question_answer

The area between the parabola \[{{y}^{2}}=4ax\]and \[{{x}^{2}}=8ay\] is                                                               [RPET 1997]

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

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

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

D)            \[\frac{16}{3}{{a}^{2}}\]

Correct Answer: C

Solution :

 Required area \[A=\int_{0}^{(a{{2}^{8/3}})}{\sqrt{4ax}}dx-\int_{0}^{a{{2}^{8/3}}}{\frac{{{x}^{2}}}{8a}}\,dx=\frac{32{{a}^{2}}}{3}\]