Ellipse Perimeter & Area Calculator
Find the perimeter (circumference) of an ellipse with Ramanujan’s accurate approximation, plus its area, eccentricity and focal distance.
How it works
Area = π · a · b · Perimeter ≈ π(a + b)·(1 + 3h ÷ (10 + √(4 − 3h))), h = (a − b)² ÷ (a + b)² · Eccentricity = √(a² − b²) ÷ a
An ellipse’s area is delightfully simple — π times the two semi-axes — but its perimeter has no formula in elementary functions; it is an elliptic integral. Ramanujan’s second approximation, used here, is accurate to better than one part in ten million for ordinary ellipses and is the standard practical choice. The eccentricity measures how stretched the ellipse is: 0 is a circle, and values approaching 1 are long and thin.
Worked example
For semi-axes 5 and 3: area = π × 5 × 3 = 15π ≈ 47.1239. With h = (5 − 3)² ÷ (5 + 3)² = 4 ÷ 64 = 0.0625, the perimeter ≈ π(8)(1 + 0.1875 ÷ (10 + √3.8125)) ≈ 25.5269. The focal distance c = √(25 − 9) = 4, so the eccentricity is 4 ÷ 5 = 0.8 and the foci are 8 apart.
Sources: Ellipse — Wikipedia
Frequently asked questions
Why is there no exact perimeter formula?
The arc length of an ellipse leads to a complete elliptic integral of the second kind, which cannot be written with a finite combination of elementary functions. That is why we rely on excellent approximations like Ramanujan’s rather than a closed form.
How accurate is Ramanujan’s approximation?
Extremely — for a moderate ellipse like 5 by 3 the relative error is smaller than one part in ten million, and it stays superb until the ellipse becomes almost a line segment. For everyday geometry it is effectively exact.
What does eccentricity tell me?
It quantifies elongation. A circle has eccentricity 0; as the ellipse stretches, eccentricity rises toward 1. Planetary orbits, for instance, have small eccentricities (Earth’s is about 0.017), meaning they are nearly circular.
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