Parabolas in the plane

An ellipse keeps two focal points and a constant sum of distances $2a$. A hyperbola keeps two focal points and a constant difference $2a$. Both stretch on two axes $a$ and $b$. A parabola in this handbook never closes and never grows a second axis. The question is which freeze turns “two foci” into one — and which extra placement then lets us write a Cartesian equation at all.

Those Cartesian axes sit with lines in the plane. A circle is the ellipse case $a=b$.

Equal distance to a point and a line

Fix a point $B$ (a focal point) and a line $L$ (the directrix). A point $P$ lies on this parabola when the two distances are equal:

$BP=\text{distance from }P\text{ to }L.$

That is the numerical eccentricity frozen at $\epsilon =1$. Drop the freeze and the same language gives an ellipse ($\epsilon <1$) or a hyperbola ($\epsilon >1$) — those ratios live on polar coordinates and conic sections. This page keeps $\epsilon =1$.

Drag $P$. The labeled segments stay the same length. $B$ does not move; $L$ does not move. Only $P$ walks the curve that the freeze allows.

Sit the vertex at the origin

The equal-distance rule does not yet name coordinates. Extra assumptions for the rest of this page: the vertex (the point on the curve closest to $L$) sits at the origin, the axis of the figure is the positive $x$-axis of the Cartesian plane, and the constant $p$ is positive. Then the same set of points $(x,y)$ satisfies

$y^2=2px.$

From $p$ one also writes the linear eccentricity and the numerical eccentricity

$e=\frac{p}{2},\epsilon =1.$

The point $(e,0)$ is the focal point $B$. The live figure uses $p=2$, so $B$ sits at $(1,0)$ and the hidden directrix is the line $x=-1$. Rotate the axis, or move the vertex off the origin, and $y^2=2px$ is the wrong formula for the same geometric freeze.

The tangent is that equation, linearized

At a point $(x_0,y_0)$ already on $y^2=2px$, the tangent line is

$y{y}_0=p(x+x_0).$

This is not a new curve. It is the Cartesian equation made linear at that point. Drag $P$. The line through $P$ is that tangent. Leave the point, and the same $p$ still names the same parabola; only the line changes.

Parallel to the axis, then through $B$

A further freeze: a light ray that is parallel to the $x$-axis and hits the parabola is reflected so that it passes through the focal point. On the last board the incoming arrow comes from the open side (the right); after the bounce it runs to $B$. Drag $P$ and both arrows follow.

Drop “parallel to the axis” and the bounce misses $B$. That is the extra assumption this board is spending — the same special placement as $y^2=2px$, not a claim about a rotated parabola.

The object is the $\epsilon =1$ freeze: one focus, one directrix, equal distances. Under the extra placement (vertex at the origin, axis along $+x$) that freeze is the Cartesian equation $y^2=2px$, with $B$ at $(e,0)$ and $e=p/2$. The tangent and the mirror are that equation at a point, and that equation for rays parallel to the axis.

Thaw $\epsilon$ and the same focus-directrix language writes one polar equation for the whole family. That is the next page: polar coordinates and conic sections.