The area (in sq. units) bounded by the parabola $y^2 = 4ax$, its latus rectum and the x-axis in the first quadrant is:

$\frac{4}{3}a^2$

The correct answer is Option (3) → $\frac{4}{3}a^2$

Given parabola:

\(y^2 = 4ax\)

Latus rectum length = \(4a\)

Focus at \((a, 0)\), endpoints of latus rectum are at \(\left(a, 2a\right)\) and \(\left(a, -2a\right)\).

Considering the first quadrant, the latus rectum segment is from \(\left(a, 0\right)\) to \(\left(a, 2a\right)\).

The area bounded by the parabola, the latus rectum, and the x-axis in the first quadrant is:

\[ \text{Area} = \int_0^a y \, dx \]

From parabola, \(y = 2\sqrt{ax}\) (positive root in first quadrant).

Integral:

\[ \text{Area} = \int_0^a 2 \sqrt{ax} \, dx = 2 \sqrt{a} \int_0^a \sqrt{x} \, dx = 2 \sqrt{a} \left[ \frac{2}{3} x^{3/2} \right]_0^a = 2 \sqrt{a} \times \frac{2}{3} a^{3/2} = \frac{4}{3} a^2 \]