Eccentricity Calculator
An ellipse's eccentricity e measures how stretched it is, on a scale from 0 (a perfect circle) up to just under 1 (a very flattened ellipse). Enter the semi-major axis a (the longer half) and semi-minor axis b (the shorter half) to get e = √(1 − (b/a)²).
From Semi-Axes to Eccentricity
Every ellipse has a semi-major axis a (half of its longest diameter) and a semi-minor axis b (half of its shortest diameter), with a ≥ b. The distance from the center to each focus is c = √(a² − b²), and eccentricity is simply e = c/a, which simplifies to e = √(1 − (b/a)²).
Worked Example
For a = 5 and b = 3: c = √(25 − 9) = √16 = 4, so e = 4/5 = 0.8 — a visibly elongated ellipse. For a = 5 and b = 4: c = 3 and e = 0.6, a rounder shape. As b approaches a, e approaches 0 and the ellipse approaches a circle.
Why Eccentricity Matters
Eccentricity shows up beyond geometry class: planetary orbits are ellipses with eccentricity describing how circular or elongated the orbit is (Earth's is about 0.017, nearly circular; Halley's Comet's is about 0.967, extremely elongated). The same formula applies whether you're graphing a conic section or checking an orbital path.
Frequently Asked Questions
What does an eccentricity of 0 mean?
e = 0 means the semi-major and semi-minor axes are equal — the "ellipse" is actually a circle. The closer e gets to 1, the more stretched and flattened the ellipse looks.
What is linear eccentricity?
Linear eccentricity c is the distance from the ellipse's center to either focus, found from c = √(a² − b²). Eccentricity e is just that distance scaled by the semi-major axis: e = c/a.
Can eccentricity be negative or greater than 1 for an ellipse?
No. For any real ellipse, 0 ≤ e < 1 because b can never exceed a by definition — a is always the longer semi-axis. This calculator flags an input as invalid if b is entered larger than a.
How is this different from a hyperbola's eccentricity?
A hyperbola's eccentricity is always greater than 1, computed as e = √(1 + (b/a)²) instead of 1 minus that ratio, because its two branches curve away from each other rather than closing into a loop.
Do I need the full ellipse equation to use this?
No — only the two axis lengths. If you have the ellipse in the form x²/a² + y²/b² = 1, a and b are exactly the semi-major and semi-minor axes this calculator asks for.