Ellipse Calculator
Graph an ellipse from its center and semi-axes or its equation, and find the vertices, co-vertices, foci, eccentricity, area and perimeter step by step.
Try an example
Ellipse
(x − 2)²/25 + (y + 1)²/9 = 1
Horizontal major axis, center (2, −1), foci (−2, −1) and (6, −1), eccentricity 0.8.
Ellipse properties
- Standard equation
- (x − 2)²/25 + (y + 1)²/9 = 1
- Orientation
- ↔ Horizontal major axis
- Center
- (2, −1)
- Semi-major axis a
- 5
- Semi-minor axis b
- 3
- Major axis length 2a
- 10
- Minor axis length 2b
- 6
- Vertices
- (−3, −1), (7, −1)
- Co-vertices
- (2, −4), (2, 2)
- Foci
- (−2, −1), (6, −1)
- Focal distance c
- 4
- Eccentricity e
- 0.8
- Area πab
- 15π ≈ 47.1239 square units
- Approximate perimeter
- ≈ 25.527 units
Eccentricity: e = 0.8
For every ellipse 0 ≤ e < 1. This ellipse is moderately elongated. An eccentricity of exactly 1 would be a parabola, not an ellipse.
Step-by-Step Calculation
Step 1 — Identify the center, orientation and semi-axes
The larger denominator, a² = 25, is under the x-term, so the major axis is horizontal.
(x − 2)²/25 + (y + 1)²/9 = 1
Center (h, k) = (2, −1)
a² = 25 → a = 5
b² = 9 → b = 3
Step 2 — Focal distance c
For an ellipse the focal distance satisfies c² = a² − b² (minus, not plus).
c = √(a² − b²)
= √(25 − 9)
= √16
= 4
Step 3 — Eccentricity
e = c/a measures how stretched the ellipse is: 0 for a circle, approaching 1 as it gets longer and thinner.
e = c / a
= 4 / 5
= 0.8
Step 4 — Vertices, co-vertices and foci
The vertices are a from the center along the major axis, the co-vertices b along the minor axis, and the foci c along the major axis.
Vertices: (h ± a, k) = (2 ± 5, −1) → (−3, −1) and (7, −1)
Co-vertices: (h, k ± b) = (2, −1 ± 3) → (2, −4) and (2, 2)
Foci: (h ± c, k) = (2 ± 4, −1) → (−2, −1) and (6, −1)
Step 5 — Area and perimeter
The area has an exact formula. The perimeter does not have one in elementary terms, so Ramanujan’s approximation is used.
A = πab
= π(5)(3)
= 15π ≈ 47.1239
P ≈ π[3(a + b) − √((3a + b)(a + 3b))]
≈ 25.527
Advanced ellipse properties: directrices
Each directrix is a line at distance a/e from the center, perpendicular to the major axis. For any point on the ellipse, its distance to a focus is e times its distance to the matching directrix.
x = −4.25 and x = 8.25
Standard ellipse equation
Horizontal major axis ↔ ← this ellipse
(x − h)²/a² + (y − k)²/b² = 1
The larger denominator a² is under x, e.g. (x − h)²/25 + (y − k)²/9 = 1.
Vertical major axis ↕
(x − h)²/b² + (y − k)²/a² = 1
The larger denominator a² is under y, e.g. (x − h)²/9 + (y − k)²/25 = 1.
- (h, k)
- center
- a
- semi-major axis (a ≥ b)
- b
- semi-minor axis
- c
- focal distance, c² = a² − b²
Other formulas: e = c/a, area = πab, and perimeter ≈ π[3(a + b) − √((3a + b)(a + 3b))] (Ramanujan’s approximation; there is no exact elementary formula).
Ellipse vs circle
| Ellipse (a > b) | Circle (a = b = r) | |
|---|---|---|
| Semi-axes | Two different lengths | Equal: the radius r |
| Eccentricity | 0 < e < 1 | e = 0 |
| Foci | Two distinct points, c from the center | Both at the center (c = 0) |
| Area | πab | πr² |
| Perimeter | Approximation only | Exactly 2πr |
Common mistakes
- Semi-axis vs full axis. a is half the major axis. The major axis length is 2a, like radius vs diameter.
- Using a² and b² as lengths. The denominators are a² and b². For /25 and /9, a = 5 and b = 3, not 25 and 9.
- Ignoring which semi-axis is larger. The larger of a and b is always the semi-major axis a, whichever variable it sits under.
- Larger denominator under the wrong variable. Under x → horizontal major axis; under y → vertical. Swapping them rotates the ellipse by 90°.
- c = √(a² + b²). That is the hyperbola formula. For an ellipse, c = √(a² − b²).
- Confusing foci with vertices. Vertices are a from the center, on the curve; foci are c from the center, inside it.
- Eccentricity as b/a or c/b. e = c/a. It is always between 0 and 1 for an ellipse.
- Treating the perimeter formula as exact. Ramanujan’s formula is a very close approximation; the exact perimeter needs an elliptic integral.
- Sign errors in the center. (x + 3)² means h = −3, and (y − 4)² means k = 4.
When a = b, the Circle Calculator covers area, circumference and the circle equation in more depth. The Distance Formula Calculator checks distances such as PF₁ and PF₂, and the other conic sections have their own tools: the Parabola Calculator and the Hyperbola Calculator.
This ellipse calculator finds every key property of an ellipse and draws it on a coordinate plane. Enter the center (h, k), the semi-major axis a and the semi-minor axis b and choose a horizontal or vertical major axis, or type a standard-form equation such as (x − 2)²/25 + (y + 1)²/9 = 1.
The calculator gives the center, vertices, co-vertices, foci, focal distance c, eccentricity e, major and minor axis lengths, area and an approximate perimeter, with exact values such as 15π and 2√5 where possible. The graph marks each point and both axes, zooms to fit the ellipse wherever it is, and shows the focal property PF₁ + PF₂ = 2a. When a = b, it recognises the shape as a circle.
Worked Calculation Examples
| Scenario | Result | Calculation Step |
|---|---|---|
| Horizontal ellipse: (x − 2)²/25 + (y + 1)²/9 = 1 | Foci (−2, −1) and (6, −1), e = 0.8, area 15π | Center (2, −1); 25 > 9 and 25 is under x, so the major axis is horizontal with a = 5, b = 3. Vertices (−3, −1), (7, −1); co-vertices (2, −4), (2, 2). c = √(25 − 9) = 4, e = 4/5 = 0.8, area = 15π ≈ 47.1239. |
| Vertical ellipse: (x + 3)²/16 + (y − 4)²/36 = 1 | Vertices (−3, −2) and (−3, 10) | 36 is under y, so the major axis is vertical: a = 6, b = 4, center (−3, 4). Co-vertices (−7, 4), (1, 4). c = √(36 − 16) = √20 = 2√5 ≈ 4.4721, so the foci are (−3, 4 ± 2√5). e = √5/3 ≈ 0.7454, area = 24π ≈ 75.3982. |
| Circle: x²/16 + y²/16 = 1 | c = 0, e = 0 | a = b = 4, so the ellipse is a circle of radius 4. c = √(16 − 16) = 0: both foci are at the center (0, 0), and e = 0. Area = 16π ≈ 50.2655. |
| Negative center: (x + 4)²/49 + (y + 2)²/16 = 1 | Center (−4, −2), foci (−4 ± √33, −2) | h = −4, k = −2. Vertices (−11, −2), (3, −2); co-vertices (−4, −6), (−4, 2). c = √(49 − 16) = √33 ≈ 5.7446, e = √33/7 ≈ 0.8207, area = 28π ≈ 87.9646. |
| Decimal axes: a = 2.5, b = 1.5 (vertical) | c = 2, e = 0.8, area 3.75π | Equation x²/2.25 + y²/6.25 = 1. c = √(6.25 − 2.25) = √4 = 2, so the foci are (0, ±2). e = 2/2.5 = 0.8, area = π(2.5)(1.5) = 3.75π ≈ 11.781. |
What Is an Ellipse?
An ellipse is the set of points whose distances to two fixed points, the foci, add up to the same constant. That constant is 2a, the length of the major axis: for every point P on the ellipse, PF₁ + PF₂ = 2a. It looks like a stretched circle, and a circle is the special case where the two foci merge at the center.
The longest diameter is the major axis, with the vertices at its ends; the shortest is the minor axis, with the co-vertices at its ends. The two axes cross at the center (h, k).
a, b and c: Semi-Axes and Focal Distance
a is the semi-major axis, half the length of the major axis, and b is the semi-minor axis, half the minor axis, with a ≥ b. The vertices are a units from the center along the major axis, and the co-vertices are b units along the minor axis.
The foci are c units from the center along the major axis, where c² = a² − b². For a = 5 and b = 3, c = √(25 − 9) = 4. Note the minus sign: c² = a² + b² belongs to the hyperbola, not the ellipse. Because c < a, the foci always lie inside the ellipse.
Horizontal or Vertical: Reading the Equation
Compare the two denominators. The larger one is a², and the variable it sits under tells you the direction of the major axis. In (x − 2)²/25 + (y + 1)²/9 = 1, 25 is under x, so the major axis is horizontal and the vertices are (2 ± 5, −1). In (x + 3)²/16 + (y − 4)²/36 = 1, 36 is under y, so the major axis is vertical and the vertices are (−3, 4 ± 6).
The signs inside the brackets are opposite to the center coordinates: (x + 3) means h = −3, and (y − 4) means k = 4.
Eccentricity, Area and Perimeter
Eccentricity e = c/a describes the shape. It is always at least 0 and less than 1 for an ellipse: e = 0 is a circle, and values close to 1 give a long, thin ellipse. Planetary orbits are ellipses with small eccentricities, which is why they look nearly circular.
The area is exactly πab; for a = 5 and b = 3, that is 15π ≈ 47.12 square units. The perimeter has no exact elementary formula, so it is approximated. Ramanujan’s formula P ≈ π[3(a + b) − √((3a + b)(a + 3b))] is accurate to within a tiny fraction of a percent for most ellipses, and it gives exactly 2πr for a circle.
How to Use the Ellipse Calculator
- Enter the center (h, k), the semi-major axis a and the semi-minor axis b, with a ≥ b > 0. Or switch to Equation and type a standard-form equation such as (x − 2)²/25 + (y + 1)²/9 = 1.
- Choose whether the major axis is horizontal or vertical. The equation preview updates as you type, putting a² under x or y.
- Read the graph: the center C, the vertices V₁ and V₂ on the major axis, the co-vertices B₁ and B₂ on the minor axis, and the foci F₁ and F₂.
- Check the properties panel for the axis lengths, foci, focal distance c, eccentricity e, area and approximate perimeter.
- Follow the step-by-step working for c = √(a² − b²), e = c/a, the key points and the area.
- If a = b, the calculator recognises a circle: c = 0, e = 0 and both foci sit at the center.
Frequently Asked Questions
What is an ellipse?
The set of points whose distances to two fixed points (the foci) always add up to the same value, 2a. It is an oval shape with two axes of symmetry; a circle is the special case with both foci at the center.
What is the standard equation of an ellipse?
(x − h)²/a² + (y − k)²/b² = 1 for a horizontal major axis, and (x − h)²/b² + (y − k)²/a² = 1 for a vertical one, where (h, k) is the center and a ≥ b > 0.
How do you find the center of an ellipse?
Read h and k from the brackets, changing their signs: (x − 2)² and (y + 1)² give the center (2, −1). If a term is just x² or y², that coordinate is 0.
How do you find the vertices of an ellipse?
Move a units from the center along the major axis: (h ± a, k) for a horizontal ellipse, (h, k ± a) for a vertical one. For (x − 2)²/25 + (y + 1)²/9 = 1, the vertices are (−3, −1) and (7, −1).
How do you find the foci of an ellipse?
Compute c = √(a² − b²), then move c units from the center along the major axis. For a = 5, b = 3 and center (2, −1), c = 4 and the foci are (−2, −1) and (6, −1).
What is the eccentricity of an ellipse?
e = c/a, a number from 0 up to (but not including) 1 that measures how stretched the ellipse is. For a = 5 and c = 4, e = 0.8. A circle has e = 0.
How do you calculate the area of an ellipse?
Area = πab, using the semi-axes (not the full axis lengths). For a = 5 and b = 3, the area is 15π ≈ 47.12 square units.
What is the difference between a horizontal and vertical ellipse?
A horizontal ellipse is wider than it is tall, with the larger denominator under the x-term; a vertical ellipse is taller than it is wide, with the larger denominator under the y-term.
What happens when the two semi-axes are equal?
The ellipse is a circle of radius a. Then c = √(a² − a²) = 0, so both foci coincide with the center, and the eccentricity is 0.
How do you find the perimeter of an ellipse?
There is no exact elementary formula. A standard approximation is Ramanujan’s: P ≈ π[3(a + b) − √((3a + b)(a + 3b))]. For a = 5 and b = 3, P ≈ 25.53.
What is the relationship between a, b and c in an ellipse?
c² = a² − b², so c = √(a² − b²). Equivalently, a² = b² + c²: the distance from a co-vertex to either focus equals a.
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Last updated: September 27, 2026.