Question

The area cut off by the parabola $y^2=4ax$ and its latus rectum is

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

Answer 1

The correct option is C $\frac{8a^2}{3}$


Area = $2{\int }_0^{a}2\sqrt{a}\sqrt{x}dx=4\sqrt{a}\frac{2}{3}{x}^{\frac{3}{2}}{]}_0^a$
$=\frac{8a^2}{3}sq.unit$