Question

Find an equation of normal line to the curve y = x³ + 2x + 6 which is parallel to the line x + 14y + 4 = 0.

Answer 1

Let (x₁, y₁) be a point on the curve where we need to find the normal.
Slope of the given line = $\frac{-1}{14}$
$$\begin{gathered} \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},y_1={x_1}^3+2x_1+6 \\ \mathrm{Now},y=x^3+2x+6 \\ \Rightarrow \frac{dy}{dx}=3x^2+2 \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=3}{{\mathit{\text{x}}}_1}^2\text{+2} \\ \text{Slope of the normal=}\frac{-1}{\text{slope of the tangent}}\text{=}=\frac{-1}{3{x_1}^2+2} \\ \text{Given that,} \\ \text{slope of the normal=slope of the given line} \\ \Rightarrow \frac{-1}{3{x_1}^2+2}=\frac{-1}{14} \\ \Rightarrow 3{x_1}^2+2=14 \\ \Rightarrow 3{x_1}^2=12 \\ \Rightarrow {x_1}^2=4 \\ \Rightarrow x_1=\pm 2 \\ \text{Case-1:}{x}_1=2 \\ {y}_1={x_1}^3+2x_1+6=8+4+6=18 \\ \therefore \left(x_1,y_1\right)=\left(2,18\right) \\ \text{Slope of the normal,}\mathit{\text{m}}\text{=}\frac{-1}{14} \\ \text{Equation of normal is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-18=\frac{-1}{14}\left(x-2\right) \\ \Rightarrow 14y-252=-x+2 \\ \Rightarrow x+14y-254=0 \\ \text{Case-2:}{x}_1=-2 \\ {y}_1={x_1}^3+2x_1+6=-8-4+6=-6 \\ \therefore \left(x_1,y_1\right)=\left(-2,-6\right) \\ \text{Slope of the normal,}\mathit{\text{m}}\text{=}\frac{-1}{14} \\ \text{Equation of normal is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y+6=\frac{-1}{14}\left(x+2\right) \\ \Rightarrow 14y+84=-x-2 \\ \Rightarrow x+14y+86=0 \end{gathered}$$