What is the equation of the Lemniscate of Bernoulli?

Polar equation: r^2 = a^2 cos 2θ; Cartesian equation: (x^2 + y^2)^2 = a^2(x^2 - y^2). a is the distance from the center to each tip; ϖ ≈ 2.6221 is the lemniscate constant.

Polar curves

Lemniscate of Bernoulli

Q: What is the area of the Lemniscate of Bernoulli?

Area (both loops): A = a^2. a is the distance from the center to each tip; ϖ ≈ 2.6221 is the lemniscate constant.

The figure-eight curve r² = a²·cos 2θ — the set of points whose distances to two foci multiply to a constant.

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Formulas and key values

Polar equation: $r^2 = a^2\cos 2\theta$
Cartesian equation: $(x^2 + y^2)^2 = a^2(x^2 - y^2)$
PerimeterIn the graph ≈ 5.244: $L = 2\varpi a \approx 5.2441\,a$
Area · both loopsIn the graph = 1: $A = a^2$
Foci: $\left(\pm\frac{a}{\sqrt{2}},\ 0\right)$

a is the distance from the center to each tip; ϖ ≈ 2.6221 is the lemniscate constant.

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01 · Definition

What is the Lemniscate of Bernoulli?

The lemniscate of Bernoulli is the locus of points whose distances to two foci have a constant product — the multiplicative cousin of the ellipse, which fixes the sum. Jacob Bernoulli described it in 1694 and named it after the Latin lemniscus, a hanging ribbon. Its polar form r² = a²·cos 2θ shows why the curve only exists where cos 2θ ≥ 0: two symmetric lobes meeting at the origin.

The curve became far more than a pretty figure eight: attempts to compute its arc length led Fagnano and Euler to the lemniscatic integrals, a direct ancestor of the theory of elliptic functions.

02 · Properties

Key properties

Total enclosed area: a² (both lobes together).
Foci sit at (±a/√2, 0); the distance product equals a²/2.
The origin is a crossing point where the two tangents are y = ±x.
It is the inverse of the rectangular hyperbola with respect to its center.
Arc length involves the lemniscate constant ϖ ≈ 2.6221, computed via elliptic integrals.

03 · Questions