The area of the region bounded by the parabola $y^2 = 8x$ and its latus rectum in the first quadrant, is

$\frac{16}{3}$ sq. units

The correct answer is Option (4) → $\frac{16}{3}$ sq. units

Given: Parabola $y^2 = 8x$

⇒ $y = \sqrt{8x}$ in the first quadrant

Latus rectum: $x = 2$

Calculate the area between $x = 0$ and $x = 2$ under the curve in terms of $dx$:

Area = $\int_{x=0}^{2} y \, dx = \int_{0}^{2} \sqrt{8x} \, dx$

$= \int_{0}^{2} \sqrt{8} \cdot \sqrt{x} \, dx = \sqrt{8} \int_{0}^{2} x^{1/2} \, dx$

$= \sqrt{8} \cdot \frac{2}{3} x^{3/2} \Big|_0^2$

$= \sqrt{8} \cdot \frac{2}{3} \cdot (2)^{3/2}$

$= \frac{2}{3} \cdot \sqrt{8} \cdot \sqrt{8} = \frac{2}{3} \cdot 8 = \frac{16}{3}$