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

Equation of the curve is, $y=x^3+2x+6$
Let $(h,h^3+2h+6)$ be any point on the curve
Slope of a tangent to the curve is $\frac{dy}{dx}$
$\frac{dy}{dx}=3x^2+2$
Slope of tangent at $(h,h^3+2h+6)$ is
$\Rightarrow \frac{dy}{dx}=3h^2+2$
We know that
$\text{Slope of tangent}\times \text{Slope of normal}=-1$
Slope of normal $=\frac{-1}{3h^2+2}$
Given that normal is parallel to the line
$x+14y+4=0$
$\Rightarrow y=\left(\frac{-1}{14}\right)x-\left(\frac{4}{14}\right)$
Here, slope of line is $-\frac{1}{14}$
Slopes of Normal $=-\frac{1}{14}$
$\Rightarrow \frac{-1}{3h^2+2}=\frac{-1}{14}$
$\Rightarrow 3h^2+2=14$
$\Rightarrow 3h^2=12$
$\Rightarrow h^2=4$
$\Rightarrow h=\pm 2$
When $h=-2$
$h^3+2h+6$
$=-8-4+6=-6$
Point is $(-2,-6)$
When $h=2$
$h^3+2h+6$
$=8+4+6=18$
Point is $(2,18)$
Now, the equation of normal at $(2,18)\&$ having slope $\frac{-1}{14}$ is
$(y-18)=\frac{-1}{14}(x-2)$
$\Rightarrow 14(y-18)=-(x-2)$
$\Rightarrow 14y-252=-x+2$
$\Rightarrow x+14y-254=0$
Equation if normal at $(-2,-6)\&$ having slope $\frac{-1}{14}$ is $(y-(-6\left)\right)=\frac{-1}{14}(x-(-2\left)\right)$
$\Rightarrow y+6=\frac{-1}{14}(x+2)$
$\Rightarrow 14y+84=-x-2$
$\Rightarrow x+14y+86=0$
Hence, the required equations of normal are $x+14y-254=0$ and $x+14y+86=0$.