Question

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

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

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 $(e^{-2},-2e^{-2}).$
$\therefore$ Required equation of he line is $<br>y+2e^{-2}=1(x-e^{-2})$
$\Rightarrow$ $x-y=3e^{-2}$