find the equation of a normal to the curve y=xlogx which is parallel to the line 2x-2y+3=0
 

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$$\begin{gathered} \mathrm{Slope}\left(\mathrm{m}\right) \mathrm{of} \mathrm{line} 2\mathrm{x}-2\mathrm{y}+3=-\frac{\mathrm{Coefficient} \mathrm{of} \mathrm{x}}{\mathrm{Coefficient} \mathrm{of} \mathrm{y}}=-\frac{2}{\left(-2\right)}=1 \\ \mathrm{As} \mathrm{normal} \mathrm{is} \mathrm{parallel} \mathrm{to} \mathrm{that} \mathrm{line}, \mathrm{hence} \mathrm{slope} \mathrm{wof} \mathrm{normal} \mathrm{will} \mathrm{also} \mathrm{be} \mathrm{m}=1 \\ \mathrm{y}=\mathrm{x}.\mathrm{logx} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}.\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{logx}\right)+\left(\mathrm{logx}\right).\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\right) \\ =\mathrm{x}.\frac{1}{\mathrm{x}}+\left(\mathrm{logx}\right).1 \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1+\log \left(\mathrm{x}\right) \\ \Rightarrow {\frac{\mathrm{dy}}{\mathrm{dx}}}_{\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)}=1+\log \left({\mathrm{x}}_1\right) \\ \mathrm{Slope} \mathrm{of} \mathrm{normal}=-\frac{1}{{{\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}}_{\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)}}=-\frac{1}{1+\log \left({\mathrm{x}}_1\right)} \mathrm{which} \mathrm{should} \mathrm{equal} \mathrm{to}1. \\ \Rightarrow -\frac{1}{1+\log \left({\mathrm{x}}_1\right)}=1 \\ \Rightarrow 1+\log \left({\mathrm{x}}_1\right)=-1 \\ \Rightarrow \log \left({\mathrm{x}}_1\right)=-2 \\ \Rightarrow {\mathrm{x}}_1={\mathrm{e}}^{-2}=\frac{1}{{\mathrm{e}}^2} \\ {\mathrm{y}}_1={\mathrm{x}}_1\log \left({\mathrm{x}}_1\right)=\left(\frac{1}{{\mathrm{e}}^2}\right)\left(-2\right)=-\frac{2}{{\mathrm{e}}^2} \\ \mathrm{Equation} \mathrm{of} \mathrm{normal} \mathrm{will} \mathrm{be}: \\ \mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right) \\ \Rightarrow \mathrm{y}-\left(-\frac{2}{{\mathrm{e}}^2}\right)=1\left(\mathrm{x}-\frac{1}{{\mathrm{e}}^2}\right) \\ \Rightarrow \mathrm{y}+\frac{2}{{\mathrm{e}}^2}=\mathrm{x}-\frac{1}{{\mathrm{e}}^2} \\ \Rightarrow \mathrm{x}-\mathrm{y}=\frac{2}{{\mathrm{e}}^2}+\frac{1}{{\mathrm{e}}^2} \\ \Rightarrow \mathbf{x}\mathbf{-}\mathbf{y}\mathbf{=}\frac{\mathbf{3}}{{\mathbf{e}}^{\mathbf{2}}} \end{gathered}$$

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