Question

The area of the region bounded by the parabola $y^2=4ax$ and its latus rectum is:

• A. $\frac{8a^2}{3}$
• B. $\frac{4a^2}{3}$
• C. $\frac{2a^2}{3}$
• D. $\frac{a^2}{3}$

Answer 1

Answer: (A)

Given parabola is $y^2=4ax$
$\Rightarrow y=2\sqrt{a}\sqrt{x}$
Vertex of this parabola is (0, 0) and focus is (a, 0)
Axis of this parabola is x-axis and is symmetric about x-axis.
Latus-rectum is LL' which is perpendicular to x-axis and passing through focus.
$\therefore$ Required area is area of the region OLL' = 2 $\times$ area of region OSL
$2{\int }_0^{a}ydx=2{\int }_0^{a}2\sqrt{a}\sqrt{x}dx$
$=4\sqrt{a}{\left(\frac{2}{3}{x}^{\frac{3}{2}}\right)}_0^a=\frac{8\sqrt{a}}{3}\left[a^{\frac{3}{2}}\right]=\frac{8}{3}{a}^2$