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Dual curve

Curve in the dual projective plane made from all lines tangent to a given curve

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Through Tangent, Conic section and Parametric equation

The construction of the dual curve is the geometrical underpinning for the Legendre transformation in the context of Hamiltonian mechanics.

Equations

Let f(x, y, z) = 0 be the equation of a curve in homogeneous coordinates on the projective plane. Let Xx + Yy + Zz = 0 be the equation of a line, with (X, Y, Z) being designated its line coordinates in a dual projective plane. The condition that the line is tangent to the curve can be expressed in the form F(X, Y, Z) = 0 which is the tangential equation of the curve.

At a point (p, q, r) on the curve, the tangent is given by

x∂f∂x(p,q,r)+y∂f∂y(p,q,r)+z∂f∂z(p,q,r)=0.

So Xx + Yy + Zz = 0 is a tangent to the curve if

X=λ∂f∂x(p,q,r),Y=λ∂f∂y(p,q,r),Z=λ∂f∂z(p,q,r).

Eliminating p, q, r, and λ from these equations, along with Xp + Yq + Zr = 0, gives the equation in X, Y and Z of the dual curve.

Conic

For example, let C be the conic ax2 + by2 + cz2 = 0. The dual is found by eliminating p, q, r, and λ from the equations

X=2λap,Y=2λbq,Z=2λcr,Xp+Yq+Zr=0.

The first three equations are easily solved for p, q, r, and substituting in the last equation produces

X22λa+Y22λb+Z22λc=0.

Clearing 2λ from the denominators, the equation of the dual is

X2a+Y2b+Z2c=0.

General algebraic curve

Consider a parametrically defined curve (x,y)=(x(t),y(t)), in projective coordinates (x,y,z)=(x(t),y(t),1). Its projective tangent line is a linear plane spanned by the point of tangency and the tangent vector, with linear equation coefficients given by the cross product:

(X,Y,Z)=(x,y,1)×(x′,y′,0)=(−y′,x′,xy′−yx′),

which in affine coordinates (X,Y,1) is:

X=−y′xy′−yx′,Y=x′xy′−yx′.

The dual of an inflection point will give a cusp and two points sharing the same tangent line will give a self-intersection point on the dual.

Dual of the dual

From the projective description, one may compute the dual of the dual:

(x(x′y″−y′x″),y(x′y″−y′x″),x′y″−y′x″)=(x,y,1)(x′y″−y′x″),

which is projectively equivalent to the original curve (x(t),y(t),1).

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