Question

If the area included between two parabolas $y^2=4a(x+a)$ and $y^2=4b(b-x)$ is $\frac{16}{3}$ then product of $A.M$ and the G.M of $a$ and $b$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

The 2 curves intersect at $(b-a,2\sqrt{ab}),(b-a,-2\sqrt{ab})$

$Area=2\sqrt{a}{\int }_{-a}^{b-a}\sqrt{x+a}dx+2\sqrt{b}{\int }_{b-a}^b\sqrt{b-x}dx$

$\Rightarrow \frac{2}{3}.2\sqrt{a}\left[\right(x+a{)}^{3/2}{]}_{-a}^{b-a}-\frac{2}{3}.2\sqrt{b}\left[\right(b-x{)}^{3/2}{]}_{b-a}^b$

$\Rightarrow \frac{4}{3}\sqrt{a}\left[b^{3/2}\right]-\frac{4}{3}\sqrt{b}[-a^{3/2}]$

$\Rightarrow \frac{4}{3}b\sqrt{ab}+\frac{4}{3}a\sqrt{ab}$

$\Rightarrow Area=\frac{4}{3}\sqrt{ab}(a+b)$

Given that, $Area=\frac{16}{3}$

$\Rightarrow \frac{4}{3}\sqrt{ab}(a+b)=\frac{16}{3}$

$\Rightarrow \sqrt{ab}(a+b)=4$

$\Rightarrow \sqrt{ab}\frac{(a+b)}{2}=2$

Therefore, $G.M\times A.M=2$