The equation of the given curve is y = x3 + 2x + 6.
The slope of the tangent to the given curve at any point (x, y) is given by,
∴ Slope of the normal to the given curve at any point (x, y)
The equation of the given line is x + 14y + 4 = 0.
x + 14y + 4 = 0 ∴ (which is of the form y = mx + c)
If the normal is parallel to the line, then we must have the slope of the normal being equal to the slope of the line.
When x = 2, y = 8 + 4 + 6 = 18.
When x = −2, y = − 8 − 4 + 6 = −6.
Therefore, there are two normals to the given curve with slope -1/14 and passing through the points (2, 18) and (−2, −6). Thus, the equation of the normal through (2, 18) is given by,]
Hence, the equations of the normals to the given curve (which are parallel to the given line) are 
