Hyperbolas in the plane

An ellipse freezes the sum of distances from two focal points at $2a$, and the curve closes. A parabola freezes one focus against a directrix at $\epsilon =1$, and there is no second branch. A hyperbola also never closes — yet it splits in two and runs to both infinities. The question is which freeze does that, 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 $a=b$.

A constant difference, two branches

Fix two points $B_{-}$ and $B_{+}$ (the focal points). A point $P$ lies on this hyperbola when the difference of distances is constant, equal to $2a$:

$|P{B}_{-}-P{B}_{+}|=2a.$

That is not the ellipse string. Swap the freeze from a sum to a difference and the locus cannot close: $P$ is forced onto one branch or the other. Drag $P$. The two segments change length; their difference does not. $Bm$ and $Bp$ stay put.

The numerical eccentricity for this shape is $\epsilon >1$. The ratio to a directrix that makes $\epsilon$ visible lives on polar coordinates and conic sections. This page keeps the two-focus difference.

Sit the center at the origin

The difference rule does not yet name coordinates. Extra assumptions for the rest of this page: the center sits at the origin, the figure opens along the $x$-axis of the Cartesian plane, and the constants $a$ and $b$ are positive. Then the same set of points $(x,y)$ satisfies

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$

The curve meets the $x$-axis at the vertices $(\pm a,0)$. From $a$ and $b$ one also writes the linear eccentricity, the numerical eccentricity, and the half-parameter

$e=\sqrt{a^2+b^2},\epsilon =\frac{e}{a},p=\frac{b^2}{a}.$

Linear eccentricity here is $\sqrt{a^2+b^2}$, not the ellipse formula with a minus — that plus is why $\epsilon >1$. The two points $(\pm e,0)$ are $B_{\pm }$. On the next board they are $Bm$ and $Bp$; the unmarked dots on the $x$-axis are the vertices $\pm a$. Rotate the axis, or move the center off the origin, and $x^2/a^2-y^2/b^2=1$ is the wrong formula for the same geometric freeze.

One branch at a time

The plus-or-minus already sitting in $(\pm a,0)$ is two branches. The right-hand branch is traced by

$x=a\cosh t,y=b\sinh t,-\infty <t<\infty .$

This math handbook measures $t$ in radians. As $t$ runs through all real numbers, a point on that branch travels it once. The starting value $t=0$ is the vertex $(a,0)$. The left-hand branch is the same formulas with a minus in $x$:

$x=-a\cosh t,y=b\sinh t,-\infty <t<\infty .$

An ellipse can walk both “sides” with one $t$ in $[0,2\pi )$. Here one formula never crosses to the other branch. That is the extra freeze these two displays are spending: pick a sign in $x$, then let $t$ run.

Far away, two lines

A parabola has one infinity and no pair of lines it hugs. This figure has two branches, so far from the origin they run closer and closer to

$y=\pm \frac{b}{a}x.$

Those two lines are the asymptotes. The slope $\pm b/a$ is the same $a$ and $b$ already in the Cartesian equation. Change that ratio and the dashed lines tilt; drop the two-branch freeze and there is nothing to hug.

The tangent is that equation, linearized

At a point $(x_0,y_0)$ already on $x^2/a^2-y^2/b^2=1$, the tangent line is

$\frac{x{x}_0}{a^2}-\frac{y{y}_0}{b^2}=1.$

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 $a$ and $b$ still name the same hyperbola; only the line changes.

Leave one focus, aim at the other

A further freeze: a light ray that leaves $B_{-}$ and reflects off the hyperbola has its backward extension through $B_{+}$. On the last board the solid segment is the incoming ray from $Bm$; the dashed segment is that extension through $Bp$. Drag $P$ and both follow.

This is not the parabolic mirror (parallel to the axis, then through one focus). Drop “the ray started at a focus” and the bounce does not line up with the other. That is the extra assumption this board is spending.

The object is the difference freeze: two foci, constant $2a$, two branches. Under the extra placement (center at the origin, axis along $x$) that freeze is the Cartesian equation $x^2/a^2-y^2/b^2=1$, with $B_{\pm }$ at $(\pm e,0)$ and $e=\sqrt{a^2+b^2}$. The asymptotes, the tangent, and the two cosh formulas are that equation at infinity, at a point, and along one branch.

Thaw $\epsilon$ and the same family writes one polar equation. That is the next-but-one page after the parabola: polar coordinates and conic sections.