The normal y = mx - 2am - $am^3$ passes through (h,k): k = mh - 2am - $am^3$ => $am^3$ + (2a-h)m + k = 0. This cubic in m can have 1 or 3 real roots. Three normals exist from (h,k) when h > 2a (roughly). The sum of slopes m1+m2+m3=0, sum of pairwise products = $\frac{2a-h}{a}$, product m1m2m3 = -k/a. The feet of three concurrent normals are called co-normal points.