Find the equations of the normals to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Equation of the curve is  y= x³ + 2x + 6 

Slope of the normal at point  ( x, y ) = $\frac{-1}{\left({\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3{\mathrm{x}}^2 + 2$

on substitution, we get

$\mathrm{Slope} \mathrm{of} \mathrm{the} \mathrm{normal} = \frac{-1}{3{\mathrm{x}}^2 + 2} ..........\left(\mathrm{i}\right)$

Normal to the curve  is parallel to the line  x + 14y + 4 = 0,

$\mathrm{i}.\mathrm{e}. \mathrm{y} = -\frac{1}{14} \mathrm{x} - \frac{4}{14}$

So the slope of the line is the slope of the normal.

$$\begin{gathered} \mathrm{Slope} \mathrm{of} \mathrm{the} \mathrm{line} \mathrm{is} - \frac{1}{14} = \frac{-1}{3{\mathrm{x}}^2 + 2} \\ \Rightarrow 3{\mathrm{x}}^2 + 2 = 14 \\ \Rightarrow 3{\mathrm{x}}^2 = 12 \\ \Rightarrow {\mathrm{x}}^2 = 4 \\ \Rightarrow \mathrm{x} = \pm 2 \end{gathered}$$

When  x = 2,  y = 18  and when  x = -2,  y = -6

Therefore, there are two normals to the curve   y = x³ + 2x + 6.

Equation of  normal through point ( 2, 18 ) is given by:

$$\begin{gathered} \mathrm{y} - 18 = -\frac{1}{14} \left( \mathrm{x} - 2 \right) \\ \Rightarrow 14\mathrm{y} - 252 = -\mathrm{x} + 2 \\ \Rightarrow \mathrm{x} + 14\mathrm{y} - 254 = 0 \end{gathered}$$

Equation of normal through point ( -2, -6 ) is given by:

$$\begin{gathered} \mathrm{y} - \left( - 6 \right) = -\frac{1}{14} \left( \mathrm{x} - \left( - 2 \right)\right) \\ \Rightarrow 14\mathrm{y} + 84 = -\mathrm{x} - 2 \\ \Rightarrow \mathrm{x} + 14\mathrm{y} + 86 = 0 \end{gathered}$$

Therefore, the equation of normals to the curve are  x + 14y - 254 = 0  and x + 14y + 86 = 0.