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

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

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

\[\text { Hence }, y_1 = x_1 \log_e x_1 ......... \left( 1 \right)\]

\[\text { Now,} y = x \log_e x \]

\[ \Rightarrow \frac{dy}{dx} = x \times \frac{1}{x} + \log_e x \left( 1 \right) = 1 + \log_e x\]

\[\text { Slope of tangent }=1 + \log_e x_1 \]

\[\text { Slope of normal } =\frac{- 1}{\text { Slope of tangent }}=\frac{- 1}{1 + \log_e x_1}\]

\[\text { Given that }\]

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

\[\frac{- 1}{1 + \log_e x_1} = 1\]

\[ \Rightarrow - 1 = 1 + \log_e x_1 \]

\[ \Rightarrow - 2 = \log_e x_1 \]

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

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

\[ \therefore \left( x_1 , y_1 \right) = \left( \frac{1}{e^2}, \frac{- 2}{e^2} \right)\]

\[\text { Equation of normal is },\]

\[y + \frac{2}{e^2} = 1 \left( x - \frac{1}{e^2} \right)\]

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

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

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