What is the equation of the Kampyle of Eudoxus?

Cartesian equation: x^4 = a^2(x^2 + y^2); Polar equation: r = a sec^2θ; Parametric equations: x = a sec t, y = a tan t sec t. a is the distance from the centre to each vertex (±a, 0).

Classic Cartesian curves

Kampyle of Eudoxus

x⁴ = a²(x² + y²): two branches opening outward from (±a, 0), traditionally linked to Eudoxus and the doubling of the cube.

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a (vertex distance)1

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Formulas and key values

Cartesian equation: $x^4 = a^2(x^2 + y^2)$
Polar equation: $r = a\sec^2\theta$
Parametric equations: $x = a\sec t$ $y = a\tan t\sec t$

a is the distance from the centre to each vertex (±a, 0).

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01 · Definition

What is the Kampyle of Eudoxus?

The kampyle (Greek for “curved”) is traditionally credited to Eudoxus of Cnidus in the fourth century BC, in connection with doubling the cube; the ancient sources are thin, so the attribution is conventional rather than certain. Its equation is the quartic x⁴ = a²(x² + y²).

In polar form it is simply r = a·sec²θ, and with x = a·sec t, y = a·tan t·sec t it is easy to plot. Each branch starts at a vertex (±a, 0) and opens outward, growing roughly like y ≈ ±x²/a far from the origin.

02 · Properties

Key properties

Polar form r = a·sec²θ.
Parametric form x = a·sec t, y = a·tan t·sec t.
Vertices at (±a, 0); no points with |x| < a.
Far from the origin, y ≈ ±x²/a.
Symmetric about both axes.

03 · Questions

Kampyle of Eudoxus: common questions

What is the equation of the Kampyle of Eudoxus?: Cartesian equation: x^{^4} = a^{^2}(x^{^2} + y^{^2}); Polar equation: r = a sec^{^2θ}; Parametric equations: x = a sec t, y = a tan t sec t. a is the distance from the centre to each vertex (±a, 0).