Parametric Point on y² = 4ax

Every point on y² = 4ax can be written as (at², 2at) for a unique t. Parametric form simplifies tangent, normal, and chord problems considerably.

For $y^2 = 4ax$, let $y = 2at$ for a real parameter $t$. Substituting:

$$(2at)^2 = 4ax \implies 4a^2t^2 = 4ax \implies x = at^2$$

So every point on $y^2 = 4ax$ can be written as $(at^2, 2at)$ for a unique $t \in \mathbb{R}$.

Why this parametrisation is natural:

• $t = 0$: vertex $(0, 0)$
• $t = 1$: point $(a, 2a)$ — upper end of latus rectum
• $t = -1$: point $(a, -2a)$ — lower end of latus rectum
• $t \to \pm\infty$: the curve extends to infinity along the axis

Slope at $(at^2, 2at)$: Differentiating $y^2 = 4ax$ implicitly: $2y\,dy = 4a\,dx$, so $dy/dx = 2a/y = 2a/(2at) = 1/t$.

The slope of the curve at parameter $t$ is $1/t$. This determines the tangent directly.

Two distinct points on the parabola are $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$. The slope of the chord joining them:

$$\frac{2at_1 - 2at_2}{at_1^2 - at_2^2} = \frac{2(t_1 - t_2)}{(t_1-t_2)(t_1+t_2)} = \frac{2}{t_1 + t_2}$$