Eccentricity, Foci, Directrices and the Latus Rectum

Once $a$ and $b$ are fixed, every other measurement of the hyperbola follows automatically — none of them are independent choices.

Eccentricity. Since $b^2 = a^2(e^2-1)$ by construction, solving for $e$ gives

Because $b^2>0$, this always comes out greater than $1$, consistent with the defining property of a hyperbola.

Foci. Working through the same construction that located the focus during the derivation shows the two foci sit on the transverse axis at

Every hyperbola in standard form has two foci (and correspondingly two directrices) — one for each branch, even though a single branch was used to set up the equation.

Directrices. The two directrices are the vertical lines

Because $e>1$, $a/e<a$, so both directrices sit inside the vertices, between the two branches — a directrix never touches the branch it's paired with.

The focal-distance theorem. For any point $P$ on the hyperbola, the two distances to the foci don't just vary independently — their difference is always the same constant, $2a$ (in contrast to the ellipse, where it's the sum that's constant). Concretely, $S'P - SP = 2a$ for a point on the branch nearer $S$. This gives an equivalent, coordinate-free definition of a hyperbola: the locus of a point whose distances from two fixed points differ by a constant.

Latus rectum. The latus rectum is the focal chord perpendicular to the transverse axis. Substituting $x=ae$ into the equation and solving for $y$ shows its endpoints are $\left(\pm ae,\; \pm \dfrac{b^2}{a}\right)$, so its length is

A point's position relative to the curve. Writing $S_{11} \equiv \dfrac{x_1^2}{a^2}-\dfrac{y_1^2}{b^2}-1$ for a point $P(x_1,y_1)$: $P$ lies on the curve when $S_{11}=0$; it lies in the region not containing the centre (i.e., genuinely "inside" a branch) when $S_{11}>0$; and it lies in the region containing the centre (the wide middle strip, effectively "outside" both branches) when $S_{11}<0$.

A special case worth naming — the rectangular hyperbola. When the transverse and conjugate axes are equal in length ($a=b$), the equation reduces to $x^2-y^2=a^2$, and its eccentricity is fixed at $e=\sqrt2$ regardless of the size of $a$.

Worked example. Continuing with $\dfrac{x^2}{9}-\dfrac{y^2}{4}=1$ ($a=3$, $b=2$): here $a^2+b^2 = 9+4=13$, so

$e=\sqrt{\dfrac{13}{9}} = \dfrac{\sqrt{13}}{3}.$ …