A) \[x+y=3{{e}^{-2}}\]
B) \[x-y=6{{e}^{-2}}\]
C) \[x-y=3{{e}^{-2}}\]
D) \[x-y=6{{e}^{2}}\]
Correct Answer: C
Solution :
Given curve is \[y=x\,{{\log }_{e}}\,\,x\Rightarrow \frac{dy}{dx}=\frac{x}{x}+\log x=1+\log x\] \[\therefore \] Slope of normal \[=\frac{-1}{\frac{dy}{dx}}=\frac{-1}{1+\log x}\] and given line is \[2x-2y+3=0\] On differentiating w.r.t.x, we get \[2-2\frac{dy}{dx}=0\,\,\,\,\,\,\,\,\Rightarrow \,\,\frac{dy}{dx}=1\] Since, normal line is parallel to the given line. \[\therefore \] Slopes are equal ie, \[\frac{-1}{1+\log x}=1\] \[\Rightarrow \] \[{{\log }_{e}}x=-2\] \[\Rightarrow \] \[x={{e}^{-2}}\] Now, intersecting point of given curve and \[x={{e}^{-2}}\]is \[({{e}^{-2}},\,-2{{e}^{-2}}).\] \[\therefore \] Required equation of he line is \[y+2{{e}^{-2}}=1(x-{{e}^{-2}})\] \[\Rightarrow \] \[x-y=3{{e}^{-2}}\]
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