Question

The area bounded by the curve $x{y}^2=a^2(a-x)$ and $y$-axis is

• A. $\pi a$ sq.unit
• B. $a^2$ sq.unit
• C. $\pi a^2$ sq.unit
• D. None of these

Answer 1

The correct option is C $\pi a^2$ sq.unit

The given curve is symmetrical about $x$-axis and meet it at $(a,0)$.

The line $x=0$ is asymptote.
Required area $=2{\int }_0^{a}ydx$
$=2]{\int }_0^{a}a\sqrt{\frac{a-x}{x}}dx$
Put $x=a{\sin }^2\theta$
$\Rightarrow dx=2a\sin \theta \cos \theta d\theta$
Therefore, area $=2{\int }_0^{\pi /2}a\frac{\cos \theta }{\sin \theta }\times 2a\sin \theta \cos \theta d\theta$
$=2a^2{\int }_0^{\pi /2}2{\cos }^2\theta d\theta =2a^2{\int }_0^{\pi /2}(1+\cos 2\theta )d\theta$
$=2a^2{[\theta -\frac{\sin \theta }{2}]}_0^{\pi /2}=2a^2\left(\frac{\pi }{2}\right)=\pi a^2$