# Eccentricity - Definition, Formula, and Examples

## What Is Eccentricity?
**Eccentricity** is the ratio of the distance from any point on a conic to a fixed point (the **focus**) to the distance from that same point to a fixed line (the **directrix**). It is a pure number with no units, and it measures how much a conic section departs from being a circle.

e=\frac{\text{distance from point to focus}}{\text{distance from point to directrix}}

a **conic section** is any curve you get by slicing a cone with a flat plane - a [circle](/content/math/geometry/area-of-a-circle/index.html), an [ellipse](/content/math/geometry/ellipse/index.html), a [parabola](/content/math/geometry/parabola/index.html), or a hyperbola. The **focus** is a special interior point, and the **directrix** is a line outside the curve; together they define the conic. As the eccentricity grows, the curvature of the shape decreases and the conic opens out.

## What Are The Eccentricity Formulas For Each Conic?
Each conic has its own eccentricity value and formula. Here aaa is the semi-major axis (or semi-transverse axis for a hyperbola), bbb is the semi-minor axis, and ccc is the distance from the centre to a focus.

| Conic | Eccentricity | Formula |
| --- | --- | --- |
| Circle | e=0 | (focus at the centre) |
| Ellipse | 0<e<1 | e=\frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}} |
| Parabola | e=1 | (equal focus and directrix distances) |
| Hyperbola | e>1 | e=\frac{c}{a} = \sqrt{1 + \frac{b^2}{a^2}} |

Where the symbols come from: for both the ellipse and the hyperbola, e=c/ae=c/a. The two square-root forms follow because in an ellipse c2=a2−b2, while in a hyperbola c2=a2+b2. The single sign flip - minus for the ellipse, plus for the hyperbola - is the whole difference between the two formulas, and it is worth noting rather than memorising two unrelated expressions.

- **Circle:** the "focus" sits at the centre, so every point is the same distance from it, and e=0.
- **Parabola:** a point on the curve is always equidistant from the focus and the directrix, so the ratio is exactly 1.

## Examples Of Eccentricity
Six worked cases, from a direct ratio to standard-form conversions. The problem statement is bolded; the working is not. One multiplication symbol, ×, is used throughout.

### Example 1
**An ellipse has semi-major axis a = 5 and a focus at distance c = 3 from the centre. Find its eccentricity.**

Use e=c/a directly:

e=\frac{c}{a} = \frac{3}{5} = 0.6

Since 0<0.6<1, the value is consistent with an ellipse.

Final answer: e=0.6.

### Example 2
**A student sees the ellipse equation \dfrac{x^2}{25} + \dfrac{y^2}{9} = 1 and writes e=\sqrt{1+\frac{9}{25}}**, expecting a value above 1. Find the correct eccentricity.

The tempting move is to reach for the plus sign inside the root, carrying it over from the hyperbola formula. Watch it break: 1+9/25=\sqrt{\frac{34}{25}}\approx 1.17, which is greater than 1 and would mean a hyperbola - but the equation has a plus between the two terms, so it is an ellipse, and no ellipse has e>1. The result is obviously wrong.

The correct formula for an ellipse uses a minus sign. Here a2=25 and b2=9:

e=\sqrt{1 - \frac{9}{25}} = \frac{4}{5} = 0.8

Final answer: e=0.8 - the minus sign is what keeps an ellipse's eccentricity below 1.

### Example 3
**Find the eccentricity of the hyperbola \dfrac{x^2}{4} - \dfrac{y^2}{9} = 1.**

For a hyperbola, a2=4 and b2=9, and the formula carries a plus sign:

e=\sqrt{1 + \frac{9}{4}} = \frac{\sqrt{13}}{2}.

As a decimal, e≈1.80, which is greater than 1 as expected.

Final answer: e=\dfrac{\sqrt{13}}{2} \approx 1.80.

### Example 4
**An ellipse is given by 9x^2 + 25y^2 = 225. Find its eccentricity.**

First convert to standard form by dividing through by 225:
\frac{9x^2}{225} + \frac{25y^2}{225} = 1

Now a2=25 and b2=9:

e=\sqrt{1 - \frac{9}{25}} = \frac{4}{5} = 0.8

Final answer: e=0.8.

### Example 5
**State the eccentricity of a parabola, and explain why it is fixed.**

A parabola is the set of points equidistant from a focus and a directrix. Because the two distances are always equal, their ratio is always 1:

e=\frac{\text{distance to focus}}{\text{distance to directrix}} = 1

Final answer: e=1, with no dependence on how wide or narrow it looks.

### Example 6
**Find the eccentricity of the hyperbola 16x^2 - 25y^2 = 400.**

Divide through by 400 to reach standard form:
\frac{16x^2}{400} - \frac{25y^2}{400} = 1

Now a2=25 and b2=16:

e=\sqrt{1 + \frac{16}{25}} = \frac{\sqrt{41}}{5} \approx 1.28.

Final answer: e=\dfrac{\sqrt{41}}{5} \approx 1.28.

## Where Eccentricity Earns Its Keep
Eccentricity does real work far beyond the page. Astronomers describe every planetary and cometary orbit by its eccentricity - a comet on a near-parabolic path (e close to 1) may visit the inner solar system once and never return, while a planet's small eccentricity keeps it in a stable near-circular loop. Satellite engineers pick an orbit's eccentricity to trade coverage against altitude. Optical and antenna designers use the reflective property of parabolas (e=1) to focus signals to a single point. The number that started as a slicing ratio ends up steering spacecraft.

Kepler published the elliptical-orbit result in _Astronomia Nova_ (1609); you can read a short account of [Kepler's first law](https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion) and how eccentricity fixes the shape of each orbit.

## Common Mistakes With Eccentricity
### Mistake 1: Swapping the sign between ellipse and hyperbola
**Where it slips in:** Any problem where you compute e from a standard-form equation.
**Don't do this:** Use \sqrt{1 + \frac{b^2}{a^2}} for an ellipse or \sqrt{1 - \frac{b^2}{a^2}} for a hyperbola. The single most common source of wrong answers here is carrying the plus sign from the hyperbola formula into the ellipse case.

**The correct way:** Ellipse uses a minus, hyperbola uses a plus. Sanity-check the result: an ellipse must land between 0 and 1, a hyperbola above 1. A value on the wrong side of 1 flags a sign error.

### Mistake 2: Skipping the conversion to standard form
**Where it slips in:** Equations that are not yet equal to 1.
**Don't do this:** Read straight off the un-normalised equation. The rusher plugs the raw coefficients in and gets a meaningless answer.
**The correct way:** Divide through so the right side equals 1 first, then read a2 and b2 from the denominators. The larger denominator is a2 for an ellipse.

### Mistake 3: Mixing up which axis is major
**Where it slips in:** Ellipses where the larger number sits under y^2 instead of x^2.
**Don't do this:** Always assume a2 is under x^2. The second-guesser flips back and forth and loses track of which is the semi-major axis.
**The correct way:** For an ellipse, a2 is the _larger_ of the two denominators, wherever it sits. Identify the larger one first, call it a2, and the eccentricity formula follows.

## Conclusion
- **Eccentricity** measures how stretched a conic section is, as a ratio of focus-distance to directrix-distance.
- The values are fixed by type: circle e=0, ellipse 0<e<1, parabola e=1, hyperbola e>1.
- Ellipse uses e=\sqrt{1 - \frac{b^2}{a^2}}; hyperbola uses e=\sqrt{1 + \frac{b^2}{a^2}}.
- Convert any conic equation to standard form (right side = 1) before reading a2 and b2.
- Sanity-check every eccentricity against the type's expected range around 1.