Ellipse Area & Perimeter

Area and perimeter of an ellipse

Two semi-axes pin down an ellipse: the semi-major axis a and the semi-minor axis b, with a ≥ b. Area is the easy part, since it has a clean closed form, A = π·a·b. Set a = b and you get back the circle area π·r². The perimeter is where things get awkward. There's no elementary closed-form expression for it, so this tool falls back on Ramanujan's second approximation, P ≈ π[3(a + b) - √((3a + b)(a + 3b))], which stays well under 1 ppm of error for the eccentricities you'll normally run into. Speaking of which, the eccentricity is e = √(1 - b²/a²). At e = 0 you have a circle; push e toward 1 and the ellipse stretches out toward a parabolic shape. Plug in a = 5 and b = 3 and you get A = 15π ≈ 47.12 with P ≈ 25.53.

Applications

You'll find ellipses behind planetary orbits, since Kepler's first law has planets circling the Sun in ellipses with the Sun sitting at one focus. They also show up in the running lanes of athletics tracks (two straights capped by two semicircular ends, which engineers approximate as a stadium curve), in elliptical gears, in optical mirrors, and in the cross-section of ellipsoids when you move into 3D. Architects reach for elliptical arches too, since they spread load smoothly.

FAQ

Why is there no exact formula for the perimeter? Working out the arc length lands you on an elliptic integral of the second kind, and that one simply can't be written with elementary functions. For anything you'd do in engineering, Ramanujan's approximation is close enough to call exact.

What if a = b? Then the ellipse is just a circle of radius a, so A = π·a² and P = 2π·a, with eccentricity sitting at zero.

How do I find the foci? Both sit on the major axis, a distance c = √(a² - b²) out from the centre. So when the major axis runs horizontally, the foci land at (±c, 0).