What is the difference between a and b in the ellipse equation?

a is the semi-axis in the horizontal direction (how far left and right from center), and b is the semi-axis in the vertical direction (how far up and down from center). The larger value determines the major axis: if a > b, the ellipse is wider than tall; if b > a, it is taller than wide.

How do I find the vertices of an ellipse from its equation?

The vertices are the endpoints of the major axis. From \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, the vertices are at (h \pm a, k) if a > b (horizontal major axis) or at (h, k \pm b) if b > a (vertical major axis).

What is an ellipse and why is it useful?

An ellipse is an oval shape formed by all points where the sum of distances to two fixed points (called foci) is constant. Ellipses model planetary orbits, arches, and real-world structures much better than circles when objects are not perfectly round.

Can I find the foci of an ellipse from its equation?

Yes. The foci lie on the major axis at distance c from the center, where c = \sqrt{|a^2 - b^2|}. If a > b, the foci are at (h \pm c, k); if b > a, they are at (h, k \pm c).

Equation of an Ellipse

Equation of an Ellipse describes an oval-shaped curve given its center, horizontal stretch, and vertical stretch; use it to model or graph any elliptical shape.

\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1

Solve a problem with Equation of an Ellipse

Type the problem. The solver will use Equation of an Ellipse where Equation of an Ellipse is the right tool, and tell you when it is not.

What each symbol means

What Equation of an Ellipse takes

Symbol	Meaning
$a$	The semi-axis in the horizontal direction (how far the ellipse extends left and right from its center); confusing it with the full width will double your result.
$b$	The semi-axis in the vertical direction (how far the ellipse extends up and down from its center); not the same as the full height of the ellipse.
$h$	The x-coordinate of the center point of the ellipse; if you change this value, the entire ellipse shifts horizontally left or right.
$k$	The y-coordinate of the center point of the ellipse; if you change this value, the entire ellipse shifts vertically up or down.
$x$	The horizontal coordinate of any point you are testing; plug in the actual $x$ -coordinate to check whether a point lies on, inside, or outside the ellipse.
$y$	The vertical coordinate of any point you are testing; plug in the actual $y$ -coordinate along with $x$ to verify whether the equation holds.

When to use it

When you need to write, identify, or analyze the equation of an oval shape in the coordinate plane.

Level

Usually taught in: Pre-Calculus

Worked examples

1. Write an ellipse equation centered at the origin

Problem

Write the equation of an ellipse with center (0, 0), semi-axis of length 5 (horizontal), and semi-axis of length 3 (vertical).

$h = 0, \quad k = 0, \quad a = 5, \quad b = 3$
We identify the center coordinates and both semi-axes from the problem statement.
$\frac{(x - 0)^2}{5^2} + \frac{(y - 0)^2}{3^2} = 1$
We substitute the identified values into the standard form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ .
$\frac{x^2}{25} + \frac{y^2}{9} = 1$
We simplify by removing the zero subtractions and calculating $5^2 = 25$ and $3^2 = 9$ .

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

This is the standard form of an ellipse centered at the origin. Any point $(x, y)$ on this ellipse will satisfy the equation. The larger denominator 25 under the $x$ term confirms the major axis is horizontal with semi-axis length 5.

2. Find an ellipse equation with a translated center

Problem

Find the equation of an ellipse centered at

(-2, 3)

with semi-axis 4 (horizontal) and semi-axis 2 (vertical).

$h = -2, \quad k = 3, \quad a = 4, \quad b = 2$
We extract the center $(h, k) = (-2, 3)$ and the semi-axes $a = 4$ and $b = 2$ from the given information.
$\frac{(x - (-2))^2}{4^2} + \frac{(y - 3)^2}{2^2} = 1$
We substitute into the standard form, being careful with the sign when $h = -2$ is negative.
$\frac{(x + 2)^2}{16} + \frac{(y - 3)^2}{4} = 1$
We simplify: $(x - (-2))$ becomes $(x + 2)$ , and we calculate $4^2 = 16$ and $2^2 = 4$ .

Answer: $\frac{(x + 2)^2}{16} + \frac{(y - 3)^2}{4} = 1$

The negative center coordinate $h = -2$ causes the $(x + 2)$ term; subtracting a negative flips the sign. This ellipse is centered at $(-2, 3)$ , shifted 2 units left and 3 units up from the origin.

3. Determine an ellipse equation for a construction archway

Problem

An architect designs an arched doorway where the outer edge follows an ellipse. The arch's center is 1 unit to the right of the left wall and 4 units above ground. The arch spans 10 units wide and 8 units tall. Write the equation of the archway.

$h = 1, \quad k = 4, \quad \text{width} = 10, \quad \text{height} = 8$
We identify the center $(h, k) = (1, 4)$ and note that the full width is 10 units and full height is 8 units.
$a = \frac{10}{2} = 5, \quad b = \frac{8}{2} = 4$
Since width 10 is the full distance across horizontally, the semi-axis is $a = 5$ ; similarly, $b = 4$ from the height of 8.
$\frac{(x - 1)^2}{5^2} + \frac{(y - 4)^2}{4^2} = 1$
We substitute $h = 1, k = 4, a = 5, b = 4$ into the standard form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ .
$\frac{(x - 1)^2}{25} + \frac{(y - 4)^2}{16} = 1$
We calculate the squared values: $5^2 = 25$ and $4^2 = 16$ .

Answer: $\frac{(x - 1)^2}{25} + \frac{(y - 4)^2}{16} = 1$

This real-world example shows how the standard form adapts when the ellipse is shifted from the origin. The center location directly determines the $(x - h)$ and $(y - k)$ terms. An architect uses this equation to ensure the archway has the correct dimensions and positioning.

Common mistakes

Where Equation of an Ellipse usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\frac{x^2}{10^2} + \frac{y^2}{6^2} = 1$ when an ellipse has width 10 and height 6.	The width 10 and height 6 represent the full diameters across the ellipse, not the semi-axes; you must divide each by 2 first.	Divide each measurement by 2: $a = 10 / 2 = 5$ and $b = 6 / 2 = 3$ , giving the correct equation $\frac{x^2}{25} + \frac{y^2}{9} = 1$ .
Using the circle equation $(x - h)^2 + (y - k)^2 = r^2$ when the shape is an ellipse.	The circle equation assumes equal distance in all directions, which only works when $a = b$ ; an ellipse with $a \neq b$ requires the fractional form to account for different horizontal and vertical stretches.	Use the ellipse form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ whenever $a$ and $b$ are different values.
Writing $\frac{x^2}{9} + \frac{y^2}{4} = 1$ for an ellipse centered at $(3, -2)$ instead of at the origin.	Forgetting to adjust the numerators for the center shifts the entire ellipse to the origin; the translation must be included as $(x - h)$ and $(y - k)$ in the numerators.	The correct equation is $\frac{(x-3)^2}{9} + \frac{(y-(-2))^2}{4} = 1$ , which simplifies to $\frac{(x-3)^2}{9} + \frac{(y+2)^2}{4} = 1$ .

Tips and when to use something else

Remember: $a$ and $b$ are the semi-axes (half-widths), not the full diameters; if given the total width and height, always divide each by 2 before substituting.
A circle is a special case of an ellipse where $a = b$ ; if all you have is a circle, use the simpler Equation of a Circle $(x - h)^2 + (y - k)^2 = r^2$ instead.
The larger of $a$ and $b$ determines the major axis direction: if $a > b$ , the major axis is horizontal; if $b > a$ , it is vertical.
If the equation has a minus sign between the terms instead of a plus—like $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ —you are working with an Equation of a Hyperbola, which describes open unbounded curves.

Related formulas

Equation of a CircleDescribes a circle using its center and radius; essential for finding circle equations, checking if points lie on circles, or solving circle geometry problems.Equation of a HyperbolaThe equation of a hyperbola describes all points forming a hyperbola with center at (h, k), used when modeling two-branch curves in coordinate geometry.Equation of a ParabolaThe equation (x - h)^2 = 4p(y - k) describes a parabola with vertex at (h, k) and tells you which direction it opens and how wide it is.Translating a GraphTranslate graphs using y = f(x - h) + k to shift any function horizontally by h and vertically by k units, needed when repositioning known curves.

Frequently asked questions

What is the difference between $a$ and $b$ in the ellipse equation?: $a$ is the semi-axis in the horizontal direction (how far left and right from center), and $b$ is the semi-axis in the vertical direction (how far up and down from center). The larger value determines the major axis: if $a > b$ , the ellipse is wider than tall; if $b > a$ , it is taller than wide.
How do I find the vertices of an ellipse from its equation?: The vertices are the endpoints of the major axis. From $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ , the vertices are at $(h \pm a, k)$ if $a > b$ (horizontal major axis) or at $(h, k \pm b)$ if $b > a$ (vertical major axis).
What is an ellipse and why is it useful?: An ellipse is an oval shape formed by all points where the sum of distances to two fixed points (called foci) is constant. Ellipses model planetary orbits, arches, and real-world structures much better than circles when objects are not perfectly round.
Can I find the foci of an ellipse from its equation?: Yes. The foci lie on the major axis at distance $c$ from the center, where $c = \sqrt{|a^2 - b^2|}$ . If $a > b$ , the foci are at $(h \pm c, k)$ ; if $b > a$ , they are at $(h, k \pm c)$ .

Need a different method?

The full solver is not scoped to one formula — type any problem and it will pick the method.

Open the math solver

Reviewed 2026-09-18