Last updated: 2026-07-02

The foci of an ellipse

An ellipse in standard position is described by

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

where $a$ is the semi-axis along the x-direction and $b$ the semi-axis along the y-direction. Every ellipse has two special interior points, its foci. They are what make an ellipse an ellipse: for any point on the curve, the sum of the distances to the two foci is constant. Stretch a loop of string around two pins and trace it taut with a pencil, and the pencil draws an ellipse whose foci are the pins.

This calculator takes $a$ and $b$ and returns the focal distance, the foci themselves, the eccentricity, the vertices, and the lengths of the major and minor axes.

Semi-major and semi-minor axes

The longer of the two axes is the major axis; the shorter is the minor axis. Half of each is a semi-axis. To keep the formulas orientation-free, call the larger semi-axis $A$ and the smaller $B$:

$A = \max(a, b) \qquad B = \min(a, b)$

The foci always lie on the major axis. If $a > b$ the major axis is horizontal; if $b > a$ it is vertical.

Focal distance

The distance from the centre to each focus, written $c$, comes from the semi-axes:

$c = \sqrt{A^2 - B^2}$

Because $A \ge B$, the quantity under the root is never negative. For a horizontal ellipse the foci are at $(\pm c, 0)$; for a vertical one they are at $(0, \pm c)$.

Eccentricity

Eccentricity measures how far the ellipse departs from a circle:

$e = \frac{c}{A}$

It always satisfies $0 \le e < 1$. When $a = b$ the two semi-axes are equal, $c = 0$, and $e = 0$ — the ellipse is a circle. As one axis grows relative to the other, $e$ climbs toward $1$ and the ellipse becomes long and thin.

Vertices and axis lengths

The vertices are the endpoints of the major axis, at a distance $A$ from the centre: $(\pm A, 0)$ for a horizontal ellipse or $(0, \pm A)$ for a vertical one. The endpoints of the minor axis are the co-vertices. The full axis lengths are

$\text{major axis} = 2A \qquad \text{minor axis} = 2B$

Worked example

Problem: Find the foci, eccentricity, and axis lengths of $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$.

Step 1 — read off the semi-axes: $a^2 = 25$ so $a = 5$, and $b^2 = 9$ so $b = 3$. Since $a > b$, the major axis is horizontal with $A = 5$ and $B = 3$.

Step 2 — focal distance:

$c = \sqrt{A^2 - B^2} = \sqrt{25 - 9} = \sqrt{16} = 4$

Step 3 — foci: on the horizontal major axis, at $(\pm 4, 0)$, so $F_1 = (-4, 0)$ and $F_2 = (4, 0)$.

Step 4 — eccentricity:

$e = \frac{c}{A} = \frac{4}{5} = 0.8$

Step 5 — vertices and axes: vertices at $(\pm 5, 0)$; major axis $2A = 10$, minor axis $2B = 6$.

So this ellipse has foci $(-4, 0)$ and $(4, 0)$, eccentricity $0.8$, and axes of length $10$ and $6$.

Vertical ellipses

When $b > a$ the ellipse is taller than it is wide, so the major axis runs vertically. The roles simply swap: $A = b$, $B = a$, and the foci move to the y-axis at $(0, \pm c)$. For example, $\dfrac{x^2}{9} + \dfrac{y^2}{25} = 1$ has $A = 5$ and $B = 3$, the same $c = 4$ as before, but foci at $(0, -4)$ and $(0, 4)$. The calculator picks the orientation automatically from whichever semi-axis is larger.

Why the sum of distances is constant

The defining property $r_1 + r_2 = 2A$ links directly to $c$. At a vertex on the major axis, one focal radius is $A + c$ and the other is $A - c$, and their sum is $2A$. At a co-vertex the two radii are equal by symmetry, each running from a focus a distance $\sqrt{B^2 + c^2}$; setting that equal to $A$ recovers $A^2 = B^2 + c^2$, which rearranges to $c = \sqrt{A^2 - B^2}$. That single relationship ties the foci, the axes, and the eccentricity together.