A) (1, 1)
B) (1/4, 2)
C) \[({{2}^{1/6}},{{2}^{-1/3}})\]
D) \[({{2}^{-1/3}},{{2}^{1/6}})\]
E) None of the above
Correct Answer: D
Solution :
Let\[({{t}^{2}},1/t)\] be a point on the curve. If its distance from origin is\[\lambda ,\]then \[{{\lambda }^{2}}={{t}^{4}}+\frac{1}{{{t}^{2}}}\] \[\Rightarrow \] \[\frac{d}{dt}({{\lambda }^{2}})=4{{t}^{3}}-\frac{2}{{{t}^{3}}}\] \[\Rightarrow \] \[\frac{{{d}^{2}}}{d{{t}^{2}}}({{\lambda }^{2}})=12{{t}^{3}}+\frac{6}{{{t}^{4}}}>0\] Now, put\[\frac{d}{dt}({{\lambda }^{2}})=0\Rightarrow t={{2}^{-1/6}}\] Hence, required point is\[({{2}^{-1/3}},{{2}^{1/6}})\].
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