Question

The equation of the normal line to the curve $ y=x\,{{\log }_{e}}x $ parallel to $ 2x-2y+3=0 $ is

• A. $ x+y=3{{e}^{-2}} $
• B. $ x-y=6{{e}^{-2}} $
• C. $ x-y=3{{e}^{-2}} $
• D. $ x-y=6{{e}^{2}} $

Answer 1

Answer: (C)

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}} $