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

We have
$y = {x^3} + 2x + 6$ …(i)

The given line is $14y + x + 4 = 0$ …(ii)

The slope of line (ii) is $- \cfrac{1}{{14}}$.

Differentiating (i) w.r.t. $x$, we get $\cfrac{{dy}}{{dx}} = 3{x^2} + 2$

Let $P({x_1},\;{y_1})$ be a point on (i).

Therefore the slope of tangent at $P({x_1},{y_1})$ to (i) is

${\left( {\cfrac{{dy}}{{dx}}} \right)_{({x_1},{y_1})}} = 3{x_1}^2 + 2$

$\Rightarrow$ The slope of normal at $P({x_1},{y_1})$ to (i) is $- \cfrac{1}{{3x_1^2 + 2}}$

Since normal at $P({x_1},{y_1})$ to (i) is parallel to the line (ii), we get $- \cfrac{1}{{3x_1^2 + 2}} = - \cfrac{1}{{14}} \Rightarrow 3{x_1}^2 + 2 = 14$
$\Rightarrow 3{x_1}^2 = 12 \Rightarrow x_1^2 = 4 \Rightarrow {x_1} = \pm \,2$

As $P({x_1},{y_1})$ lies on the curve (i) , we get ${y_1} = {x_1}^3 + 2{x_1} + 6$and so when ${x_1} = 2,{y_1} = {2^3} + 2 \cdot 2 + 6 = 18$ and
when ${x_1} = - 2,{y_1} = {( - 2)^3} + 2 \cdot ( - 2) + 6 = - 6$

Thus, there are two points $(2,\;18)$ and $( - 2,\; - 6)$ on (i) at which the normals are parallel to (ii).

Therefore, the equations of the required normals are
$y - 18 = - \cfrac{1}{{14}}(x - 2)$ and

$y + 6 = - \cfrac{1}{{14}}(x + 2)$

or, $14y - 252 = - x + 2$ and $14y + 84 = - x - 2$

or, $x + 14y - 254 = 0$and $x + 14y + 86 = 0$