The area (in sq. units) bounded by the parabola y^2 = 16x and its lat…

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

Comparing \(y^2 = 16x\) with \(y^2 = 4ax\) gives \(a = 4\). The focus is (4, 0), so the latus rectum is the line \(x = 4\), from (4, −8) to (4, 8).

The region is symmetric about the x-axis, so the area is twice the area above the x-axis, where \(y = 4\sqrt{x}\):

    \( \displaystyle \text{Area} = 2\int_{0}^{4} 4\sqrt{x}\,dx \)

    \( \displaystyle = 8\left[\frac{2}{3}x^{3/2}\right]_{0}^{4} = 8 \cdot \frac{2}{3} \cdot 8 = \frac{128}{3} \)

Shortcut: The area cut off from a parabola by a chord perpendicular to its axis is \( \displaystyle\frac{2}{3} \) of the enclosing rectangle. Here the rectangle is 4 wide and 16 high, so the area is \( \displaystyle\frac{2}{3} \times 64 = \frac{128}{3} \). For \(y^2 = 4ax\) and its latus rectum this is \( \displaystyle\frac{8a^2}{3} \).