Equation of the curve is y = $x^3$ + 2x + 6 Slope of the normal at point (x, y) = $\frac{-1}{\frac{dy}{dx}}$ $\frac{dy}{dx}$ = $3x^2$ + 2 On substitution, we get Slope of the normal = $\frac{-1}{3x^2+2}$ ... (1) Normal to the curve is parallel to the line x + 14y + 4 = 0, i.e. y = - $\frac{1}{14} \times \frac{-4}{14}$ So the slope of the line is the slope of the normal. Slope of the line is - $\frac{1}{14}$ = $\frac{-1}{3x^2+2}$ ⇒ $3x^2$ + 2 = 14 ⇒ $3x^2$ = 12 ⇒ $x^2$ = 4 ⇒ x = ± 2 When x = 2, y = 18 and when x = - 2, y = - 6 Therefore, there are two normals to the curve y = $x^3$ + 2x + 6. Equation of normal through point (2, 18) is given by: y - 18 = $\frac{-1}{14}$ (x - 2) ⇒ 14y - 252 = - x + 2 ⇒ x + 14y - 254 = 0 Equation of normal through point (-2,-6) is given by: y - (- 6) = $\frac{-1}{14}$ (x - (- 2)) ⇒ 14y + 84 = - x - 2 ⇒ x + 14y + 86 = 0 Therefore, the equation of normals to the curve are x + 14y – 254 = 0 and x + 14y + 86 = 0