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

Equation of given curve is
$y = x^3 + 2x + 6$
Parellel to line $x + 14y + 4 = 0 \Rightarrow y = \frac{-x}{14} -\frac{4}{14}$ means slope of normal and line is equal
We know that, equation of line
y= mx + c
on comparing it with  our given equation. we get,
$m = \frac{-1}{14}$
Slope of tangent = $\frac{dy}{dx} = 3x^2+2$
We know that
$Slope \ of \ normal = \frac{-1}{Slope \ of \ tangent} = \frac{-1}{3x^2+2}$
$\frac{-1}{3x^2+2} = \frac{-1}{14}$
$3x^2+2 = 14\\ 3x^2 = 12 \\ x^2 = 4\\ x = \pm 2$
Now, when x = 2,  $y = (2)^3 + 2(2) + 6 = 8+4+6 =18$
and
When x = -2 , $y = (-2)^3 + 2(-2) + 6 = -8-4+6 =-6$
Hence, the coordinates are (2,18) and (-2,-6)
Now, the equation of at point (2,18) with slope $\frac{-1}{14}$
$y-y_1=m(x-x_1)\\ y-18=\frac{-1}{14}(x-2)\\ 14y - 252 = -x + 2\\ x+14y = 254$
Similarly,  the equation of at point (-2,-6) with slope $\frac{-1}{14}$

$y-y_1=m(x-x_1)\\ y-(-6)=\frac{-1}{14}(x-(-2))\\ 14y + 84 = -x - 2\\ x+14y + 86= 0$
Hence, the equation of the normals to the curve$y = x^3 + 2x + 6$ which are parallel
to the line $x + 14y + 4 = 0.$

are x +14y - 254 = 0  and  x + 14y +86 = 0

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