The area (in sq. units) bounded by the parabola $y^2 = 16x$ and its latus rectum is

$\frac{128}{3}$ Sq. units

The correct answer is Option (4) → $\frac{128}{3}$ Sq. units

Given: Parabola is $y^2 = 16x$

Standard form: $y^2 = 4ax$ ⟹ $4a = 16$ ⟹ $a = 4$

Latus rectum: Line segment through focus perpendicular to the axis with length $4a = 16$

Focus = $(a, 0) = (4, 0)$

Latus rectum equation: $x = 4$ from $y = -8$ to $y = 8$

Area bounded between parabola and latus rectum:

Required area = Area between curve $y^2 = 16x$ and line $x = 4$ from $y = -8$ to $y = 8$

Express $x$ in terms of $y$: $x = \frac{y^2}{16}$

Area = $\int_{-8}^{8} \left(4 - \frac{y^2}{16} \right) dy$

Split and integrate:

$= \int_{-8}^{8} 4\,dy - \int_{-8}^{8} \frac{y^2}{16}\,dy$

$= 4[y]_{-8}^{8} - \frac{1}{16} \left[ \frac{y^3}{3} \right]_{-8}^{8}$

$= 4(8 - (-8)) - \frac{1}{16} \cdot \left( \frac{8^3 - (-8)^3}{3} \right)$

$= 4 \cdot 16 - \frac{1}{16} \cdot \left( \frac{512 + 512}{3} \right)$

$= 64 - \frac{1}{16} \cdot \frac{1024}{3}$

$= 64 - \frac{64}{3}$

$= \frac{192 - 64}{3} = \frac{128}{3}$