SOLVETUTORMATH SOLVER

Instrument MI-01-199 · Mathematics

Ellipse Calculator

An ellipse is a stretched circle with two different radii. Enter both semi-axes, and this sheet returns the area and an accurate perimeter estimate.

Instrument MI-01-199
Sheet 1 OF 1
Rev A
Verified
Type 05 — Geometry SER. 2026-01199

Area

47.12388980

A = πab

25.52699886 Perimeter (Ramanujan approximation)
The working Every figure verified twice
  1. area = π·5·3 = 47.12388980
  2. perimeter = π·(5 + 3)·(1 + 3·((5 − 3) ⁄ (5 + 3))^2 ⁄ (10 + √(4 − 3·((5 − 3) ⁄ (5 + 3))^2))) = 25.52699886
Worksheet log
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How this instrument works

An ellipse generalizes a circle by allowing two different radii instead of one: a semi-major axis a (the longer reach) and a semi-minor axis b (the shorter reach). Its area follows directly from a circle's own πr² formula with the two different radii multiplied instead of one squared: A = πab — set a and b equal and it reduces exactly to the familiar circle area formula.

The perimeter is a genuinely harder problem. Unlike the area, an ellipse's exact perimeter has no elementary closed-form formula at all — it requires an elliptic integral, a function that can't be written using ordinary algebraic operations. This calculator instead uses a highly accurate approximation devised by the mathematician Srinivasa Ramanujan, which stays within a tiny fraction of a percent of the true value across the whole range of possible ellipse shapes.

The more elongated an ellipse becomes — the bigger the gap between a and b — the more the perimeter departs from what a naive average of the two circle formulas (using a or using b) would suggest. Ramanujan's formula accounts for that elongation through a correction term built from the ratio (a−b) ⁄ (a+b), which vanishes entirely when a equals b, correctly collapsing the whole approximation back to a plain circle's exact circumference.

A=πabA = \pi a bPπ(a+b)(1+3h10+43h)P \approx \pi(a+b)\left(1+\frac{3h}{10+\sqrt{4-3h}}\right)
a — the semi-major axis (longer radius); b — the semi-minor axis (shorter radius); A — area; P — perimeter, via Ramanujan's approximation.
  • Enter the ellipse's longer radius into the Semi-major axis (a) field.
  • Enter its shorter radius into the Semi-minor axis (b) field.
  • Read Area: computed exactly as πab.
  • Read Perimeter: computed via Ramanujan's approximation, accurate to a small fraction of a percent for any ellipse shape.

Worked example — semi-axes 5 and 3

An ellipse has a semi-major axis of 5 and a semi-minor axis of 3. Its area is π × 5 × 3 = 15π ≈ 47.12 square units, and its perimeter, via Ramanujan's approximation, comes out to approximately 25.53 units — a figure no simple formula could produce exactly, since a genuine elliptic integral sits underneath it.

Set both axes equal, 4 and 4: the ellipse becomes a plain circle of radius 4, area 16π ≈ 50.27, and the approximation correctly reduces to exactly 2π(4) ≈ 25.13, the ordinary circumference formula — Ramanujan's correction term vanishing entirely once a equals b.

Questions

What is the formula for an ellipse's area?

A = πab, where a is the semi-major axis and b is the semi-minor axis. This is an exact formula, a direct generalization of a circle's πr² with two independent radii multiplied instead of one squared.

Why is the perimeter only an approximation?

An ellipse's exact perimeter has no elementary closed-form formula — it requires evaluating an elliptic integral, a type of function that cannot be expressed using ordinary algebraic operations. Ramanujan's approximation gets extremely close (within a small fraction of a percent) without needing that integral.

What happens when the two semi-axes are equal?

The ellipse becomes a plain circle. The area formula reduces exactly to πr², and Ramanujan's perimeter approximation reduces exactly to 2πr, the ordinary circle circumference, since the correction term built from (a−b)⁄(a+b) vanishes completely when a equals b.

How accurate is Ramanujan's approximation?

Extremely — it stays within roughly 0.04% of the true perimeter for even highly elongated ellipses, and gets closer still for ellipses nearer to circular. For nearly every practical purpose, the approximation is indistinguishable from the exact elliptic-integral result.

Can the semi-minor axis be larger than the semi-major axis?

No — by definition, a (the semi-major axis) is the longer of the two radii and b (the semi-minor axis) is the shorter. If the labels were swapped, the formulas would still work, but the convention keeps a as the larger value.

References