Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
It is given that the equation of the normal to the curve y = x3 + 2x + 6
Then, the slope of the tangent to the given curve at any point (x, y) is given by:
Then, slope of normal to the given curve at any point (x,y)
=
Mediumy = 
⇒ Slope of the given line = 
We know that 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.
⇒ 
⇒ 3x2 + 2 = 14
⇒ 3x2 = 12
⇒ x2 = 4
⇒ 
So, when x = 2 then y = 18
and x -2 then y = -6
Hence, there are two normals to the given curve with the slope 
and passing through the points (2, 18) and (-2,-6).
Then, the equation of the normal through (2, 18) is:
y -18 = 
⇒ 14y -252 = -x +2
⇒ x +14y-254 = 0
And, the equation of the normal through (-2, -6) is:
y –(-6) = 
⇒ y + 6 = 
⇒ 14y +84 = -x -2
⇒ x +14y+86 = 0
Therefore, the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0 are x +14y-254 = 0 and x +14y+86 = 0.