question_answer

The area bounded by the parabola \[{{y}^{2}}=4ax\] and\[{{x}^{2}}=4ay\]is:

A) \[\frac{8{{a}^{2}}}{3}sq\,\,unit\]                              

B) \[\frac{16{{a}^{2}}}{3}sq\,\,unit\]

C) \[\frac{32{{a}^{2}}}{3}sq\,\,unit\]                            

D)  \[\frac{64{{a}^{2}}}{3}sq\,\,unit\]

Correct Answer: B

Solution :

Given equation of curves are                 \[{{y}^{2}}=4ax\]and\[{{x}^{2}}=4ay\] The point of intersection of above curves are\[(0,\,\,0)\]and\[(4a,\,\,4a)\]. \[\therefore \]Required area \[=\] Area of shaded curve                 \[=\int_{a}^{4a}{\left( \sqrt{4ax}-\frac{{{x}^{2}}}{4a} \right)dx}\]                 \[=\left[ \sqrt{4a}\frac{{{x}^{3/2}}}{3/2}-\frac{{{x}^{2}}}{12a} \right]_{0}^{4a}\]                 \[=\frac{32{{a}^{2}}}{3}-\frac{16{{a}^{2}}}{3}\]                 \[=\frac{16{{a}^{2}}}{2}sq\,\,unit\]