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, an ellipse, a parabola, 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 $a$ is the semi-major axis (or semi-transverse axis for a hyperbola), $b$ is the semi-minor axis, and $c$ is the distance from the centre to a focus.

Where the symbols come from: for both the ellipse and the hyperbola, $e = c/a$. The two square-root forms follow because in an ellipse $c^2 = a^2 - b^2$, while in a hyperbola $c^2 = a^2 + b^2$. 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, $\times$, 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 $e = \sqrt{1 + \dfrac{9}{25}}$, 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: $\sqrt{1 + 9/25} = \sqrt{34/25} \approx 1.17$, 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 $a^2 = 25$ and $b^2 = 9$:

$$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} = 0.8$$

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, $a^2 = 4$ and $b^2 = 9$, and the formula carries a plus sign:

$$e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$$

As a decimal, $e \approx 1.80$, 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:

$$\frac{9x^2}{225} + \frac{25y^2}{225} = 1$$

$$\frac{x^2}{25} + \frac{y^2}{9} = 1$$

Now $a^2 = 25$ and $b^2 = 9$:

$$e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} = 0.8$$

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: every parabola has $e = 1$, 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:

$$\frac{16x^2}{400} - \frac{25y^2}{400} = 1$$

$$\frac{x^2}{25} - \frac{y^2}{16} = 1$$

Now $a^2 = 25$ and $b^2 = 16$:

$$e = \sqrt{1 + \frac{16}{25}} = \sqrt{\frac{41}{25}} = \frac{\sqrt{41}}{5}$$

As a decimal, $e \approx 1.28$.

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 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 $\sqrt{1 + b^2/a^2}$ for an ellipse or $\sqrt{1 - b^2/a^2}$ 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 like $9x^2 + 25y^2 = 225$ that are not yet equal to 1.

Don't do this: Read $a^2 = 9$ and $b^2 = 25$ 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 $a^2$ and $b^2$ from the denominators. The larger denominator is $a^2$ 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 $a^2$ 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, $a^2$ is the larger of the two denominators, wherever it sits. Identify the larger one first, call it $a^2$, 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 $e = \sqrt{1 - b^2/a^2}$; hyperbola uses $e = \sqrt{1 + b^2/a^2}$.

• Convert any conic equation to standard form (right side = 1) before reading $a^2$ and $b^2$.

• Sanity-check every eccentricity against the type's expected range around 1.

To take eccentricity further with a teacher, explore Bhanzu's geometry tutor or high school math tutor sessions, or browse math tutoring.

A Practical Next Step

Work through the exercises below to lock in the sign rule. Find the eccentricity of the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{7} = 1$ (Answer to Question 1: $3/4 = 0.75$), then the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$ (Answer to Question 2: $5/3 \approx 1.67$). If you get stuck choosing the sign, return to the formula table above. Want a live Bhanzu trainer to talk through conic sections? Book a free demo class -

Read More

• Ellipse — definition and properties

• Parabola explained

• Conic sections overview

• Eccentricity of an ellipse

• Latus rectum of parabola, ellipse, and hyperbola

• Foci of a hyperbola