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Ellipse Calculator

Area equals pi a b

Solution

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Calculate Ellipse Area from Axes

Use this form when the semi-major axis (a, half the longer diameter) and semi-minor axis (b, half the shorter diameter) are known. The formula generalizes circle area πr².

A = π a b

Calculate Ellipse Perimeter (Approximation)

Ramanujan's first approximation for the ellipse perimeter. Accurate to better than 0.4% for any eccentricity. The exact formula requires an elliptic integral — this gives a closed-form result good enough for engineering.

P ≈ π[3(a+b) − √((3a+b)(a+3b))]

Calculate Ellipse Semi-Major Axis from Area and Minor

Use this rearrangement when the area and minor axis are known and you need the semi-major axis.

a = A / (π b)

Calculate Ellipse Semi-Minor Axis from Area and Major

Use this rearrangement when the area and major axis are known and you need the semi-minor axis.

b = A / (π a)

How It Works

This ellipse calculator solves A = πab for area and uses Ramanujan's first approximation for the perimeter (the exact closed form requires an elliptic integral). Inverse solves cover semi-major axis from A and b, and semi-minor axis from A and a. Eccentricity e = √(1 − b²/a²) is always shown — 0 for a circle, approaching 1 as the ellipse becomes very elongated. Inputs accept any supported length and area units.

Example Problem

An elliptical garden bed has semi-major axis 5 m and semi-minor axis 3 m. Compute its area, perimeter, and eccentricity.

  1. Knowns: a = 5 m, b = 3 m
  2. Area: A = π a b = π · 5 · 3 = 15π ≈ 47.124 m²
  3. Perimeter (Ramanujan 1st): P ≈ π[3(a+b) − √((3a+b)(a+3b))] = π[24 − √252] ≈ 25.527 m
  4. Eccentricity: e = √(1 − b²/a²) = √(1 − 9/25) = √(16/25) = 0.8
  5. Sanity check (inverse): from A = 15π and b = 3, a = A/(πb) = 15π/(3π) = 5 m, recovering the major axis.

Eccentricity 0.8 is fairly elongated. Earth's orbital eccentricity is about 0.0167 (nearly circular); Halley's Comet is about 0.967 (very elongated).

When to Use Each Variable

  • Solve for Area — when both axes are known — garden beds, tracks, stadium shapes, planetary orbits.
  • Solve for Perimeter — when you need the elliptical circumference — track length, fencing, trim for elliptical shapes.
  • Solve for Semi-Major Axis — when area and minor axis are known and you need the major dimension.
  • Solve for Semi-Minor Axis — when area and major axis are known and you need the minor dimension.

Key Concepts

An ellipse is the set of all points where the sum of distances from two fixed points (the foci) is constant. The semi-major axis a is half the longer diameter; the semi-minor axis b is half the shorter diameter. When a = b the ellipse becomes a circle. Eccentricity e = √(1 − b²/a²) measures how much the ellipse deviates from circular: 0 for a circle, approaching 1 for highly elongated shapes. Planetary orbits, oval racetracks, and elliptical machine paths all use ellipse geometry.

Applications

  • Astronomy: planetary orbits (Kepler's first law: planets follow elliptical orbits with the Sun at one focus)
  • Engineering: elliptical gears, cams, and machine components with non-uniform motion
  • Architecture and design: oval rooms, stadium shapes, decorative arches
  • Sports: running tracks, where the curves are typically circular arcs joined by straights but elliptical tracks are also common

Common Mistakes

  • Using the full axis lengths instead of semi-axes — the formula needs HALF the diameters
  • Treating the perimeter formula as exact — Ramanujan's approximation is excellent but not exact (exact requires an elliptic integral, which is implemented numerically)
  • Confusing eccentricity with elongation — a circle has e = 0 and aspect ratio 1; e is non-linear in aspect ratio
  • Forgetting that eccentricity computation requires a ≥ b — if you enter axes in either order, this calculator picks the larger as the major automatically

Frequently Asked Questions

How do you calculate the area of an ellipse?

Multiply π by the two semi-axes: A = π a b. For an ellipse with a = 5 m and b = 3 m, A = 15π ≈ 47.12 m². The formula generalizes circle area πr², where r = a = b.

What is the formula for the perimeter of an ellipse?

There's no exact closed-form formula — the perimeter is given by an elliptic integral. Ramanujan's first approximation P ≈ π[3(a+b) − √((3a+b)(a+3b))] is accurate to better than 0.4% for any eccentricity and matches the exact value for circles.

What is the eccentricity of an ellipse?

e = √(1 − b²/a²), where a is the semi-major axis. Eccentricity 0 means a circle (a = b). As e → 1 the ellipse becomes very elongated. For a = 5 and b = 3, e = 0.8.

How is an ellipse different from a circle?

A circle is a special ellipse where the two semi-axes are equal (a = b = r). Ellipses have two different axis lengths and two foci; circles have one center point. Every circle is an ellipse, but only ellipses with a = b are circles.

How do you find the semi-major axis of an ellipse given the area?

Rearrange A = πab to a = A / (π b). For A = 47.12 m² and b = 3 m, a = 47.12 / (3π) ≈ 5 m.

What are the foci of an ellipse?

Two fixed points on the major axis, at distance c = √(a² − b²) from the center. Every point on the ellipse has a constant sum of distances to the two foci. For a = 5 and b = 3, c = √(25 − 9) = 4, so the foci are 4 m from the center along the major axis.

Why is the ellipse perimeter formula an approximation?

The exact circumference is C = 4a · E(e), where E is the complete elliptic integral of the second kind — there's no elementary closed form. Ramanujan derived two approximations in 1914; the second is even more accurate than the first one used here. For most engineering work, the first approximation is well within measurement precision.

Can an ellipse have an eccentricity greater than 1?

No — by definition the eccentricity of an ellipse is in [0, 1). e = 1 gives a parabola (degenerate case), and e > 1 gives a hyperbola. The square root inside √(1 − b²/a²) requires b ≤ a, which limits e to [0, 1).

Reference: Weisstein, Eric W. "Ellipse." MathWorld — A Wolfram Web Resource. https://mathworld.wolfram.com/Ellipse.html

Worked Examples

Garden Bed

How much soil does an elliptical garden bed need?

An elliptical garden bed measures 10 m across (a = 5 m) by 6 m across (b = 3 m). Compute its area to size the soil order.

  • Knowns: a = 5 m, b = 3 m
  • Formula: A = π a b
  • A = π · 5 · 3 = 15π ≈ 47.12 m²

Area = 47.12 m² (about 507 ft²)

An elliptical bed of the same major/minor diameters as a rectangle has π/4 ≈ 0.785 of the rectangle's area — useful for comparison.

Running Track

What is the perimeter of an elliptical 400 m track?

A running track is an ellipse with semi-major axis 60 m and semi-minor axis 40 m. Find the lap length.

  • Knowns: a = 60 m, b = 40 m
  • Formula: P ≈ π[3(a+b) − √((3a+b)(a+3b))]
  • P ≈ π[300 − √(220 · 180)] = π[300 − √39600] = π[300 − 198.99]
  • P ≈ π · 101.01 ≈ 317.36 m

Lap length ≈ 317.4 m

Real running tracks use straight sections joined by semicircles, not pure ellipses. A pure-ellipse 400 m track would have different dimensions.

Inverse Solve

What semi-minor axis does a 100 m² ellipse with 6 m major need?

An elliptical pond must enclose 100 m² with semi-major axis 6 m. Find the required semi-minor axis.

  • Knowns: A = 100 m², a = 6 m
  • Formula: b = A / (π a)
  • b = 100 / (6π) ≈ 5.305 m

Semi-minor axis ≈ 5.305 m

Note b < a (5.3 < 6), so the ellipse is only mildly elongated — eccentricity ≈ 0.47. For a perfect circle of the same area, r = √(100/π) ≈ 5.642 m.

Ellipse Formulas

An ellipse is defined by two axes: the semi-major axis a (half the longer diameter) and the semi-minor axis b (half the shorter diameter).

Area equals pi a bPerimeter approximately equals pi times the quantity 3 a plus b minus the square root of 3 a plus b times a plus 3 bSemi-major axis a equals A over pi bSemi-minor axis b equals A over pi a
Ellipse with semi-major axis a (horizontal) and semi-minor axis b (vertical)ab

Where:

  • A — area (m², ft², in²)
  • P — perimeter (Ramanujan 1st approximation; exact requires an elliptic integral)
  • a — semi-major axis: half the LONGER diameter
  • b — semi-minor axis: half the SHORTER diameter
  • e — eccentricity = √(1 − b²/a²); 0 for circles, approaches 1 for elongated ellipses

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