Question

The area of the loop of the curve $a{y}^2=x^2(a-x)$ is

• A. $4a^2$ sq. units
• B. $\frac{8a^2}{15}$ sq. units
• C. $\frac{16a^2}{9}$ sq. units
• D. None of these

Answer 1

Answer: (B)

$a{y}^2=x^2\left(a-x\right)$ or $y=\pm x\sqrt{\frac{a-x}{a}}$
Curve tracing: $y=x\sqrt{\frac{a-x}{a}}$
We must have $x\le a$
For $0<x\le a,y>0$ and for $x<0,y<0$
Also $y=0$
$\Rightarrow x=0,a$
Curve is symmetrical about $x$-axis
when $x\to -\infty ,y\to -\infty$
Also, it can be verified that y has only one point of maxima for $0<x<a$

Area $=2\underset{0}{\overset{a}{\int }}x\sqrt{\frac{a-x}{a}}dx\sqrt{\frac{a-x}{a}}=t$
$\Rightarrow 1-\frac{x}{a}=t^2$
or $x=a\left(1-t^2\right)$
$\Rightarrow A=2\underset{1}{\overset{0}{\int }}a\left(1-t^2\right)t\left(-2at\right)dt$
$=4a^2\underset{0}{\overset{1}{\int }}\left(t^2-t^4\right)dt$
$=4a^2{\left[\frac{t^3}{3}-\frac{t^5}{5}\right]}_0^1$
$=4a^2\left[\frac{1}{3}-\frac{1}{5}\right]$
$=\frac{8a^2}{15}$ sq. units