Poles, polars and duality -the algebraic version

A plane conic has an equation of the form ax²+bxy +cy²+fx+gy+h=0.
In terms of homogeneous coordinates, this becomes ax²+bxy +cy²+fxz+gyz+hz²=0
which can be written as x^TMx=0 where x=(x,y,z), and M is a symmetric 3x3 matrix.

For a non-degenerate conic, M must be non-singular and have eigenvalues of different sign.

Note that, if a conic contains three (distinct) collinear points, then it must be degenerate.

Definition
If C: x^TMx=0 is a non-degenerate conic and U=[u] is any point,
then the algebraic polar of U with respect to C is the line u^TMx=0.

Note that, as M is non-singular, we cannot have u^TM=0, so that the line always exists.

A line L has an equation a^Tx=0. Now, u^TMx=0 and a^Tx=0 give the same line if and only if [u]=[M^-1a].
Thus L is the polar of a unique point U=[u].

Definition
If C: x^TMx=0 is a non-degenerate conic and L is any line,
then the algebraic pole of L with respect to C is the point U=[u] such that L has equation u^TMx=0.

Remark
If L has equation a^Tx=0, then, as we have seen, the pole of L is U=[M^-1a].

Theorem 1
If C: x^TMx=0 is a non-degenerate conic and U is any point on C,
then the algebraic polar of U with respect to C is the tangent to C at U .

Proof
Let U=[u], so that the algebraic polar is L: u^TMx=0.
Suppose that L cuts C again at V=[v].
As U lies on C, u^TMu=0.
As V lies on C, v^TMv=0.
As V lies on L, u^TMv=0.
Transposing this, and using the fact that M is symmetric, v^TMu=0.
Then, by expanding the left hand side, we see that
for any real a and b, (au+bv)^TM(au+bv)=0.
Thus W=[au+bv] lies on C.
If U and V are distinct, then C contains the collinear points U, V and W,
but then C would be degenerate.
Thus U=V, so that L meets C just once, and hence is a tangent.

We now show that the algebraic polar coincides with the geometrical polar.

Theorem 2
Suppose that C: x^TMx=0 is a non-degenerate conic and that VW is a chord of C passing through a fixed point U.
Then the tangents at V and W meet on the algebraic polar of U with respect to C.

Proof
By Theorem 1, the tangents at V=[v] and W=[w] have equations v^TMx=0 and w^TMx=0, respectively.
Suppose that these meet in T=[t]. By definition, T is on the geometrical polar of U.
As VW passes through U=[u], u = av + bw, for some a, b.
Then u^TMt=av^TMt+w^TMt=0 since T lies on each tangent.
Thus T lies on u^TMx=0,the algebraic polar of U.

It follows readily that the algebraic and geometric poles of a line coincide.

We now have the idea of duality defined in algebraic terms.
The fundamental result is that duality preserves incidence. This is now trivial.

Theorem 3 La Hire's Theorem
Suppose that C: x^TMx=0 is a non-degenerate conic and that U and V are any points.
Then U lies on the polar of V if and only if V lies on the polar of U.

Proof
Suppose that U=[u] and that V=[v].
Then each condition is clearly equivalent to u^TMv=0.

Theorem 4
Suppose that C: x^TMx=0 is a non-degenerate conic and that U is any point.
Then the pair of tangents to C from U has equation u^TMu x^TMx =u^TMx u^TMx .

Proof
Suppose that the tangent at V=[v] on C passes through U=[u].
Then v^TMu=0, and, transposing, u^TMv=0.
Also, as V is on C, v^TMv=0.
Any point T on the line UV may be written as T=[t], where t=au+bv, for some a, b.
Expanding each side, and using the equalities above,
we see that t satisfies the equation u^TMu x^TMx =u^TMx u^TMx .
Similarly, if the tangent at W passes through U, then each point of UW satisfies the equation.
Finally, the given equation is quadratic in x, so can represent no more than two lines.

Of course, by La Hire's Theorem, the chord VW in the above theorem is the polar of U.

Together, Theorems 1, 2 and 4 establish

Joachimsthal's Formulae Tell me about Joachimsthal
Suppose that C: x^TMx=0 is a non-degenerate conic. Then

• The polar of U=[u] has equation u^TMx=0.
• If U=[u] is on C, then the tangent to C at U has equation u^TMx=0.
• The pair of tangents to C from U=[u] has equation u^TMu x^TMx =u^TMx u^TMx .

We now show that the dual of a conic with respect to a fixed conic is actaully another conic.
There is a macro to draw the dual conic in the conics toolbar.

Theorem 5
Suppose that C: x^TMx=0 and D: x^TNx=0 are non-degenerate conics.
Then the dual of D with respect to C is the non-degenerate conic with equation x^TMN^-1Mx=0.

Proof
We think of D as the line-conic (the envelope of its tangents), {u^TNx=0 : u^TNu=0}.
The dual of u^TNx=0 with respect to C is the point V=[v], where v= M^-1(Nu).
Then u= N^-1(Mv).
Since [u] is on D, u^TNu=0, and so v^TMN^-1NN^-1Mv=0, i.e. v^TMN^-1Mv=0.
Thus the dual of D with respect to C has equation x^TMN^-1Mx=0, as required.