Question

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

Answer 1

Given equation of curve is

$y=x^3+2x+6$ .....(1)

Differentiating w.r.t $x$, we get

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

Let $P(x_1,y_1)$ be any point on the curve.
Slope of the tangent to the given curve at $P(x_1,y_1)$ is

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

Slope of the normal to the given curve at point $P(x_1,y_1)$ is
$\frac{-1}{\text{Slope of the tangent at the point}(x_1,y_1)}$
$=\frac{-1}{3x^2+2}$

The equation of the given line is $x+14y+4=0$ which can be written as

$y=-\frac{1}{14}x-\frac{4}{14}$ (which is of the form $y=mx+c$)
So, Slope of this line = $-\frac{1}{14}$
Since, the normal is parallel to this line.

So, slope of normal = slope of the given line.
$\therefore \frac{-1}{3x_1^2+2}=\frac{-1}{14}$
$\Rightarrow 3x_1^2+2=14$
$\Rightarrow 3x_1^2=12$
$x_1^2=4$
$x_1=\pm 2$

Since, $P(x_1,y_1)$ lies on the curve

$y_1=x_1^3+2x_1+6$ (by (1))
When $x_1=2$,

$\Rightarrow y_1=8+4+6=18$

When $x_1=-2$,

$\Rightarrow y_1=-8-4+6=-6$
Therefore, there are two normals to the given curve with slope $\frac{-1}{14}$ and passing through the points $(2,18)$ and $(-2,-6)$.
Thus, the equation of the normal through $(2,18)$ is given by,
$y-18=\frac{-1}{14}(x-2)$
$\Rightarrow 14y-252=-x+2$
$\Rightarrow x+14y-254=0$
And, the equation of the normal through $(-2,-6)$ is given by,
$y-(-6)=\frac{-1}{14}[x-(-2\left)\right]$
$\Rightarrow y+6=\frac{-1}{14}(x+2)$
$\Rightarrow 14y+84=-x-2$
$\Rightarrow x+14y+86=0$
Hence, the equations of the normals to the given curve (which are parallel to the given line) are$x+14y-254=0$ and $x+14y+86=0$.