}=\frac{4}{3}\sqrt{a}(\frac{4a}{m^2}{)}^{\frac{3}{2}}-\frac{32}{3}.\frac{a^2}{m^3}-\frac{8a^2}{m^3}=\frac{8a^2}{3m^3}$