Given slope m, find the equation of normal having slope m to the parabola y2=4ax
y=mx−2am−am3
Given value of normal is m. we know at point (h,k) slope of tangent is 2ak. Therefore slope of normal would be −k2a.
We have, k2=4ah and m=−k2a
Substituting for c in y = mx + c at point (h,k)
k = m(h) + c
−2a(m)=m(k24a)+c
−2am−m((−2am)24a)=c
we get c=−2am−am3
Therefore equation of normal with slope m for the given parabola would be y=mx−2am−am3