question_answer

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

A) \[4{{a}^{2}}\]sq. units 

B) \[\frac{8{{a}^{2}}}{15}\]sq. units

C) \[\frac{16{{a}^{2}}}{9}\]sq. units

D) None of these

Correct Answer: B

Solution :

[b] \[a{{y}^{2}}={{x}^{2}}(a-x)\] 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 \[x<0,y<0\] 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 point of maxima for \[0<x<a.\] Area \[=2\int\limits_{0}^{a}{x\sqrt{\frac{a-x}{a}}dx\sqrt{\frac{a-x}{a}}=t}\] \[\Rightarrow \] \[1-\frac{x}{a}={{t}^{2}}\]  or  \[x=a\,(1-{{t}^{2}})\] \[\Rightarrow A=2\int\limits_{1}^{0}{a\,(1-{{t}^{2}})\,t(-2at)\,dt}\] \[=4{{a}^{2}}\int\limits_{0}^{1}{({{t}^{2}}-{{t}^{4}})dt}\] \[=4{{a}^{2}}{{\left[ \frac{{{t}^{3}}}{3}=\frac{{{t}^{5}}}{5} \right]}^{1}}_{0}\] \[=4{{a}^{2}}\left[ \frac{1}{3}-\frac{1}{5} \right]\] \[=\frac{8{{a}^{2}}}{15}\] sq. units.