Find the area of the parabola $y^2 = 4ax$ bounded by its latus rectum.

$\frac{8}{3} a^2$

The correct answer is Option (2) → $\frac{8}{3} a^2$

From Figure, the vertex of the parabola $y^2 = 4ax$ is at the origin $(0, 0)$. The equation of the latus rectum $LSL'$ is $x = a$. Also, the parabola is symmetrical about the $x$-axis.

The required area of the region $OLL'O$:

$= 2 (\text{area of the region } OLSO)$

$= 2 \int\limits_{0}^{a} y \, dx = 2 \int\limits_{0}^{a} \sqrt{4ax} \, dx$

$= 2 \times 2 \sqrt{a} \int\limits_{0}^{a} \sqrt{x} \, dx$

$= 4 \sqrt{a} \left[ \frac{2}{3} x^{\frac{3}{2}} \right]_0^a$

$= \frac{8}{3} \sqrt{a} \left[ a^{\frac{3}{2}} \right] = \frac{8}{3} a^2$