# 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.

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