Ellipse — Equation, Foci, Eccentricity & Graph

An ellipse is a graph shaped like a stretched circle. Geometrically, it is the set of points whose distances to two fixed points, the foci, add up to a constant, which is why the graph has a center, a longer direction, and a shorter direction. In coordinate geometry you usually identify it from a standard equation, then read off the center, the half-axes, the foci, and the eccentricity.

When to use the standard-form method

The standard forms below are the fastest to read because the ellipse is axis-aligned, not rotated. Reach for this method when an ellipse equation is already in, or can be put into, axis-aligned standard form. For a non-circular ellipse centered at the origin with a horizontal major axis,

where $a$ is the semi-major axis and $b$ is the semi-minor axis. More generally, with center $(h, k)$:

For these axis-aligned forms, the larger denominator tells you the major-axis direction. The foci lie on the major axis at distance $c$ from the center, where $c^2 = a^2 - b^2$, so the focus coordinates are $(h \pm c, k)$ for a horizontal major axis and $(h, k \pm c)$ for a vertical one. The eccentricity is $e = \frac{c}{a}$, with $0 < e < 1$ for a non-circular ellipse: smaller values mean closer to a circle, values closer to $1$ mean more stretched. If the foci are close to the center, the ellipse looks rounder; if they are farther apart, it looks narrower.

The procedure, step by step

1. Identify the center. In $(x-h)^2/a^2 + (y-k)^2/b^2 = 1$, the center is $(h, k)$.
2. Find the major axis. For an axis-aligned ellipse in standard form, the larger denominator tells you which direction stretches farther from the center.
3. Mark vertices. Move $a$ units from the center along the major axis and $b$ units along the minor axis.
4. Compute the foci. Use $c^2 = a^2 - b^2$ with $a > b > 0$, then place the foci on the major axis.
5. Read eccentricity. Use $e = c/a$ to measure how stretched the ellipse is.

A full example: graph $x^2/25 + y^2/9 = 1$

Start with the equation

Because $25 > 9$, the major axis is horizontal. Read off

Now find $c$:

so $c = 4$. The important points are:

• Center: $(0, 0)$
• Vertices: $(\pm 5, 0)$
• Co-vertices: $(0, \pm 3)$
• Foci: $(\pm 4, 0)$

The eccentricity is

To sketch the graph, plot the center first, then the vertices and co-vertices, and draw a smooth curve through those four endpoints. Since the major axis is horizontal, the ellipse should be wider than it is tall. If the ellipse is centered at $(h, k)$ instead of the origin, the same steps work after shifting every key point by $(h, k)$.

Where students get stuck, and how to check each step

• Mixing up $a$ and $b$. For a non-circular ellipse in standard form, $a$ is the semi-major axis, so $a > b$. Assigning $a$ to the $x$ term automatically is only correct when the major axis is horizontal.
• Using the wrong relation for the foci. For an ellipse, $c^2 = a^2 - b^2$, not $a^2 + b^2$. The wrong sign gives the wrong foci and eccentricity.
• Confusing vertices with foci. The vertices are the endpoints of the major axis; the foci are inside the ellipse. They are not the same points.
• Overusing the denominator shortcut. The larger denominator identifies the major axis only after the equation is in standard axis-aligned form. A rotated ellipse does not read that way directly.

This is also why Step 1 matters so much: put the equation into standard form first, because the shortcuts only work cleanly for axis-aligned standard form. Ellipses appear throughout analytic geometry and conic sections, and in physics models such as orbital paths in the idealized two-body model, where one focus sits at the central body.

Practice this procedure

Run all five steps on

and find the center, vertices, foci, and eccentricity before you sketch it. For one more check, compare your graph with the example above and see exactly how the shift changes the key points without changing the overall shape.