Q.Find the area of the region bounded by the parabola $y^2 = 16x$ and its latus rectum.

For a standard parabola $y^2=4ax$ together with a vertical line $x=h$ (region symmetric about the x-axis), the enclosed area is $2\int_0^h 2\sqrt{ax}\,dx=\tfrac{8\sqrt a}{3}h^{3/2}$; at the latus rectum ($h=a$) this reduces to the standard result $\tfrac83a^2$. When a region is instead bounded by the y-axis and a horizontal line $y=k$, it is more natural to solve for $x=y^2/4a$ and integrate horizontal …