The area of the region bounded by the parabola $y^{2}=8 x$ and its latus rectum is
The area of the region bounded by the parabola $y^{2}=8 x$ and its latus rectum is
$\frac{16}{3}$ sq. units
$\frac{8}{3}$ sq. units
$\frac{32}{3}$ sq. units
$\frac{4}{3}$ sq. units
Solution
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.
$\begin{aligned}
\therefore A &=2 \int_{0}^{2} \sqrt{8 x} d x=4 \sqrt{2} \int_{0}^{2} x^{\frac{1}{2}} d x \\
&=4 \sqrt{2}\left[\frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]_{0}^{2}=\frac{8 \sqrt{2}}{3}(2 \sqrt{2})=\frac{32}{3}
\end{aligned}$
