Question

Coordinates of a point on the curve $y=x\log \left(x\right)$ at which the normal is parallel to the line $2x-2y=3$ are

• A. $\left(0,0\right)$
• B. $\left(e,e\right)$
• C. $\left(e^2,2e^2\right)$
• D. $\left(e^{-2},-2e^{-2}\right)$

Answer 1

The correct option is D

$\left(e^{-2},-2e^{-2}\right)$

Explanation for correct option:

Step-1: Finding slop of normal to the curve.

Given, curve $y=x\log \left(x\right)$

Differentiate with respect to $x$.

$\frac{dy}{dx}=x\times \frac{1}{x}+1\times \log \left(x\right)$

$\Rightarrow$$\frac{dy}{dx}=\left(1+\log \left(x\right)\right)$

Slop of the curve $\left(1+\log \left(x\right)\right)$.

Therefore, slop of normal to the curve is

$\left(-\frac{1}{\left({\displaystyle \frac{dy}{dx}}\right)}\right)=\left(-\frac{1}{1+\log \left(x\right)}\right)$

Step-2: Finding slop of the given curve.

Given, equation of line $2x-2y=3$

Slope of the line is $-\frac{2}{\left(-2\right)}=1$

Step-3: Equate the slope of normal to the curve.

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

$\Rightarrow$$\log \left(x\right)=-2$

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

put value of $x$in equation of curve,

$\Rightarrow$$y=e^{-2}\log \left(e^{-2}\right)$

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

Therefore, required point is $\left(e^{-2},-2e^{-2}\right)$.

Hence, correct answer is option $\left(D\right)$.