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\[(0,0)\] |
\[Area=\int\limits_{0}^{1/k}{\left( \sqrt{\frac{x}{k}}-k{{x}^{1}} \right)}dx=1\] |
\[\Rightarrow \] \[{{\left( \frac{1}{\sqrt{k}}\frac{{{x}^{3/2}}}{3/2}-\frac{k{{x}^{3}}}{3} \right)}^{1/k}}=1\] |
\[\Rightarrow \] \[\frac{2}{3{{k}^{2}}}=1\] |
\[\Rightarrow \] \[{{k}^{2}}=\frac{1}{3}\] |
\[k=\frac{1}{\sqrt{3}}.\] |
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