Find the equations of the normal to the curve `y=x^3+2x+6` which are parallel to the line `x+14 y+4=0.`

Explanation

Given equation of curve is
`y=x^3+2x+6`
and the given equation of line is
`x+14y+4=0`
On differentiating both sides
`dy/dx=3x^2+2`
Slope of normal `=-1/(dy/dx)=-1/(3x^2+2)`
and slope of line is `x+14y+4=0` is `-1/14`
We know that if two lines are parallel, then their slopes are equal.
`-1/(3x^2+2)=-1/14`
`3x^2+2=14`
`x=pm2`
When `x=2`then `y=(2)^3+2(2)+6=8+4+6=18`
and when `x=-2`
`y=(-2)^3+2(-2)+6=-6`
Normal passes through `(2,18)` and `(-2,-6)`
and slope of normal is `-1/14`
Hence, equation of normal at a point `(-2,-6)`
`y+6=-1/14(x+2)`
`14y+84=-x-2`
`x+14y=-86`
Hence, the two equations of a normal are `x+14y=254` and `x+14y=-86`