Equation of the curve is y = x3 + 2x + 6
Slope of the normal at point (x, y) =

−1

( dy dx )

dy

dx

= 3x2 + 2
On substitution, we get
Slope of the normal =

−1

3x2+2

... (1)
Normal to the curve is parallel to the line x + 14y + 4 = 0,
i.e. y = -

1

14

×−

4

14

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

1

14

=

−1

3x2+2

⇒ 3x2 + 2 = 14
⇒ 3x2 = 12
⇒ x2 = 4
⇒ x = ± 2
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:
y - 18 = −

1

14

(x - 2)
⇒ 14y - 252 = - x + 2
⇒ x + 14y - 254 = 0
Equation of normal through point (-2,-6) is given by:
y - (- 6) = −

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