Hyperbolas
A LORAN navigation system fixes a ship’s position by timing radio signals from two stations. The set of points whose distances to the two stations differ by a constant is one branch of a hyperbola, and the ship sits somewhere on that curve. The same shape shows up in the path of a comet that swings past the sun once and never returns.
By the end of this page you can:
- Read the vertices, foci, and asymptotes straight from a hyperbola’s standard equation.
- Find the relationship between a, b, and c using c squared equals a squared plus b squared.
- Sketch a hyperbola by drawing its central box and asymptotes first, then the two branches.
The floor. A hyperbola in standard form is the difference of two squared terms set equal to 1. For a horizontal one, x squared over a squared minus y squared over b squared equals 1: vertices at plus or minus a on the major axis, asymptotes y equals plus or minus (b over a) x, and foci found from c squared equals a squared plus b squared.
Quick Reference
| Field | Value |
|---|---|
| Textbook | Stewart, Redlin, Watson 7e |
| Chapter | 11 (Conic Sections) |
| Section | 11.3 Hyperbolas |
| Open alternate | OpenStax Precalculus 2e, 12.2 The Hyperbola |
| Difficulty | Core |
| Time | About 30 minutes |
Before You Start
The Idea
A hyperbola is the set of all points in the plane the difference of whose distances from two fixed points (the foci) is a constant. This is the ellipse’s definition with the sum traded for a difference, and that single change splits the curve into two separate branches that open away from each other.
In standard form the equation is a difference of squared terms equal to 1. The term that is positive tells you which way the hyperbola opens: if the x term is positive it opens left and right, if the y term is positive it opens up and down. The value a sets the vertices, b sets the height of a guiding box, and the two diagonals of that box are the asymptotes that the branches approach but never touch. The distance to the foci comes from c squared equals a squared plus b squared, which is a plus, not the minus you used for the ellipse.
Key Formulas
Horizontal hyperbola centered at the origin, opening left and right:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Vertices at \((\pm a, 0)\), foci at \((\pm c, 0)\), and asymptotes
\[ y = \pm \frac{b}{a}\, x \]
Vertical hyperbola centered at the origin, opening up and down:
\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]
with vertices at \((0, \pm a)\) and asymptotes \(y = \pm \frac{a}{b}\, x\).
For either orientation the focal distance satisfies
\[ c^2 = a^2 + b^2 \]
Source: Stewart 7e, Section 11.3 Hyperbolas.
Worked Example
Find the vertices, foci, and asymptotes of the hyperbola
\[ \frac{x^2}{9} - \frac{y^2}{16} = 1 \]
then describe its graph.
Work through it before opening the answer. Identify a squared and b squared, take square roots, then use c squared equals a squared plus b squared.
Show the answer
The x term is positive, so the hyperbola opens left and right. Here a squared equals 9 and b squared equals 16, so a equals 3 and b equals 4.
Vertices: \((\pm 3, 0)\), that is \((3, 0)\) and \((-3, 0)\).
Focal distance: \(c^2 = a^2 + b^2 = 9 + 16 = 25\), so \(c = 5\). Foci: \((\pm 5, 0)\).
Asymptotes: \(y = \pm \dfrac{b}{a}\, x = \pm \dfrac{4}{3}\, x\).
To sketch it, draw the box from \(x = \pm 3\) and \(y = \pm 4\), draw the two diagonals through its corners as the asymptotes, then draw a branch through each vertex hugging those diagonals.
A common slip is writing \(c^2 = a^2 - b^2\), the ellipse rule, which would give \(c = \sqrt{9 - 16}\) and a nonsense negative under the root. For a hyperbola the foci sit farther out than the vertices, so c is larger than a, and the plus is what makes that true.
Connections
This builds on ellipses, since the hyperbola shares the focus idea but uses the difference of distances rather than the sum, and it reuses the same a and b labels in a new c relationship. It builds on reading a standard conic equation and plotting intercepts.
It leads to shifted conics, where the center moves off the origin and x and y are replaced by x minus h and y minus k. Together with parabolas and ellipses it completes the family of conic sections.
Resources
- Stewart, Redlin, Watson, Precalculus 7e, Section 11.3 Hyperbolas (primary).
- OpenStax Precalculus 2e, Section 12.2 The Hyperbola (open alternate).
Last updated: 2026-06-24