Question

The area bounded by the curve $y^2\left(a^2+x^2\right)=x^2\left(a^2-x^2\right)$ is

• A. $a^2(\pi -2)$ sq unit
• B. $a^2(\pi +2)$ sq unit
• C. $a^2(\pi -1)$ sq unit
• D. $a^2(\pi +1)$ sq unit

Answer 1

Answer: (A)

Since, the curve is symmetrical about both the axes. Therefore, the required area

$=4\underset{0}{\overset{a}{\int }}ydx=4\underset{0}{\overset{a}{\int }}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx=4\underset{a}{\overset{\sqrt{2}a}{\int }}\sqrt{(\sqrt{2}a{)}^2-z^2}dz$
Let $x^2+a^2=z^2$
$\Rightarrow xdx=zdz.$
Also, $a^2-x^2=2a^2-z^2$
$=4{\left[\frac{z}{2}\sqrt{(\sqrt{2}a{)}^2-z^2}+\frac{(\sqrt{2}a{)}^2}{2}{\sin }^{-1}\frac{z}{\sqrt{2}a}\right]}_a^{\sqrt{2}a}$
$=a^2(\pi -2)$ sq. unit