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

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$\frac{a^{2}}{3}$

$\frac{16 a^{2}}{3}$

$\frac{4 a^{2}}{3}$

$\frac{8 a^{2}}{3}$

answer is A.

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Detailed Solution

The required area= $= 4 \int_{0}^{a} \sqrt{4 a (a - x)} d x = 4 (2 \sqrt{a}) {(\frac{- 2 {(a - x)}^{3 / 2}}{3})}_{0}^{a} = \frac{16 a^{2}}{3} s q . u n i t s$

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