Last updated: 2026-07-02

The equation of a hyperbola

A hyperbola is the set of all points whose distances to two fixed points — the foci — differ by a constant amount. A horizontal hyperbola centered at $(h, k)$ has the standard form:

$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$

The minus sign is what separates a hyperbola from an ellipse. It produces two disconnected branches that open left and right, each racing off toward a pair of straight asymptotes. This calculator takes the semi-transverse axis $a$, the semi-conjugate axis $b$, and the center $(h, k)$, and returns the focal distance, vertices, foci, eccentricity, and asymptote slopes.

Vertices, foci and the focal distance

The vertices are the two points where the branches are closest together. They lie on the transverse axis at a distance $a$ from the center:

$V = (h \pm a,\ k)$

The foci lie farther out, at a distance $c$ from the center, where

$c = \sqrt{a^2 + b^2} \qquad\Longrightarrow\qquad F = (h \pm c,\ k)$

Notice that $c > a$ always, because $b^2 > 0$. This is the reverse of the ellipse relationship $c = \sqrt{a^2 - b^2}$: for a hyperbola the foci sit beyond the vertices, not inside them.

Asymptotes and eccentricity

Far from the center each branch hugs one of two straight lines through the center with slopes $\pm b/a$:

$y = k \pm \frac{b}{a}(x - h)$

A quick way to draw them by hand is to sketch the central rectangle of width $2a$ and height $2b$ centered at $(h, k)$; the asymptotes are its extended diagonals.

The eccentricity measures how sharply the branches open:

$e = \frac{c}{a}$

Because $c > a$, a hyperbola always has $e > 1$. (An ellipse has $0 \le e < 1$, and a parabola sits exactly at $e = 1$.)

Worked example

Problem: Describe the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.

Step 1 — read off the axes: $a^2 = 9$ and $b^2 = 16$, so $a = 3$ and $b = 4$. The center is $(0, 0)$.

Step 2 — vertices: at $(\pm a, 0) = (3, 0)$ and $(-3, 0)$.

Step 3 — focal distance:

$c = \sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

so the foci are $(5, 0)$ and $(-5, 0)$.

Step 4 — eccentricity:

$e = \frac{c}{a} = \frac{5}{3} \approx 1.667$

Step 5 — asymptotes: slope $\pm b/a = \pm \tfrac{4}{3}$, giving $y = \pm \tfrac{4}{3}x$.

So this hyperbola has vertices $(\pm 3, 0)$, foci $(\pm 5, 0)$, eccentricity $5/3$, and asymptotes $y = \pm\tfrac{4}{3}x$.

The semi-latus rectum

A useful checkpoint is the height of the branch directly above a focus. Substituting $x = c$ into the equation gives $y = \pm b^2/a$, a quantity called the semi-latus rectum. For the example above it is $16/3 \approx 5.33$ — the graph passes through $(5, \pm 16/3)$, which is why the branch is already quite tall by the time it reaches the focus.

How the transverse and conjugate axes differ

The transverse axis connects the two vertices and passes through the foci; its half-length is $a$. The conjugate axis is perpendicular to it through the center; its half-length is $b$. Unlike an ellipse, there is no requirement that $a > b$ — the branches cross the transverse axis regardless of which of $a$ or $b$ is larger. When $a = b$ the asymptotes are the perpendicular lines $y = \pm(x - h) + k$ and the hyperbola is called equilateral (or rectangular), with eccentricity exactly $\sqrt{2}$.