Question

The area included between the parabolas
$y=\frac{x^2}{4a}$ and $y=\frac{8a^3}{x^2+4a^2}$ is

• A. $a^2(2\pi +\frac{2}{3})$
• B. $a^2(2\pi -\frac{8}{3})$
• C. $a^2(\pi +\frac{4}{3})$
• D. $a^2(\pi -\frac{4}{3})$

Answer 1

Answer: (D)

$a^2(\pi -\frac{4}{3})$

Firstly we have to find the point of intersection.

$\frac{x^2}{4a}=\frac{8a^3}{x^2+4a^2}$

On solving the equation We get $x^2=4a^2$ and $x^2=-8a^2$ (which is not possible)

We have $x=2a$ and $x=-2a$

Now area between these 2 curves is

${\int }_{-2a}^{2a}(\frac{8a^3}{x^2+4a^2}-\frac{x^2}{4a})dx$

$={\int }_{-2a}^{2a}\frac{8a^3}{x^2+4a^2}dx-{\int }_{-2a}^{2a}\frac{x^2}{4a}dx$

$=\frac{8a^3}{2a}{\tan }^{-1}\frac{x}{2a}-\frac{x^3}{12a}$

After putting the limits we get

$\frac{8a^3}{2a}{\tan }^{-1}\frac{2a}{2a}-\frac{{8a}^3}{12a}-(\frac{8a^3}{2a}{\tan }^{-1}\frac{-2a}{2a}-\frac{{-8a}^3}{12a})$

$=\pi a^2-\frac{4}{3}{a}^2$