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If the area included between the two parabolas $y^{2} = 4 a (x + a)$ and $y^{2} = 4 b (b - x)$ is $\frac{16}{3}$ , then $(a, b > 0)$

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$a + b = a b$

AM and GM of a and b are reciprocals of each other $\forall a, b > 0$

AM and GM of a and b are equal $\forall a, b > 0$

$\sqrt{\frac{a^{2} + b^{2}}{2}} = a + b$

answer is B.

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Detailed Solution

$\frac{16}{3} = 2 \int_{- a}^{b - a} 2 \sqrt{a} \sqrt{x + a} d x + 4 \int_{b - a}^{b} \sqrt{b} \sqrt{b - x} d x$

$\Rightarrow \frac{8}{3} \sqrt{a b} (a + b) = \frac{16}{3} \Rightarrow (\frac{a + b}{2}) \sqrt{a b} = 1$

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