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

Given parabola: $y^2 = 16x$

This is a standard parabola opening to the right with form $y^2 = 4ax$.

Comparing: $4a = 16$

$a = 4$

Key points:

• Vertex = $(0, 0)$
• Focus = $(a, 0) = (4, 0)$
• Latus rectum is a vertical line at $x = 4$

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The latus rectum is a vertical line passing through the focus, perpendicular to the axis of symmetry.

For this parabola, latus rectum is the line $x = 4$

Finding the endpoints by substituting $x = 4$ in $y^2 = 16x$:

$y^2 = 16(4) = 64$

$y = \pm 8$

The latus rectum goes from $(4, -8)$ to $(4, 8)$ with length 16 units.

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The area needed is the region between the curved parabola and the straight latus rectum line.

From the parabola: $y^2 = 16x$

$y = 4\sqrt{x}$ (taking positive value)

Since the parabola is symmetric about the x-axis, the area above the x-axis can be found and multiplied by 2.

Area $= 2 \times \int_0^4 y \, dx$

Area $= 2 \times \int_0^4 4\sqrt{x} \, dx$

Area $= 8 \int_0^4 \sqrt{x} \, dx$

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$\int \sqrt{x} \, dx = \int x^{1/2} \, dx$

$= \dfrac{x^{3/2}}{3/2}$

$= \dfrac{2}{3}x^{3/2}$

Area $= 8 \times \left[\dfrac{2}{3}x^{3/2}\right]_0^4$

Area $= 8 \times \dfrac{2}{3} \times [4^{3/2} - 0]$

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$4^{3/2} = (4^{1/2})^3$

$= 2^3$

$= 8$

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Area $= 8 \times \dfrac{2}{3} \times 8$

$= \dfrac{16}{3} \times 8$

$= \dfrac{128}{3}$ sq. units

Therefore, the area bounded by the parabola and its latus rectum is $\dfrac{128}{3}$ square units.