PROJECTIVE GEOMETRY COURSE

§ 22: *Theory of pole and polar line*

Let k be a non-degenerated conic with homogeneous equation x^{t}Ax = 0 (so A is symmetric and regular).

Now define for this conic k a bijective mapping from the set of the points in *P*^{2} onto the set of the lines in *P*^{2} by

*P*:λp →^{γ} γ(*P*):λAp .

We have: *P* ∈ γ(*Q*) ↔ *Q* ∈ γ(*P*), because p^{t}Aq = 0 ↔ q^{t}Ap = 0

So, if we extend γ by γ(*P*) →^{γ} *P*, then we have a bijection from *P*^{2} to *P*^{2} that maps points to lines,
lines to points, preserves incidence and cross ratio (see O37 and O41), and is involutory, so a polarity.

Henceforth we denote a point and its image line with the same letter: X and x.

In projective coordinates, the polar line of *P*:λp becomes *p*:λAp, so the equation of the polar line is *p*: x^{t}Ap = 0. So we find the equation of the
polar line by 'honest distribution'

Therefore we see from the second half of §20 that if *P* lies on k then the polar line is tangent line. It will appear from the text below that if *P* is not lying on k then the polar
line doesn't go through *P*. So, apparently, *P* is selfconjugated if and only if *P* lies on k.

If we can draw from *P* two tangents to k, with contact points *Q*_{1} and *Q*_{2}, we see:

(*P* ∈ *q*_{1} → *Q*_{1} ∈ *p*), en (*P* ∈ *q*_{2} → *Q*_{2} ∈ *p*), so *p* =
*Q*_{1}*Q*_{2}.

Now look at the sketch below and the corresponding proposition thereafter.

*Proposition:* If we draw through a point *P*, not on k, a line *l* that intersects k in *R*_{1} and *R*_{2}, and a line *m* that intersects k in
*S*_{1} and *S*_{2}, then the diagonal points *P*, *Q* and *U* of the complete quadrangle
*R*_{1}*R*_{2}*S*_{1}*S*_{2} form a polar triangle.

*Proof:* According to the last proposition of §19, γ induces on *l* an involution *X* → *x.l* with fixed points *R*_{1} and *R*_{2}, and also
on *m* an involution with fixed points *S*_{1} and *S*_{2}.

According to the second proposition of §14, *P* and *p.l* are then harmonically situated with respect to *R*_{1} and *R*_{2}, and likewise
*P* and *p.m* with respect to *S*_{1} and *S*_{2}.

Hence, from the proposition of §12 it follows that *p* coincides with *UQ*.

Likewise we prove *u* = *PQ*, so *q* = *UP*.

Now there are some interesting consequences in the picture: the contact points from the tangents to k through *P* are lying on *QU*, the tangents in *R*_{1} and *S*_{2}
intersect on *q*, etcetera.

*O83* Construct the pole of a given line *l* with respect to a given conic k that has no point in common with *l*. Taking the intersection point of a line and a conic
is allowed in this problem.