Question

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

Answer 1

Slop of normal $=-\frac{dx}{dy}=\frac{-1}{14}$

$\Rightarrow \frac{dy}{dx}=14$

$3x^2+2=14$

$x^2=4\Rightarrow x=\pm 2$

$\therefore (2,18)and(-2,-6)$

the required equation of normal are

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