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

The correct option is C

$x–y=3e^{-2}$

Explanation of the correct option.

Step $1$: Find the points from where the given line passes.

Given: $2x-2y+3=0$

It can be written as $y=x+\frac{3}{2}$

Slope of the line is $1$.

Line of normal is parallel to given line.

Therefore, slope of the tangent is $-1$.

$\frac{dy}{dx}=-1$

$\Rightarrow$$x\left(\frac{1}{x}\right)+{\log }_{e}x=-1$

$\Rightarrow$ ${\log }_{e}x=-2$

$\Rightarrow$ $x=e^{-2}$

Substituting the value of $x$ in the given equation of cure,

$y=e^{-2}{\log }_e{e}^{-2}$

$\Rightarrow$$y=-2e^{-2}$

Now, we have the points from where the line will pass.

Step $2$: Find the equation of the line.

We know that equation of line passing through point $\left(x_1,y_1\right)$ is given by

$y-y_1=m(x-x_1)$

Therefore the line passing through $\left(e^{-2},-2e^{-2}\right)$ is,

$$\begin{gathered} y+2e^{-2}=1(x-e^{-2}) \\ \Rightarrow x-y=3e^{-2} \end{gathered}$$ [ $m=1$, because the line is parallel to $2x-2y+3=0$]

Hence, Option $\left(\mathrm{C}\right)$ is the correct answer.