The area bounded by the parabolas $y^2=4 a(x+a)$ and $y^2=-4 a(x-a)$ is

• Correct answer
  A

  $\frac{16}{3} a^2$
• B

  $\frac{8}{3} a^2$
• C

  $\frac{4}{3} a^2$
• D

  None of these

Solution

Required area $=4 \int_0^a \sqrt{4 a(a-x)} d x$

$\begin{aligned} & =4 \cdot 2 \sqrt{a} \cdot\left[\frac{-2(a-x)^{3 / 2}}{3}\right]_0^a \\ & =\frac{16}{3} a^2\end{aligned}$

Hence, the correct answer is option: 'A'.