Area bounded by y2 = 4ax and latus rectum is

The area bounded by the parabola y^2 = 4ax and its latus rectum is given by the integral of the function defining the parabola from the x-coordinate of the vertex to the x-coordinate of the end of the latus rectum.

Step 1: Identify the function defining the parabola. In this case, it is y = sqrt(4ax).

Step 2: Identify the x-coordinate of the vertex. For the parabola y^2 = 4ax, the vertex is at x = 0.

Step 3: Identify the x-coordinate of the end of the latus rectum. The latus rectum of a parabola y^2 = 4ax is a line segment parallel to the x-axis, passing through the focus, with its endpoints on the parabola. The length of the latus rectum is 4a, so its x-coordinates are a and -a.

Step 4: Set up the integral to find the area. The area A under the curve y = f(x) from x = a to x = b is given by the integral ∫ from a to b of f(x) dx. In this case, we want to find the area from x = 0 to x = a, so the integral is ∫ from 0 to a of sqrt(4ax) dx.

Step 5: Evaluate the integral. This requires knowledge of integral calculus. The integral of sqrt(4ax) with respect to x from 0 to a is (2/3) * (4a)^(3/2).

So, the area bounded by the parabola y^2 = 4ax and its latus rectum is (2/3) * (4a)^(3/2).