Find the equation of normal to the curve $y=x^3-3 x$, which is parallel to the line $2 x+18 y=9$ ?

Given, line is $2 x+18 y=9$ $$ \begin{gathered} \text { slope }(m)=\frac{-a}{b} \\ m=\frac{-2}{18}=-\frac{1}{9} \end{gathered} $$ Given equation of curve is $$ y=x^3-3 x $$ Differentiate w.r.to ' $x$ ', we get $$ \frac{d y}{d x}=3 x^2-3 $$ $\therefore$ Slope of normal $=\frac{-1}{\frac{d y}{d x}}=\frac{-1}{3 x^2-3}$ Since, normal is parallel to given line $\therefore$ Slope of line $=$ slope of normal $$ \begin{aligned} \frac{-1}{9} & =\frac{-1}{3\left(x^2-1\right)} \\ \Rightarrow \quad x^2-1 & =3 \Rightarrow x= \pm 2 \end{aligned} $$ Case (i) if $x=2$ from Eq. (i) $$ \begin{aligned} & y=(2)^3-3(2) \\ \therefore \quad y & =2 \\ \quad P & =(2,2) \end{aligned} $$ Equation of normal is $y-y_1=m\left(x-x_1\right)$ $$ \begin{aligned} y-2 & =\frac{-1}{9}(x-2) \\ x+9 y & =20 \end{aligned} $$ Case (ii) If $x=-2$ $$ \begin{aligned} y & =(-2)^3-3(-2)=-2 \\ \therefore \quad Q & =(-2,-2) \end{aligned} $$ Equation of Normal is $y-y_1=m\left(x-x_1\right)$ $$ \begin{aligned} y+2 & =\frac{-1}{9}(x+2) \\ x+9 y & =-20 \end{aligned} $$ Required Equation of normal is $x+9 y= \pm 20$ Hence, option (3) is correct.