What is the equation of the Trisectrix of Maclaurin?

Cartesian equation: 2x(x^2 + y^2) = a(3x^2 - y^2); Polar equation: r = a (sin 3θ)/(sin 2θ). a sets the size: the loop reaches x = 3a/2 and the asymptote is x = −a/2.

What is the area of the Trisectrix of Maclaurin?

Area (inside the loop): A = ((3√3)/4)a^2. a sets the size: the loop reaches x = 3a/2 and the asymptote is x = −a/2.

Polar curves

Trisectrix of Maclaurin

The cubic 2x(x² + y²) = a(3x² − y²): a loop and an asymptote, studied by Maclaurin in 1742 for trisecting angles.

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Formulas and key values

Cartesian equation: $2x(x^2 + y^2) = a(3x^2 - y^2)$
Polar equation: $r = a\,\frac{\sin 3\theta}{\sin 2\theta}$
Asymptote: $x = -\frac{a}{2}$
Area · inside the loopIn the graph ≈ 1.299: $A = \frac{3\sqrt3}{4}a^2$

a sets the size: the loop reaches x = 3a/2 and the asymptote is x = −a/2.

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01 · Definition

What is the Trisectrix of Maclaurin?

Colin Maclaurin studied this cubic in his Treatise of Fluxions (1742) as a tool for trisecting an angle, one of the problems that cannot be solved with straightedge and compass alone. A curve that is allowed to be drawn in advance gets around that limit.

In polar form, r = a·sin 3θ / sin 2θ, the factor sin 3θ is the trisection at work. The curve crosses itself at the origin, closes a loop that reaches x = 3a/2, and has the vertical asymptote x = −a/2.

02 · Properties

Key properties

Polar form: r = a·sin 3θ / sin 2θ.
Node at the origin; the loop reaches x = 3a/2 on the x-axis.
Vertical asymptote x = −a/2.
The loop encloses area 3√3a²/4.
Maclaurin used it to trisect angles (1742).

03 · Questions