The area bounded by the parabola $\mathrm{y}^2=4 \mathrm{ax}$ and its latus-rectum $x=a$ is
- $\frac{8}{3} a^2$ sq. units
- $\frac{2}{3} a^2$ sq. units
- $\frac{4}{3} a^2$ sq. units
- $8 \mathrm{a}^2$ aq. units
Solution
Required area is shaded.
Point of intersection of $x=a$ and $y^2=4 a x$, is
$y^2=4 a^2 \Rightarrow y= \pm 2 a \text { and } x=a \Rightarrow(a, \pm 2 a)$
$\begin{aligned}
& \therefore \mathrm{A}=2 \int_0^{\mathrm{a}}(2 \sqrt{\mathrm{a}} \sqrt{\mathrm{x}}) \mathrm{dx} \\
& =4 \sqrt{\mathrm{a}} \int_0^{\mathrm{a}} \mathrm{x}^{\frac{1}{2}} \mathrm{dx}=4 \sqrt{\mathrm{a}}\left[\frac{\mathrm{x}^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_0^{\mathrm{a}}=(4 \sqrt{\mathrm{a}})\left(\frac{2}{3}\right)(\mathrm{a} \sqrt{\mathrm{a}})=\frac{8}{3} \mathrm{a}^2 \text { sq. units }
\end{aligned}$Asked in: MHT CET 2021 (23 Sep Shift 2)