Question

If area bounded by the curve $x{y}^2=a^2\left(a-x\right)$ and the $y$ -axis is $k\pi a^2$ then find $k$ .

Answer 1

Answer: 1

$y^2=a^2\frac{\left(a-x\right)}{x}$
$y=\pm a\sqrt{\frac{a-x}{x}}$

Required area $=2{\int }_0^{a}a\sqrt{\frac{a-x}{x}}dx$
Put $x=asi{n}^2\theta$ $\Rightarrow dx=a\left(2sin\theta cos\theta \right)d\theta$
Also, $\theta =si{n}^{-1}\sqrt{\frac{x}{a}},$
Thus when, $x=0\Rightarrow \theta ={\left(sin\right)}^{-1}\left(0\right)=0$ and when, $x=a\Rightarrow \theta ={\left(sin\right)}^{-1}\left(\frac{a}{a}\right)=\frac{\pi }{2}$
Area $=2{\int }_0^{\left(\pi \right)/2}a\sqrt{\frac{a-a{\left(sin\right)}^2\theta }{a{\left(sin\right)}^2\theta }}\left(2asin\theta cos\theta \right)d\theta$
$=2{\int }_0^{\left(\pi \right)/2}a\sqrt{\frac{a\left(1-{\left(sin\right)}^2\theta \right)}{a{\left(sin\right)}^2\theta }}\left(2asin\theta cos\theta \right)d\theta$
Using, $1-si{n}^2\theta =co{s}^2\theta ,$ we get
Area $=2{\int }_0^{\left(\pi \right)/2}a\sqrt{\frac{{\left(cos\right)}^2\theta }{{\left(sin\right)}^2\theta }}\left(2asin\theta cos\theta \right)d\theta$
$=2{\int }_0^{\left(\pi \right)/2}a\frac{cos\theta }{sin\theta }\left(2asin\theta cos\theta \right)d\theta$
$=2a^2{\int }_0^{\pi /2}2co{s}^2\theta d\theta$
Now, by using $cos2\theta =2co{s}^2\theta -1,$ $\Rightarrow 1+cos2\theta =2co{s}^2\theta ,$ we have
Area $=2a^2{\int }_0^{\left(\pi \right)/2}\left(1+cos2\theta \right)d\theta$
$=2a^2{\left[\theta +\frac{sin2\theta }{2}\right]}_0^{\pi /2}$
$=2a^2\left[\frac{\pi }{2}+\frac{sin\pi }{2}-0\right]=a^2\pi$
Given area is $k\pi a^2,$ hence, on comparing, we get
$k=1$