A) \[(1,\,\,1)\]
B) \[(4,\,\,1/2)\]
C) \[(5/4,\,\,1)\]
D) \[(3/4,\,\,5/2)\]
Correct Answer: D
Solution :
Key Idea: If \[({{x}_{1}},\,\,{{y}_{1}})\] is a point on the curve, then the nearest distance in which \[\frac{dy}{d{{x}_{({{x}_{1}},\,\,{{y}_{1}})}}}=0\] Let \[P(x,\,\,y)\] be a point on the curve. Given curve is \[x\,\,{{y}^{2}}=1\] ... (i) Again, let d be the distance from origin to the curve \[\therefore \] \[{{d}^{2}}={{x}^{2}}+{{y}^{2}}\] \[\Rightarrow \] \[D={{x}^{2}}+\frac{1}{x}\] (let\[{{d}^{2}}=D)\] On differentiating w.r.t.\[x,\] we get \[\frac{dD}{dx}=2x-\frac{1}{{{x}^{2}}}\] For maxima or minima, put\[\frac{dD}{dx}=0\] \[\Rightarrow \] \[2x-\frac{1}{{{x}^{2}}}=0\] \[\Rightarrow \] \[{{x}^{3}}=\frac{1}{2}\Rightarrow x={{\left( \frac{1}{2} \right)}^{1/3}}\] Again differentiating Eq. (ii), we get \[\frac{{{d}^{2}}D}{d{{x}^{2}}}=2+\frac{2}{{{x}^{3}}}\] At\[x={{\left( \frac{1}{2} \right)}^{1/3}},\,\,\frac{{{d}^{2}}D}{d{{x}^{2}}}>0\]\[i.e.,\] minima. On putting the value of \[x\] in Eq. (i), we get \[{{y}^{2}}=\frac{1}{{{\left( \frac{1}{2} \right)}^{1/3}}}\] \[\Rightarrow \] \[y={{(2)}^{1/6}}\] \[\therefore \]Required point is\[\left( {{\left( \frac{1}{2} \right)}^{1/3}},\,\,{{(2)}^{1/6}} \right)\]
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