The area of the region bounded by the parabola $y^2 = 8x$ and its latus rectum in the first quadrant, is

Given parabola: $y^2 = 8x$

Comparing with the standard form $y^2 = 4ax$:

$4a = 8$

$a = 2$

The latus rectum is the vertical line $x = a = 2$

---

To find where the latus rectum meets the parabola, substitute $x = 2$:

$y^2 = 8(2) = 16$

$y = 4$ (in the first quadrant)

The region is bounded by the parabola from $x = 0$ to $x = 2$, the latus rectum at $x = 2$, and the x-axis.

---

From $y^2 = 8x$:

$y = \sqrt{8x} = 2\sqrt{2x}$ (positive in first quadrant)

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

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

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

$= 2\sqrt{2} \cdot \left[\frac{x^{3/2}}{3/2}\right]_0^2$

$= 2\sqrt{2} \cdot \frac{2}{3} \cdot [x^{3/2}]_0^2$

$= \frac{4\sqrt{2}}{3} \cdot (2^{3/2} - 0)$

$= \frac{4\sqrt{2}}{3} \cdot 2\sqrt{2}$

$= \frac{8 \cdot 2}{3}$

$= \frac{16}{3}$ square units