# Hyperbola: Definition, Equation, Foci, and Asymptotes

## What Is a Hyperbola?
A **hyperbola** is the set of all points in a plane for which the **difference** of the distances to two fixed points, called the **foci**, is a positive constant. Those two distances need not be equal — only their difference stays fixed, and that constant difference is what bends the curve into two separate branches.

A hyperbola is one of the four [**conic sections**](/content/math/geometry/conic-sections/index.html) — the curves you get by slicing a cone with a plane. Slice gently and you get a circle or an [ellipse](/content/math/geometry/ellipse/index.html); slice parallel to the side and you get a [parabola](/content/math/geometry/parabola/index.html); slice steeply enough to cut _both_ halves of the cone and you get a hyperbola's two branches.

The contrast with the ellipse is the cleanest way to hold the idea. An ellipse is the set of points where the _sum_ of the two focal distances is constant — a closed loop. Swap "sum" for "difference" and the loop breaks open into two mirror curves. That single word, _difference_ versus _sum_, is the whole distinction.

## How Is A Hyperbola Different From A Parabola?
Both curves open outward, so they are easy to confuse. A parabola is a _single_ open curve, defined by one focus and a directrix line; a hyperbola has _two_ branches and _two_ foci. A parabola has no asymptotes; a hyperbola's branches hug two straight asymptote lines as they run off to infinity.

## The Standard Equation of a Hyperbola
For a hyperbola centred at the origin with its **transverse axis** (the axis through both vertices) along the x-axis, the standard equation is:

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

If the transverse axis runs along the y-axis instead, the equation flips:

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

The single feature that tells the two apart is which variable carries the **positive** term. Positive $x^2$ term means the branches open left and right; positive $y^2$ term means they open up and down.

**Variable key:** $a$ is the distance from the centre to each vertex (so the vertices sit at $(\pm a,0)$ for a horizontal hyperbola); $b$ sets the height of the central rectangle that controls the asymptotes; $c$ is the distance from the centre to each focus.

### The focus relation, derived
For a hyperbola, the three lengths are tied together by:

$$c^2 = a^2 + b^2$$

This looks like the ellipse relation but with a plus sign instead of a minus — and the plus is not arbitrary. It falls straight out of the definition. Take the vertex point $(a,0)$, which lies on the curve. Its distance to the near focus is $c-a$ and to the far focus is $c+a$, so the constant difference of focal distances is $(c+a)-(c-a)=2a$. Working that constant through the distance definition for a general point and simplifying produces $c^2 = a^2 + b^2$ — which is why, for a hyperbola, the focus always sits _farther_ from the centre than the vertex.

## The Key Parts of a Hyperbola

| Part                       | For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$  |
|----------------------------|-------------------------------------------------|
| Centre                      | $(0,0)$                                        |
| Vertices                    | $(\pm a,0)$                                   |
| Foci                       | $(\pm c,0)$, where $c=\sqrt{a^2 + b^2}$       |
| Transverse axis length      | $2a$                                          |
| Conjugate axis length       | $2b$                                          |
| Asymptotes                  | $y=\pm \frac{b}{a}x$                         |
| Eccentricity                | $e=\frac{c}{a} = \sqrt{1 + \frac{b^2}{a^2}}$|

The **asymptotes** are the two straight lines the branches approach but never touch. The fastest way to draw them: mark the central rectangle of width $2a$ and height $2b$, then extend its diagonals. The **eccentricity** $e$ measures how "open" the curve is; for every hyperbola $e>1$, and the larger $e$, the wider the branches spread.

## Examples of Hyperbola
The examples move from reading values off a standard equation to building one from given parts. Coordinates are in plain units throughout.

### Example 1
**For $\frac{x^2}{9} - \frac{y^2}{16} = 1$, find the vertices.**
Here $a^2=9$, so $a=3$. The positive term is $x^2$, so the hyperbola opens horizontally and the vertices sit on the x-axis:

Vertices: $(\pm 3,0)$

### Example 2
**For $\frac{x^2}{9} - \frac{y^2}{16} = 1$, find the foci — and avoid the most common slip.**
A reader who has just met the ellipse often reaches for $c^2=a^2−b^2$ out of habit, giving $c^2=9−16=-7$. A squared length cannot be negative, so the result is impossible — the formula was the wrong one. That impossibility is the signal to stop.
For a hyperbola the relation carries a plus sign:

$$c^2 = a^2 + b^2 = 9 + 16 = 25, \quad c = 5$$
Foci: $(\pm 5,0)$

### Example 3
**For $\frac{x^2}{9} - \frac{y^2}{16} = 1$, find the asymptotes.**
With $a=3$ and $b=4$, the asymptotes are:

y=±43x

### Example 4
**Find the eccentricity of $\frac{x^2}{9} - \frac{y^2}{16} = 1$.**
Using $c=5$ from Example 2 and $a=3$:

e=ca=53≈1.67

### Example 5
**A hyperbola opens up and down with vertices at $(0,±5)$ and foci at $(0,±13)$. Find its equation.**
The vertices are on the y-axis, so the $y^2$ term is positive and $a=5$. The foci give $c=13$. Solve for $b^2$:

$$b^2 = c^2 - a^2 = 13^2 - 5^2 = 169 - 25 = 144$$

$$\frac{y^2}{25} - \frac{x^2}{144} = 1$$

### Example 6
**The difference of the distances from a point to two stations at $(±10,0)$ km is always 12 km. Find the hyperbola the point lies on.**
The constant difference equals $2a$, so $2a=12$ gives $a=6$ and $a^2=36$. The stations are the foci, so $c=10$ and:

$$b^2 = c^2 - a^2 = 100 - 36 = 64$$

$$\frac{x^2}{36} - \frac{y^2}{64} = 1$$

## Why the Hyperbola Matters
The hyperbola is the geometry of _differences_, and differences turn out to run the physical world.

- **Navigation and positioning** — systems that locate a receiver from the _difference_ in signal arrival times place it on a hyperbola; this is the principle behind hyperbolic navigation.

- **Astronomy** — an object passing the Sun with more than escape speed follows a hyperbolic orbit, swinging by once and never returning. Comets on hyperbolic paths are visitors, not residents.

- **Optics and design** — hyperbolic mirrors redirect light toward a focus, a property used in some reflecting telescopes.

The reason the curve keeps appearing is structural: **whenever a quantity is governed by a constant difference rather than a constant sum, the geometry is a hyperbola.**

## Where Hyperbola Problems Go Wrong
### Mistake 1: Using the ellipse focus relation
**Where it slips in:** Right after studying the ellipse, when $c^2=a^2−b^2$ is fresh in memory.

**Don't do this:** Apply $c^2=a^2−b^2$ to a hyperbola — it can produce a negative number for $c^2$, which is impossible.

**The correct way:** A hyperbola uses $c^2=a^2+b^2$, because the focus sits farther from the centre than the vertex.

### Mistake 2: Misreading which way the hyperbola opens
**Where it slips in:** Identifying vertices and foci from the equation.

**Don't do this:** Assume the curve always opens left-right, or read $a$ from whichever denominator is larger.

**The correct way:** The variable with the **positive** term sets the opening direction.

### Mistake 3: Flipping the asymptote slope
**Where it slips in:** Writing the asymptotes for a horizontal hyperbola.

**Don't do this:** Write $y=±abx$ from memory.

**The correct way:** For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the slopes are $±\frac{b}{a}$ — rise over run from the central rectangle.

## Conclusion
- A **hyperbola** is the set of points whose distances to two foci have a constant difference, producing two mirror-image branches.

- Its standard equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, with the positive term setting the opening direction.

- The focus relation is $c^2 = a^2 + b^2$ — a plus sign, unlike the ellipse.

- Asymptotes are $y=±\frac{b}{a}x$, and eccentricity $e=\frac{c}{a} > 1$ for every hyperbola.

- The most common mistake is borrowing the ellipse's minus-sign focus relation, which gives an impossible negative $c^2$.

## Practice and Next Steps
Work through these problems to solidify your understanding.

1. Find the vertices, foci, and asymptotes of $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

2. A hyperbola has vertices $(\pm 4,0)$ and eccentricity $\frac{5}{4}$. Find its equation.

3. Find the eccentricity of $\frac{y^2}{36} - \frac{x^2}{64} = 1$.