What Is the Eccentricity of an Ellipse?

The word eccentricity is easy to confuse with the distance between an ellipse’s foci. They are related, but they are not the same quantity: focal distance has units of length, while eccentricity is a dimensionless number describing the ellipse’s shape.

For the standard ellipse with major semiaxis $a$ and minor semiaxis $b$, the numerical eccentricity is

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

where $c$ is the distance from the center to either focus. The distance between the two foci is $2c$.

The three quantities to keep separate

Assume the ellipse is centered at the origin, with its longer axis along the $x$-axis:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a\ge b>0.$$

Here, $a$ and $b$ are semiaxis lengths. The full major and minor axis lengths are $2a$ and $2b$. The foci are at $(\pm c,0)$, with

$$c^2=a^2-b^2.$$

These definitions give three related but distinct measurements:

Quantity	Meaning	Units
$c$	Center-to-focus distance	Length
$2c$	Distance between the foci	Length
$e=c/a$	Numerical eccentricity	None

Because $c=\sqrt{a^2-b^2}$, dividing by $a$ gives

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

The square root is nonnegative. Since $a\ge b>0$, an ellipse has $0\le e<1$.

Why eccentricity measures shape

If you enlarge an ellipse by the same scale factor in every direction, its values of $a$, $b$, and $c$ all grow by that factor. Their ratio $c/a$ stays unchanged. That is why eccentricity describes shape rather than size.

A circle is the special case $a=b$. Then $c=0$, the two foci coincide at the center, and $e=0$. As the ellipse gets more elongated, $b/a$ gets smaller and $e$ gets closer to $1$. A nondegenerate ellipse never has $e=1$; that value is associated with a parabola in the usual classification of conic sections.

Worked example

Consider

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

The semiaxes are $a=5$ and $b=3$. Find the focal distance:

$$\begin{aligned} c^2&=a^2-b^2=25-9=16,\\ c&=4. \end{aligned}$$

So the foci are $(4,0)$ and $(-4,0)$. Their distance apart is $8$. The eccentricity, however, is

$$e=\frac{c}{a}=\frac{4}{5}=0.8.$$

You can check it using the other formula:

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

Thus $4$ is the center-to-focus distance, $8$ is the focus-to-focus distance, and $0.8$ is the eccentricity. None of these should be substituted for another without converting.

If the major axis is vertical

For an ellipse with its longer axis along the $y$-axis, the equation is commonly written

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

with $a\ge b$. Its foci are $(0,\pm c)$, but the same relationships hold: $c^2=a^2-b^2$ and $e=c/a$. The letter $a$ still denotes the larger semiaxis, not necessarily the denominator under $x^2$.

Common mistakes

• Calling $c$ the eccentricity. In standard ellipse notation, $c$ is a length; eccentricity is $c/a$.
• Using the full axis lengths in the formula without care. If you know the full major and minor axis lengths $A$ and $B$, use $e=\sqrt{1-(B/A)^2}$; the factor of $2$ cancels. Do not divide the focal distance by the full major axis length.
• Assuming $a$ always belongs under $x^2$. First identify which semiaxis is longer. For a vertical ellipse, the larger denominator is under $y^2$.
• Treating $e$ as a distance. Eccentricity has no units, so it cannot be a number of centimeters or meters.
• Thinking every ellipse has two distinct foci. A circle has coincident foci, which is why its eccentricity is zero.

A useful check is to make sure your answer for an ellipse’s numerical eccentricity is between $0$ and $1$, and that your formula uses the longer semiaxis in the denominator.