Eccentricity of an Ellipse: Formula, Meaning, Examples

What Does the Eccentricity of an Ellipse Measure?

The eccentricity of an ellipse measures how far the ellipse is from being a circle, in other words, how elongated or "squashed" it is. It is written $e$ and defined as the ratio of two lengths:

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

where $c$ is the distance from the center to a focus (one of the two fixed points that define the ellipse), and $a$ is the semi-major axis, half the length of the long axis. A larger $e$ means the foci sit farther from the center relative to the ellipse's size, which means a more stretched curve.

Because both $c$ and $a$ are lengths and $c$ is always smaller than $a$ for an ellipse, the ratio always lands strictly between 0 and 1:

$$0 < e < 1.$$

The two endpoints tell the story. At $e = 0$, the two foci have merged at the center and the ellipse is a perfect circle. As $e$ climbs toward 1, the foci slide out toward the vertices and the ellipse stretches into a long, thin shape.

The Eccentricity of an Ellipse Formula

There are two formulas, and they are the same statement in two outfits. The first uses the focal distance directly:

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

The second avoids $c$ entirely, expressing eccentricity from just the two axes:

$$e = \sqrt{1 - \frac{b^2}{a^2}},$$

where $b$ is the semi-minor axis, half the short axis. You reach for the first formula when you already know $c$ (or the foci), and the second when you only have the equation $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ and can read off $a^2$ and $b^2$.

The two are connected by the focal relationship $c^2 = a^2 - b^2$, which holds for every ellipse, the foci sit inside, so $c$ is always shorter than $a$.

Where the Second Formula Comes From

The formula $e = \sqrt{1 - \dfrac{b^2}{a^2}}$ is not a separate fact to memorise. It falls straight out of $e = \dfrac{c}{a}$ once you bring in the focal relationship $c^2 = a^2 - b^2$.

Start from the definition and square it to clear the square root that $c$ will introduce:

$$e = \frac{c}{a} ;\Rightarrow; e^2 = \frac{c^2}{a^2}.$$

Substitute $c^2 = a^2 - b^2$:

$$e^2 = \frac{a^2 - b^2}{a^2} = \frac{a^2}{a^2} - \frac{b^2}{a^2} = 1 - \frac{b^2}{a^2}.$$

Take the positive square root (eccentricity is a length ratio, so it is positive):

$$e = \sqrt{1 - \frac{b^2}{a^2}}.$$

This also shows why $0 < e < 1$ falls out for free. Since $b < a$, the fraction $\dfrac{b^2}{a^2}$ is a positive number less than 1, so $1 - \dfrac{b^2}{a^2}$ sits between 0 and 1, and its square root does too. When $b = a$ the ellipse is a circle and $e = 0$; the foci can never reach the vertices, so $e$ never actually hits 1.

Where the Eccentricity of an Ellipse Shows Up

Eccentricity is the number scientists and engineers actually quote when they describe an orbit, a lens, or an antenna, because it captures the shape's character in one figure, independent of size.

• Planetary orbits. Kepler's first law says planets orbit the Sun in ellipses with the Sun at one focus. Earth's orbital eccentricity is about $0.017$, almost a circle, which is why our seasons come from axial tilt, not distance. A comet like Halley's has eccentricity near $0.97$, a long, swooping orbit that swings it from the inner solar system far past Neptune.

• Satellite design. Engineers choose an orbit's eccentricity deliberately. A near-zero $e$ gives a steady-altitude circular orbit for imaging; a high $e$ gives a Molniya-type orbit that lingers over high latitudes.

• Optics and acoustics. The reflective property of an ellipse, set by its foci and so by its eccentricity, is what makes whispering galleries and lithotripsy machines work, sound or shock waves from one focus reconverge at the other.

• Telling conics apart. Eccentricity is the universal label for the whole conic family: $e = 0$ is a circle, $0 < e < 1$ an ellipse, $e = 1$ a parabola, and $e > 1$ a hyperbola. One number sorts them all.

For a Class 11 student, eccentricity is where a static equation turns into a description you can picture: hand someone $e = 0.9$ and they already know the curve is long and thin before drawing a thing.

Examples of the Eccentricity of an Ellipse

The problems below build from a clean read-and-compute up to recovering an axis from a given eccentricity. Watch how the two formulas trade places depending on what you start with.

Example 1

Find the eccentricity of the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$.

Here $a^2 = 25$ (the larger denominator) and $b^2 = 16$. Use the axis-only formula:

$$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} = 0.6.$$

The eccentricity is $0.6$, moderately stretched, well clear of a circle.

Example 2

Find the eccentricity of the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{25} = 1$.

A natural first move is to assume $a^2$ is the number under $x^2$, so take $a^2 = 9$ and $b^2 = 25$, then compute $e = \sqrt{1 - \frac{25}{9}} = \sqrt{1 - 2.78} = \sqrt{-1.78}$ and hit the square root of a negative number. That impossibility is the signal: $e$ is a real number between 0 and 1, so a negative inside the root means the labels are swapped.

The fix: $a$ is always the semi-major axis, so $a^2$ is the larger denominator. Here 25 is larger and sits under $y^2$, so the major axis is vertical, $a^2 = 25$ and $b^2 = 9$:

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

The eccentricity is $0.8$.

Example 3

An ellipse has semi-major axis $a = 13$ and focal distance $c = 5$. Find its eccentricity.

Here $c$ is given directly, so use the first formula:

$$e = \frac{c}{a} = \frac{5}{13} \approx 0.385.$$

The eccentricity is about $0.385$, fairly close to circular.

Example 4

Find the eccentricity of the ellipse $\dfrac{x^2}{36} + \dfrac{y^2}{11} = 1$.

With $a^2 = 36$ and $b^2 = 11$:

$$e = \sqrt{1 - \frac{11}{36}} = \sqrt{\frac{25}{36}} = \frac{5}{6} \approx 0.833.$$

The eccentricity is about $0.833$, a noticeably elongated ellipse.

Example 5

An ellipse has eccentricity $e = 0.6$ and semi-major axis $a = 10$. Find its semi-minor axis $b$.

Work the formula backwards. From $e = \dfrac{c}{a}$, the focal distance is $c = ea = 0.6 \times 10 = 6$. Then use $c^2 = a^2 - b^2$:

$$b^2 = a^2 - c^2 = 100 - 36 = 64 ;\Rightarrow; b = 8.$$

The semi-minor axis is $8$.

Example 6

Earth's orbit has eccentricity about $0.017$, and Halley's Comet about $0.967$. Interpret each.

No computation, just reading. Earth's $e = 0.017$ is almost 0, so its orbit is very nearly a circle, the Sun (at one focus) is only slightly off-center. Halley's $e = 0.967$ is very close to 1, so its orbit is a long, thin ellipse, the comet races close to the Sun, then drifts far out before returning.

The interpretation: eccentricity alone, with no lengths attached, tells you the character of the orbit. A single decimal separates a gentle near-circle from a dramatic, stretched sweep.

Where Students Trip Up on the Eccentricity of an Ellipse

Mistake 1: Mislabelling which axis is the major axis

Where it slips in: Reading $a^2$ and $b^2$ off the equation when the larger denominator is under $y^2$.

Don't do this: Always assign $a^2$ to the number under $x^2$.

The correct way: $a$ is the semi-major axis, so $a^2$ is always the larger denominator, wherever it sits. If $1 - \dfrac{b^2}{a^2}$ comes out negative, you have swapped them.

Mistake 2: Forgetting the square root

Where it slips in: Using the second formula and stopping at $1 - \dfrac{b^2}{a^2}$.

Don't do this: Report $e = 1 - \dfrac{b^2}{a^2}$ — that quantity is $e^2$, not $e$.

The correct way: The formula is $e = \sqrt{1 - \dfrac{b^2}{a^2}}$. The square root is not optional; dropping it gives the square of the eccentricity. The rusher, the student who computes fast and skips the final operation, loses the mark exactly here.

Mistake 3: Treating eccentricity as a length

Where it slips in: Reporting eccentricity with units, or confusing it with $c$, the focal distance.

Don't do this: Write "$e = 6$ cm" or hand back $c$ when asked for $e$.

The correct way: Eccentricity is a ratio of two lengths, so it has no units and is always a pure number between 0 and 1. The focal distance $c$ does have a length; $e = c/a$ strips the units away. If your "eccentricity" is bigger than 1 or carries a unit, you have reported the wrong quantity.

What to Remember About the Eccentricity of an Ellipse

• The eccentricity of an ellipse is $e = c/a$, a unitless number measuring how stretched the ellipse is.

• The axis-only form is $e = \sqrt{1 - b^2/a^2}$; the two formulas are linked by $c^2 = a^2 - b^2$.

• Eccentricity always satisfies $0 < e < 1$: $e = 0$ is a circle, $e$ near 1 is long and thin.

• The most common mistake is mislabelling the major axis, $a^2$ is always the larger denominator.

• One number, eccentricity, sorts the whole conic family: circle, ellipse, parabola, hyperbola.

Practice These Problems to Solidify Your Understanding

1. Find the eccentricity of the ellipse $\dfrac{x^2}{100} + \dfrac{y^2}{64} = 1$.

2. Find the eccentricity of the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{49} = 1$.

3. An ellipse has $e = 0.8$ and $a = 5$. Find its semi-minor axis $b$.

Answer to Question 1: $e = \sqrt{1 - \frac{64}{100}} = \sqrt{0.36} = 0.6$. Answer to Question 2: the larger denominator 49 is under $y^2$, so $a^2 = 49$, $b^2 = 16$, giving $e = \sqrt{1 - \frac{16}{49}} = \frac{\sqrt{33}}{7} \approx 0.84$. Answer to Question 3: $c = ea = 4$, so $b^2 = 25 - 16 = 9$ and $b = 3$. If Question 2 gave you a value above 1 or a negative under the root, revisit Mistake 1.

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