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Ellipse Calculator

Calculate area, perimeter, eccentricity, and foci of any ellipse

Area

47.12

Perimeter

25.53

Eccentricity

0.8000

The longest radius of the ellipse (half of the major axis).

The shortest radius of the ellipse (half of the minor axis). If a = b, the ellipse is a circle.

Area

47.1239

sq units

Perimeter

25.5270

units (approx)

Eccentricity (e)

0.800000

Shape

Highly elongated

0 (circle)1 (parabola)
Linear Eccentricity (c)

4.0000

Flattening (f)

0.400000

Foci Coordinates

(4.0000, 0)

Focus 1

(-4.0000, 0)

Focus 2

Semi-Latus Rectum

1.8000

Distance from focus to ellipse along perpendicular to major axis

Formulas Used

Ellipse Area

A = π × a × b

The area equals pi times the product of the semi-major and semi-minor axes.

Where:

a= Semi-major axis (longest radius)
b= Semi-minor axis (shortest radius)
π= Pi (≈ 3.14159)

Ramanujan Perimeter Approximation

P ≈ π(3(a+b) - √((3a+b)(a+3b)))

Srinivasa Ramanujan's remarkably accurate approximation for the circumference of an ellipse.

Where:

a= Semi-major axis
b= Semi-minor axis
π= Pi (≈ 3.14159)

Eccentricity

e = √(1 - b²/a²)

Eccentricity measures the elongation of the ellipse, from 0 (circle) to approaching 1 (line).

Where:

e= Eccentricity (0 ≤ e < 1)
a= Semi-major axis
b= Semi-minor axis

Example Calculations

1Standard ellipse (a=5, b=3)

Inputs

Semi-Major Axis (a)5
Semi-Minor Axis (b)3

Result

Area47.1239
Perimeter25.5270
Eccentricity0.8
Foci(±4, 0)

Area = π × 5 × 3 = 15π ≈ 47.1239. Perimeter ≈ π(24 - √252) ≈ 25.5270. Eccentricity = √(1 - 9/25) = √(16/25) = 0.8. Foci at c = √(25-9) = 4.

2Circle (a=10, b=10)

Inputs

Semi-Major Axis (a)10
Semi-Minor Axis (b)10

Result

Area314.1593
Perimeter62.8319
Eccentricity0
ShapeCircle

When a = b = 10, this is a circle. Area = π × 100 ≈ 314.16. Perimeter = 2π × 10 ≈ 62.83. Eccentricity = 0, foci both at center.

3Elongated ellipse (a=12, b=2)

Inputs

Semi-Major Axis (a)12
Semi-Minor Axis (b)2

Result

Area75.3982
Perimeter49.7836
Eccentricity0.9860
Foci(±11.8322, 0)

Area = π × 12 × 2 = 24π ≈ 75.40. Perimeter ≈ 49.78 (Ramanujan). Eccentricity = √(1 - 4/144) = √(140/144) ≈ 0.986. This is a highly elongated ellipse.

Frequently Asked Questions

Q

How do you calculate the area of an ellipse?

The area of an ellipse is calculated with the formula A = π × a × b, where a is the semi-major axis and b is the semi-minor axis. For an ellipse with a = 5 and b = 3, the area is π × 5 × 3 = 15π ≈ 47.1239 square units. When a = b, this reduces to πr², the circle area formula.

  • Formula: A = π × a × b
  • a = 5, b = 3: area = π × 15 ≈ 47.12 sq units
  • a = 10, b = 10: area = 100π ≈ 314.16 (circle)
  • Area is always less than the circumscribed rectangle (4ab)
  • When a = b, formula becomes πr² (circle)
Axes (a, b)AreaShape
a=5, b=347.12Moderate ellipse
a=10, b=262.83Highly elongated
a=7, b=7153.94Circle
a=8, b=6150.80Nearly circular
Q

What is the Ramanujan approximation for ellipse perimeter?

There is no exact closed-form formula for an ellipse perimeter. Ramanujan's approximation P ≈ π(3(a+b) - √((3a+b)(a+3b))) is remarkably accurate, with errors less than 0.04% for most ellipses. For a = 5, b = 3: P ≈ π(24 - √(18 × 14)) ≈ 25.53.

  • Exact perimeter requires elliptic integrals (no simple formula)
  • Ramanujan formula: π(3(a+b) - √((3a+b)(a+3b)))
  • Error is less than 0.04% for eccentricity < 0.95
  • a = 5, b = 3: P ≈ 25.53 (Ramanujan approximation)
  • For circles (a = b): formula gives exact 2πr
Axes (a, b)Ramanujan PEccentricity
a=5, b=325.530.80
a=10, b=1062.830 (circle)
a=10, b=140.610.995
a=8, b=644.210.661
Q

What is the eccentricity of an ellipse?

Eccentricity (e) measures how elongated an ellipse is, ranging from 0 (circle) to approaching 1 (very flat). The formula is e = √(1 - b²/a²). An eccentricity of 0.8 means the ellipse is moderately elongated. Earth's orbit has e ≈ 0.0167, nearly circular.

  • e = 0: perfect circle (a = b)
  • e = 0.0167: Earth's orbit (nearly circular)
  • e = 0.8: moderately elongated ellipse
  • e approaches 1: extremely flat, nearly a line segment
  • Formula: e = √(1 - b²/a²)
ShapeEccentricityExample
Circle0a = b
Earth orbit0.0167a = 149.6M km
Moderate ellipse0.6-0.8a=5, b=3 (e=0.8)
Very elongated0.9-0.99a=10, b=1 (e=0.995)
Q

Where are the foci of an ellipse located?

The foci are two special points on the major axis at distance c = √(a² - b²) from the center. For a = 5, b = 3: c = √(25-9) = √16 = 4, so foci are at (±4, 0). The sum of distances from any point on the ellipse to both foci equals 2a (the major axis length).

  • Foci lie on the major axis at distance c from center
  • c = √(a² - b²) = √(25-9) = 4 for a=5, b=3
  • Foci at (+4, 0) and (-4, 0)
  • Key property: d1 + d2 = 2a for any point on ellipse
  • For a circle (a = b): both foci are at the center (c = 0)
Axes (a, b)c ValueFoci Positions
a=5, b=34(±4, 0)
a=10, b=68(±8, 0)
a=13, b=512(±12, 0)
a=7, b=70(0, 0) [circle]
Q

What is flattening and semi-latus rectum?

Flattening (f = 1 - b/a) measures how squashed the ellipse is, ranging from 0 (circle) to approaching 1. The semi-latus rectum (l = b²/a) is the distance from a focus to the ellipse measured perpendicular to the major axis. Both are used extensively in orbital mechanics.

  • Flattening f = 1 - b/a: 0 for circle, near 1 for flat
  • Earth: f = 1/298.257 (very slightly flattened)
  • Semi-latus rectum l = b²/a
  • a=5, b=3: f = 0.4, l = 9/5 = 1.8
  • Used in Kepler orbital equations and geodesy
Axes (a, b)FlatteningSemi-Latus Rectum
a=5, b=30.41.8
a=10, b=100 (circle)10
a=10, b=20.80.4
a=8, b=60.254.5

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Last Updated: Mar 9, 2026

This calculator is provided for informational and educational purposes only. Results are estimates and should not be considered professional financial, medical, legal, or other advice. Always consult a qualified professional before making important decisions. UseCalcPro is not responsible for any actions taken based on calculator results.

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