Question

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

Answer 1

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

Required area $=2\displaystyle \int _{0}^{a}a\sqrt{\frac{a - x}{x}}dx$
Put $x=asin^{2}\theta $ $\Rightarrow dx=a\left(2 sin \theta cos \theta \right)d\theta $
Also, $\theta =sin^{- 1}\sqrt{\frac{x}{a}},$
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\displaystyle \int _{0}^{\left(\pi \right)/2}a\sqrt{\frac{a - a \left(sin\right)^{2} \theta }{a \left(sin\right)^{2} \theta }}\left(2 a sin \theta cos \theta \right)d\theta $
$=2\displaystyle \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(2 a sin \theta cos \theta \right)d\theta $
Using, $1-sin^{2}\theta =cos^{2}\theta ,$ we get
Area $=2\displaystyle \int _{0}^{\left(\pi \right)/2}a\sqrt{\frac{\left(cos\right)^{2} \theta }{\left(sin\right)^{2} \theta }}\left(2 a sin \theta cos \theta \right)d\theta $
$=2\displaystyle \int _{0}^{\left(\pi \right)/2}a\frac{cos \theta }{sin \theta }\left(2 a sin \theta cos \theta \right)d\theta $
$=2a^{2}\displaystyle \int _{0}^{\pi /2}2cos^{2}\theta d\theta $
Now, by using $cos2\theta =2cos^{2}\theta -1,$ $\Rightarrow 1+cos2\theta =2cos^{2}\theta ,$ we have
Area $=2a^{2}\displaystyle \int _{0}^{\left(\pi \right)/2}\left(1 + cos 2 \theta \right)d\theta $
$=2a^{2}\left[\theta + \frac{sin 2 \theta }{2}\right]_{0}^{\pi /2}$
$=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
$k=1$