Find the equation of the normal to the curve y=x ln x

Find the equation of the normal to the curve $y=x\ln x$ that is parallel to the line $2x-3y+3=0$.

Step 1: Slope of the given line

Rewriting, $y=\tfrac23x+1$, so $k_{\text{line}}=\dfrac23$.

Step 2: Required slope of the tangent

If the normal has slope $k_{n}=\dfrac23$, the tangent there has slope $k_{t}=-\dfrac1{k_n}=-\dfrac32$.

Step 3: Locate the point

Since $y'=\ln x+1$, setting $y'(x_{0})=-\tfrac32$ gives $\ln x_{0}=-\tfrac52$, so

\[ x_{0} = e^{-5/2}, \qquad y_{0} = x_{0}\ln x_{0} = -\frac52\,e^{-5/2}. \tag{1} \]

Step 4: Equation of the normal

At $\left(e^{-5/2},-\tfrac52e^{-5/2}\right)$ with slope $\tfrac23$, and combining the constants over denominator $6$ ($-\tfrac23-\tfrac52=-\tfrac{19}{6}$): \[y+\frac52e^{-5/2} = \frac23\left(x-e^{-5/2}\right) \;\Longrightarrow\; y = \frac23x - \frac{19}{6}e^{-5/2}. \tag{2}\]

Equivalently, clearing denominators: $4x-6y-19e^{-5/2}=0$.

Step 5: Verification

\[\text{Slope of the normal, } \tfrac23,\text{ equals the slope of } 2x-3y+3=0\ \checkmark\]

\[\left(e^{-5/2},-\tfrac52e^{-5/2}\right) \text{ on } y=x\ln x\text{: } e^{-5/2}\!\left(-\tfrac52\right)=-\tfrac52e^{-5/2}\ \checkmark\]