Mathematics - Area Under Curves Question with Solution | TestHub

We have parabola $y^2=16x\Rightarrow 4a=16\Rightarrow a=4$. Hence coordinates of end points of latus rectum are $(4,\pm 8)$ Required area is shaded. $$\begin{array}{rl}\text{ Area }& =4{\int }_0^4\sqrt{x}\mathrm{dx}=4{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^4\\ & =4\times \frac{2}{3}\left[4^{\frac{3}{2}}-0\right]=4\times \frac{2}{3}\times 8=\frac{64}{3}\text{ sq. units }\end{array}$$