What is the equation of the Lemniscate of Gerono?

Cartesian equation: x^4 = a^2(x^2 - y^2); Parametric equations: x = a cos t, y = a sin t cos t. a is the half-width: the curve spans −a ≤ x ≤ a.

What is the area of the Lemniscate of Gerono?

Area (both lobes): A = (4/3)a^2. a is the half-width: the curve spans −a ≤ x ≤ a.

Parametric curves

Lemniscate of Gerono

The figure eight x⁴ = a²(x² − y²), drawn by x = a·cos t, y = a·sin t·cos t: two lobes of total area 4a²/3.

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Formulas and key values

Cartesian equation: $x^4 = a^2(x^2 - y^2)$
Parametric equations: $x = a\cos t$ $y = a\sin t\cos t$
Area · both lobesIn the graph ≈ 1.333: $A = \frac{4}{3}a^2$

a is the half-width: the curve spans −a ≤ x ≤ a.

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01 · Definition

What is the Lemniscate of Gerono?

Named after the French mathematician Camille-Christophe Gerono, this is the simplest figure eight to draw: x = a·cos t and y = a·sin t·cos t, which is a Lissajous figure with frequencies 1 and 2. Eliminating t gives the quartic x⁴ = a²(x² − y²).

Unlike the lemniscate of Bernoulli, it is not defined by foci; its lobes are taller and rounder. Each lobe reaches height a/2, and the two together enclose exactly 4a²/3.

02 · Properties

Key properties

Cartesian form x⁴ = a²(x² − y²).
Same as the 1 : 2 Lissajous figure x = cos t, y = sin 2t / 2, scaled by a.
Total enclosed area 4a²/3.
Each lobe peaks at height a/2, at x = ±a/√2.
Crosses itself at the origin with tangents y = ±x.

03 · Questions