question_answer

If the area enclosed between the curves \[y=k{{x}^{2}}\] and \[x=k{{y}^{2}},\] (k > 0) is 1 square unit. Then k is:

A) \[\frac{\sqrt{3}}{2}\]                             

B)  \[\frac{1}{\sqrt{3}}\]

C) \[\sqrt{3}\]                                

D) \[\frac{2}{\sqrt{3}}\]

Correct Answer: B

Solution :

\[y=k{{x}^{2}},x=k{{y}^{2}}\]

\[\Rightarrow \]   \[x=k({{k}^{2}}{{x}^{4}})\]

\[\Rightarrow \]   \[x=0\]or\[{{x}^{3}}={{\left( \frac{1}{k} \right)}^{3}}\]

\[\Rightarrow \]   \[x=\frac{1}{k},0\]

Point of intersection are\[\left( \frac{1}{k},\frac{1}{k} \right)\]and\[(0,0)\]

\[Area=\int\limits_{0}^{1/k}{\left( \sqrt{\frac{x}{k}}-k{{x}^{1}} \right)}dx=1\]

\[\Rightarrow \]   \[{{\left( \frac{1}{\sqrt{k}}\frac{{{x}^{3/2}}}{3/2}-\frac{k{{x}^{3}}}{3} \right)}^{1/k}}=1\]

\[\Rightarrow \]   \[\frac{2}{3{{k}^{2}}}=1\]

\[\Rightarrow \]   \[{{k}^{2}}=\frac{1}{3}\]

\[k=\frac{1}{\sqrt{3}}.\]