The area bounded by the parabola $y^{2}=16 x$ and its latus - rectum in the first quadrant is
The area bounded by the parabola $y^{2}=16 x$ and its latus - rectum in the first quadrant is
$128$ sq. units
$\frac{64}{3}$ sq. units
$\frac{128}{3}$ sq. units
$64$ sq. units
Solution
We have parabola $y^{2}=16 x \Rightarrow 4 a=16 \Rightarrow a=4$.
Hence coordinates of end points of latus rectum are $(4, \pm 8)$
Required area is shaded.
$\begin{aligned}
\text { Area } &=4 \int_{0}^{4} \sqrt{x} \mathrm{dx}=4\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4} \\
&=4 \times \frac{2}{3}\left[4^{\frac{3}{2}}-0\right]=4 \times \frac{2}{3} \times 8=\frac{64}{3} \text { sq. units }
\end{aligned}$
