Find the equation of the normal line

Question

Find the equation of the normal line to curve $y=2x^2+3$ that is perpendicular to the line $8x-y+3=0$

I have already solved the $m = 8$, $m'= -\frac{1}{8}$.

But I don't know how to solve for the $x,y$ points.

Answer 1

Let the normal be at $(h,k)$, then $$y=2x^2+3 \implies y'=4x=-\frac{1}{m} \implies h=\frac{-1}{4m}$$ The equation of normal is $$y-k=m(x-h) \implies y-2h^2-3=mx-mh \implies y-\frac{1}{8m^2}-3=mx+\frac{1}{4}~~~(2)$$ For for any value of $m$ Eq. (2) represents normal to the given parabola. Let $m=-1/8$, the $y-8-3=-x/8+1/4 \implies 8y+x=90$ is the required normal to the given parabola.