# Tangents to Parabolas without Limiting Processes

Let's consider the simplest parabola, which can be represented by the space of points $$[x,y]$$ in $$\mathbb{A}^2$$ over an arbitrary field $$\mathbb{F}$$ with $$\text{char}(\mathbb{F}) \neq 2$$ that satisfies the equation $$y = x^2$$. Note that a general affine transformation on this parabola will result in a general parabola.

To compute the tangent line of the parabola at the point $P \equiv \left[ t,t^2 \right]$ we pick another point $Q \equiv \left[ u,u^2 \right]$ on the parabola, so that the line joining $P$ and $Q$ has vector equation

$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} t \\ t^2 \end{pmatrix} + \lambda \begin{pmatrix} u-t \\ u^2-t^2 \end{pmatrix},$

for $\lambda \in \mathbb{F}$. Noting that $u^2 - t^2 = (u-t)(u+t)$, we use the fact that $x - t = \lambda (u-t)$ to obtain

$y - t^2 = \lambda \left( \frac{x-t}{\lambda} \right) (u+t),$

which simplifies to

$y = (u+t)x - ut.$

This line is called the secant line or the secant chord. If $Q=P$, then $u = t$ and the secant chord has equation $y = 2tx - t^2$; this line is then called the tangent line of the parabola at the point $P$. We now have a methodology for exactly computing tangent lines to parabolas which avoids limiting processes. Note that this also implies that the vertex of the parabola, i.e. the point at which the tangent line of the parabola takes the form $y = c$ for a constant $c \in \mathbb{F}$, is $[0,0]$

As an example, we consider the general affine transformation $T:\mathbb{A}^2 \mapsto \mathbb{A}^2$ given by

$T(\begin{pmatrix} x \\ y \end{pmatrix}) = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}+ \begin{pmatrix} f \\ g \end{pmatrix}.$

The vertex $[0,0]$ is mapped by $T$ to $[f,g]$, which is the vertex of the translated parabola. I will leave it as an exercise to figure out

1. the equation of the tangent line; and
2. hence, the equation of the general parabola.

Note by A Former Brilliant Member
2 years, 2 months ago

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