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
$$\begin{gathered} \text{Let}\left(x_1,y_1\right)\text{be the point where the tangent is drawn to the curve.} \\ \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},y_1=x_1{\log }_e{x}_1...\left(1\right) \\ \mathrm{Now},y=x{\log }_{e}x \\ \Rightarrow \frac{dy}{dx}=x\times \frac{1}{x}+{\log }_{e}x\left(1\right)=1+{\log }_{e}x \\ \text{Slope of tangent=}1+{\log }_e{x}_1 \\ \text{Slope of normal =}\frac{-1}{\text{Slope of tangent}}\text{=}\frac{-1}{1+{\log }_e{x}_1} \\ \text{Given that} \\ \text{Slope of normal = slope of the given line} \\ \frac{-1}{1+{\log }_e{x}_1}=1 \\ \Rightarrow -1=1+{\log }_e{x}_1 \\ \Rightarrow -2={\log }_e{x}_1 \\ \Rightarrow x_1=e^{-2}=\frac{1}{e^2} \\ \mathrm{Now},y_1=e^{-2}\left(-2\right)=\frac{-2}{e^2}\left[\text{From (1)}\right] \\ \therefore \left(x_1,y_1\right)=\left(\frac{1}{e^2},\frac{-2}{e^2}\right) \\ \text{Equation of normal is,} \\ y+\frac{2}{e^2}=1\left(x-\frac{1}{e^2}\right) \\ \Rightarrow y+\frac{2}{e^2}=x-\frac{1}{e^2} \\ \Rightarrow x-y=\frac{3}{e^2} \\ \Rightarrow x-y=3e^{-2} \end{gathered}$$