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The equation of the parabola is $y^2 = 4ax$. The latus rectum is $x = a$.
The area bounded by the parabola and the latus rectum can be found by integrating the function $x = \frac{y^2}{4a}$ with respect to $y$ from $-2a$ to $2a$. Due to symmetry, we can integrate from $0$ to $2a$ and multiply by 2.
Area = $2 \int_{0}^{2a} (a - \frac{y^2}{4a}) dy$
Area = $2 [ay - \frac{y^3}{12a}]_{0}^{2a}$
Area = $2 [(a(2a) - \frac{(2a)^3}{12a}) - (0 - 0)]$
Area = $2 [2a^2 - \frac{8a^3}{12a}]$
Area = $2 [2a^2 - \frac{2a^2}{3}]$
Area = $2 [\frac{6a^2 - 2a^2}{3}]$
Area = $2 [\frac{4a^2}{3}]$
Area = $\frac{8a^2}{3}$
Correct Answer: $\frac{8a^2}{3}$