Question

Find the equation of a normal to the curve y = x log_e x which is parallel to the line 2x − 2y + 3 = 0 ?

Answer 1

Slope of the given line is 1

$$\text{ Let }(x_1,y_1)\text{ be the point where the tangent is drawn to the curve }.$$

$$\text{ Since, the point lies on the curve }.$$

$$\text{ Hence },y_1=x_1{\log }_e{x}_1.........\left(1\right)$$

$$\text{ Now,}y=x{\log }_{e}x$$

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

$$\text{ Slope of tangent }=1+{\log }_e{x}_1$$

$$\text{ Slope of normal }=\frac{-1}{\text{ Slope of tangent }}=\frac{-1}{1+{\log }_e{x}_1}$$

$$\text{ Given that }$$

$$\text{ Slope of normal = slope of the given line }$$

$$\frac{-1}{1+{\log }_e{x}_1}=1$$

$$\Rightarrow -1=1+{\log }_e{x}_1$$

$$\Rightarrow -2={\log }_e{x}_1$$

$$\Rightarrow x_1=e^{-2}=\frac{1}{e^2}$$

$$\text{ Now },y_1=e^{-2}(-2)=\frac{-2}{e^2}............[\text{ From }(1)]$$

$$\therefore (x_1,y_1)=(\frac{1}{e^2},\frac{-2}{e^2})$$

$$\text{ Equation of normal is },$$

$$y+\frac{2}{e^2}=1(x-\frac{1}{e^2})$$

$$\Rightarrow y+\frac{2}{e^2}=x-\frac{1}{e^2}$$

$$\Rightarrow x-y=\frac{3}{e^2}$$

$$\Rightarrow x-y=3e^{-2}$$