Question

Find the equation of the normal to the curve $y=4x^3-3x+5$ which are perpendicular to the line $9x-y+5=0$.

Answer 1

$y=4x^3-3x+5⟹\frac{dy}{dx}=12x^2-3$

Given tangent is parallel to $9x-y+5=0$

$⟹12x^2-3=9⟹x=\pm 1⟹y=6,4$

Equation of normal is $y-6=-\frac{1}{9}(x-1)⟹x+9y=55$ or $y-4=-\frac{1}{9}(x+1)⟹x+9y=35$