Mathematics - Area Under Curves Question with Solution | TestHub

We have parabola ${\mathrm{y}}^2=8\mathrm{x}\Rightarrow 4\mathrm{a}=8\Rightarrow \mathrm{a}=2$ Hence coordinates of latus rectum are $=(\mathrm{a},\pm 2\mathrm{a})\text{ i.e. }(2,4)\text{ and }(2,-4)$ Required area is shaded in figure. $\begin{array}{rl}\therefore A& =2{\int }_0^2\sqrt{8x}dx=4\sqrt{2}{\int }_0^2{x}^{\frac{1}{2}}dx\\ & =4\sqrt{2}{\left[\frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]}_0^2=\frac{8\sqrt{2}}{3}(2\sqrt{2})=\frac{32}{3}\end{array}$