question_answer

The area enclosed between the curves \[y=a{{x}^{2}}\] and \[x=a{{y}^{2}}(a>0)\] is 1 sq. unit, then the value of a is

A) \[\frac{1}{\sqrt{3}}\]

B) \[\frac{1}{2}\]

C) 1

D) \[\frac{1}{3}\]   

Correct Answer: A

Solution :

[a] \[y=a{{x}^{2}}\And x=a{{y}^{2}}\] Points of intersection are O (0, 0) & \[A\left( \frac{1}{a},\frac{1}{a} \right)\] \[\therefore Area=\int\limits_{0}^{1/a}{\left( \sqrt{\frac{x}{a}}-a{{x}^{2}} \right)dx}\] \[=\frac{2}{3{{a}^{2}}}-\frac{1}{3{{a}^{2}}}=\frac{1}{3{{a}^{2}}}=1\Rightarrow a=\frac{1}{\sqrt{3}}\]