Standard Form of the Hyperbola

A hyperbola is defined the same way an ellipse is, except for one flipped inequality: it is the path traced by a point $P$ whose distance from a fixed point $S$ (the focus) stays a constant multiple $e$ of its distance from a fixed line (the directrix), where this time $e>1$. Because the point is pulled disproportionately away from the line as it moves off to infinity, the locus splits into two mirror-image branches that never close up — that's the visual signature that separates a hyperbola from an ellipse at a glance.

To get a clean equation, place the centre of symmetry $C$ at the origin, and run the $x$-axis straight through the focus, perpendicular to the directrix. Working through the focus-directrix condition $SP = e\,(PM)$ algebraically (exactly the same style of computation used for the ellipse, just with $e>1$ this time) collapses to the tidy standard form:

Here $2a$ is the length of the transverse axis — the segment $AA'$ joining the two points where the curve actually crosses the $x$-axis — and $b$ is defined through the relation $b^2 = a^2(e^2-1)$, which is forced by the derivation. Unlike the ellipse, there's no requirement that $b<a$; $b$ can be anything positive, because it no longer measures a semi-axis the curve touches. The segment $BB'$ of length $2b$ marked off on the (vertical) $y$-axis is called the conjugate axis, and it's worth remembering that the curve itself never actually meets this axis — it's a construction axis, not a boundary the curve touches.

A quick feature-check on $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ tells you everything about its shape: setting $y=0$ gives $x=\pm a$, so it crosses the $x$-axis at $(\pm a, 0)$; setting $x=0$ gives $y^2=-b^2$, which has no real solution, confirming the curve never touches the $y$-axis. Solving for $y$ gives $y = \pm\dfrac{b}{a}\sqrt{x^2-a^2}$, which is only real when $|x|\ge a$ — so the entire strip $-a<x<a$ is empty of the curve, and each branch stretches away to infinity as $x\to\pm\infty$. The curve is symmetric about both axes, since replacing $x\to -x$ or $y\to -y$ leaves the equation unchanged. …