Question

Area included between the parabola $y=\frac{x^2}{4a}$ and the curve $y=\frac{8a^3}{(x^2+4a^2)}$ is

• A. $\frac{a^2}{3}\left(6\mathrm{\pi }-4\right)$
• B. $\frac{a^2}{3}\left(4\mathrm{\pi }+4\right)$
• C. $\frac{a^2}{3}\left(8\mathrm{\pi }+3\right)$
• D. None of these

Answer 1

The correct option is A

$\frac{a^2}{3}\left(6\mathrm{\pi }-4\right)$

Explanation for the correct option:

Step 1. Draw the graph of curves $y=\frac{x^2}{4a}$ and $y=\frac{8a^3}{(x^2+4a^2)}$:

From the figure, we have

The point of intersection of given curves,

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

$\Rightarrow$ $x^4+4a^2{x}^2-32a^4=0$

$\Rightarrow$$\left(x^2-4a^2\right)\left(x^2+8a^2\right)=0$

$\Rightarrow$ $x=\pm 2a$

Step 2. The required area $=2{\int }_0^{2a}\left[\frac{8a^3}{x^2+4a^2}-\frac{x^2}{4a}\right]dx$

$=2\left[{\int }_0^{2a}\frac{8a^3}{x^2+4a^2}dx-{\int }_0^{2a}\frac{x^2}{4a}dx\right]$

$=2{\left[\frac{8a^3}{2a}{\tan }^{-1}\frac{x}{2a}-\frac{x^3}{12a}\right]}_0^{2a}$ …

$=2\left[\frac{8a^3}{2a}{\tan }^{-1}\frac{2a}{2a}-\frac{8a^3}{12a}-\left(\frac{8a^3}{2a}{\tan }^{-1}\left(0\right)-0\right)\right]$

$=\frac{a^2}{3}\left(6\mathrm{\pi }-4\right)$

Thus, Area included between the parabola and curve is $\frac{{\mathbf{a}}^{\mathbf{2}}}{\mathbf{3}}\left(6\pi -4\right)$

Hence, Option ‘A’ is Correct.