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

Explanation

Equation of curve
` y=x^(3)+2x+6` …(1)
` rArr (dy)/(dx) = 3x^(2)+2`
` :." Slope of tangent at point "(x, y) = 3x^(2)+2`
` rArr" Slope of normal at point "(x, y) = (-1)/(3x^(2)+2)`
Equation of given line is ` x+ 14y+4= 0`
` rArr y = - 1/14 x - 4/14`
Slope of this line = `-1/14`
Slope of parallel lines are equal.
` :. (-1)/(3x^(2)+2=14)=-1/14`
` rArr 3x^(2) + 2 = 14 rArr 3x^(2) = 12`
` rArr x^(2) = 4 rArr x= pm 2`
put x = 2 in equation (1)
` y = 2^(3) + 2 xx2 + 6 = 18`
Equation of normal are point (2, 18)
`y - 18 = - 1/14 (x - 2)`
` rArr 14y - 252 = - x+ 2`
` rArr x + 14y - 254 = 0`
put x =- 2 in equation (1)
` y = (-2)^(3) + 2 (-2) + 6`
` = - 8- 4+6 = - 6`
Equation of normal at point (-2, -6)
`y+6=-1/14(x+2)`
` rArr 14y + 84 = - x - 2`
`rArr x + 14y + 86 = 0`