Condition for Three Normals from a Point

Three normals can be drawn from (h, k) to y² = 4ax with slopes m₁, m₂, m₃ satisfying this cubic. By Vieta's: m₁+m₂+m₃ = 0, m₁m₂+m₂m₃+m₃m₁ = (2a−h)/a, m₁m₂m₃ = −k/a.

Three normals can be drawn from $(h, k)$ to $y^2 = 4ax$ when the cubic:

$$f(m) = am^3 + (2a - h)m + k = 0$$

has three distinct real roots.

Discriminant condition: A depressed cubic $m^3 + pm + q = 0$ (no $m^2$ term) has three real roots iff:

$$\Delta = -4p^3 - 27q^2 > 0$$

For our cubic: $p = (2a-h)/a$, $q = k/a$.

$$-4\!\left(\frac{2a-h}{a}\right)^{\!3} - 27\!\left(\frac{k}{a}\right)^{\!2} > 0$$ $$4(h - 2a)^3 > 27ak^2$$

Geometric interpretation: The set of points from which three real normals can be drawn to $y^2 = 4ax$ is the region:

$$4(h-2a)^3 > 27ak^2 \quad (h > 2a)$$

This region lies to the right of the evolute of the parabola. The evolute (locus of the centre of curvature) has the equation $27ay^2 = 4(x-2a)^3$.

Special case — point on the axis ($k = 0$):

$$f(m) = am^3 + (2a-h)m = m[am^2 + (2a-h)] = 0$$

One root is $m = 0$ (normal along the axis). The other two are $m = \pm\sqrt{(h-2a)/a}$, real when $h > 2a$. So for points on the axis beyond $x = 2a$, all three normals are real.