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

$x+14y−254=0$ and $x+14y+86=0$

The correct answer is Option (1) → $x+14y−254=0$ and $x+14y+86=0$

The given curve is $y = x^3 + 2x+6$   ...(i)

Let $P(x_1,y_1)$ be a point on the curve (i) at which the normal is parallel to the line

$x + 14y+4=0$   ...(ii)

Slope of the line (ii) = $-\frac{1}{14}$

Diff. (i) w.r.t. x, we get

$\frac{dy}{dx}= 3x^2 + 2,$

∴ the slope of the tangent to curve (i) at $P(x_1,y_1)=3x_1^2+2$

⇒ the slope of the normal to curve (i) at $P =-\frac{1}{3x_1^2 +2}$.

Since the normal to curve (i) at P is parallel to line (ii), we get

$-\frac{1}{3x_1^2 +2}=-\frac{1}{14}⇒3x_1^2+2=14$

$⇒3x_1^2=12⇒x_1^2=4⇒x_1=2,-2$.

As $P(x_1,y_1)$ lies on the curve (i), we get $y_1 = x_1^3 + 2x_1 +6$, therefore,

when $x_1 = 2, y_1 = 8+ 4 + 6 = 18$ and

when $x_1 = -2, y_1 =-8-4+6=-6$.

Thus, there are two points (2, 18) and (-2, -6) on the curve (i) the normals at which are parallel to the line (ii), therefore, the equations of the required normals are

$y-18=-\frac{1}{14}(x-2)$ and $y-(-6)=-\frac{1}{14}(x-(-2))$

i.e. $14y-252 = -x + 2$ and $14y+84 = -x-2$

i.e. $x + 14y - 254 = 0$ and $x + 14y+ 86 = 0$.