question_answer

Area included between \[y=\frac{{{x}^{2}}}{4a}\] and\[y=\frac{8{{a}^{3}}}{{{x}^{2}}+4{{a}^{2}}}\] is

A) \[\frac{{{a}^{2}}}{3}(6\pi -4)\]                 

B) \[\frac{{{a}^{2}}}{3}(4\pi +3)\]

C) \[\frac{{{a}^{2}}}{3}(8\pi +3)\]               

D)  none of these

Correct Answer: A

Solution :

 The curve of \[y({{x}^{2}}+4{{a}^{2}})=8{{a}^{3}}\] is symmetrical about y-axis and cuts it at\[A(0,\,\,2a)\]. Tangent at \[A\] is parallel to \[x-\]axis. \[x-\]axis is asymptote. This curve meets \[{{x}^{2}}=4ay\] Where\[\frac{{{x}^{2}}}{4a}=\frac{8{{a}^{3}}}{{{x}^{2}}+4{{a}^{2}}}\Rightarrow {{x}^{4}}+4{{a}^{2}}{{x}^{2}}-32{{a}^{4}}=0\] \[\Rightarrow \]\[({{x}^{2}}-4{{a}^{2}})({{x}^{2}}+8{{a}^{2}})=0\Rightarrow x=\pm 2a\] \[\therefore \]Required area \[=2\left[ \int_{0}^{2a}{\frac{8{{a}^{3}}}{{{x}^{2}}+4{{a}^{2}}}dx-\int_{0}^{2a}{\frac{{{x}^{2}}}{4a}dx}} \right]\] \[=\frac{{{a}^{2}}}{3}(6\pi -4)\].