Condition for a Focal Chord

If (at₁², 2at₁) and (at₂², 2at₂) are the endpoints of a focal chord of y² = 4ax, then t₁t₂ = −1. Equivalently, one parameter is the negative reciprocal of the other.

Let $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ be two points on $y^2 = 4ax$. The chord joining them passes through the focus $S(a, 0)$.

Collinearity condition: $S$, $P_1$, $P_2$ are collinear, so the slope from $S$ to $P_1$ equals the slope from $S$ to $P_2$:

$$\frac{2at_1 - 0}{at_1^2 - a} = \frac{2at_2 - 0}{at_2^2 - a}$$ $$\frac{2t_1}{t_1^2 - 1} = \frac{2t_2}{t_2^2 - 1}$$

Cross-multiplying:

$$2t_1(t_2^2 - 1) = 2t_2(t_1^2 - 1)$$ $$t_1t_2^2 - t_1 = t_2t_1^2 - t_2$$ $$t_1t_2(t_2 - t_1) = t_2 - t_1$$

Since $t_1 \neq t_2$ (distinct points), divide by $t_2 - t_1$:

$$t_1t_2 = -1$$

Consequence: If one end of a focal chord has parameter $t$, the other end has parameter $-1/t$. Substituting: the other endpoint is $(a/t^2, -2a/t)$.

Alternative: The chord through $S(a,0)$ with slope $m$ meets the parabola at two points. Solving shows the product of their $t$-parameters is always $-1$, confirming the result.