Sum of Parameters at Co-normal Points

If three normals from an external point meet y² = 4ax at parameters t₁, t₂, t₃, then t₁+t₂+t₃ = 0. Equivalently, the sum of the slopes of the three normals is zero.

The normal at parameter $t$ to $y^2 = 4ax$ is $y + tx = 2at + at^3$.

If this normal passes through $(h, k)$:

$$k + th = 2at + at^3$$ $$at^3 + t(2a - h) + k = 0 \quad \cdots (*)$$

The three roots $t_1, t_2, t_3$ of this cubic are the parameters of the three co-normal points.

By Vieta's formulas for $at^3 + 0 \cdot t^2 + (2a-h)t + k = 0$:

$$t_1 + t_2 + t_3 = -\frac{0}{a} = 0$$ $$t_1t_2 + t_2t_3 + t_3t_1 = \frac{2a-h}{a}$$ $$t_1t_2t_3 = -\frac{k}{a}$$

The key result: $t_1 + t_2 + t_3 = 0$ — the sum of parameters at the three feet of the normals from any point is always zero.

Equivalent statement in slopes: Since the slope of the normal at $t$ is $-t$, and $m_i = -t_i$:

$$m_1 + m_2 + m_3 = -(t_1+t_2+t_3) = 0$$

The sum of slopes of the three normals from any point is zero.

Centroid of co-normal points: The centroid of $(at_1^2, 2at_1)$, $(at_2^2, 2at_2)$, $(at_3^2, 2at_3)$:

$$\bar{y} = \frac{2a(t_1+t_2+t_3)}{3} = 0$$

The centroid of the three co-normal points always lies on the axis of the parabola.