The equation of the normal to the curve $y=x \log x$, which is parallel to the line $2 x-2 y+3=0$, is

The equation of the normal to the curve $y=x \log x$, which is parallel to the line $2 x-2 y+3=0$, is
  1. $x+y=3 \mathrm{e}^{-2}$
  2. $\quad x-y=3 \mathrm{e}^{-2}$
  3. $x-y=3 \mathrm{e}^2$
  4. $x+y=3 \mathrm{e}^2$

Solution

$y=x \log x$
Differentiating w.r.to $x$, we get $\frac{\mathrm{d} y}{\mathrm{~d} x}=1+\log x$ $\therefore \quad$ Slope of tangent $=1+\log x$ Slope of given line $=1$ $\therefore \quad$ Line of normal is parallel to given line $\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=-1 \\ & \Rightarrow 1+\log x=-1 \\ & \Rightarrow \log x=-2 \\ & \Rightarrow x=\mathrm{e}^{-2} \end{aligned}$
Substituting $x=\mathrm{e}^{-2}$ in given equation, $\begin{aligned} y & =x \log x \\ & =\mathrm{e}^{-2} \log \mathrm{e}^{-2} \\ & =-2 \mathrm{e}^{-2} \end{aligned}$ $\therefore \quad$ Equation of line passing through $\left(\mathrm{e}^{-2},-2 \mathrm{e}^{-2}\right)$ is $\begin{aligned} & y+2 \mathrm{e}^{-2}=1\left(x-\mathrm{e}^{-2}\right) \\ & \Rightarrow x-y=3 \mathrm{e}^{-2} \end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 1)

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