Question

If the area bounded by the parabola $y^2$ = 16ax and the line y = 4mx is $\frac{a^2}{12}$ sq.units, then using integration, find the value of m.

Answer 1

Let $y^2$ = 16ax ....(i), y =4mx ....(ii)

By (i) & (ii), $(4mx{)}^2$ = 16ax $\Rightarrow x(m^{2}x-a)=0$

$\Rightarrow x=0$ or $x=\frac{a}{m^2}$

Acc. to quest., area bounded by the curves = ${\int }_0^{\frac{a}{m^2}}[y_i-y_{ii}]dx=\frac{a^2}{12}$

$\Rightarrow 4\sqrt{a}{\int }_0^{\frac{a}{m^2}}\sqrt{x}dx-4m{\int }_0^{\frac{a}{m^2}}xdx=\frac{a^2}{12}$

$\frac{8}{3}\sqrt{a}{\left[x^{\frac{3}{2}}\right]}_0^{\frac{a}{m^2}}-2m{\left[x^{\frac{3}{2}}\right]}_0^{\frac{a}{m^2}}=\frac{a^2}{12}$

$\Rightarrow \frac{8}{3}\sqrt{a}{\left[x^{\frac{3}{2}}\right]}_0^{\frac{a}{m^2}}-2m{\left[x^{\frac{3}{2}}\right]}_0^{\frac{a}{m^2}}=\frac{a^2}{12}\Rightarrow \frac{8}{3}\sqrt{a}[\frac{a^{\frac{3}{2}}}{m^3}-0]-2m[\frac{a^2}{m^4}-0]=\frac{a^2}{12}$

$\Rightarrow (\frac{8}{3}-2)\frac{a^2}{m^3}=\frac{a^2}{12}\Rightarrow \frac{2}{3}\times \frac{1}{m^3}=\frac{1}{12}\therefore m=2.$