question_answer

Area bounded by the curve \[x{{y}^{2}}={{a}^{2}}(a-x)\]and y-axis is

A) \[\pi {{a}^{2}}/2\] sq. units

B) \[\pi {{a}^{2}}\] sq. units

C) \[3\pi {{a}^{2}}\] sq. units

D) None of these

Correct Answer: B

Solution :

[b] \[x{{y}^{2}}={{a}^{2}}(a-x)\Rightarrow x=\frac{{{a}^{3}}}{{{y}^{2}}+{{a}^{2}}}\] The given curve is symmetrical about x-axis, and meets it at (a, 0). The line \[x=0\], i.e., y-axis is an asymptote. Area \[=\int\limits_{0}^{\infty }{x\,\,dy}=2\int\limits_{0}^{\infty }{\frac{{{a}^{3}}}{{{y}^{2}}+{{a}^{2}}}dx}\] \[=2{{a}^{2}}\frac{1}{a}\left[ {{\tan }^{-1}}\frac{y}{a} \right]_{0}^{\infty }\] \[=2{{a}^{2}}\frac{\pi }{2}=\pi {{a}^{2}}\] sq. units