I'm stuck on a small detail in Proposition 1.2.8 in Geiges' Introduction to Contact Topology.
Let $C$ be a conic in $\mathbb{R}P^2$ given by $q^tAq=0$, where $A$ is a nonsingular, symmetric 3x3 matrix. Assume $A$ is index 2, that is, $C$ is an ellipse.
Define the correlation $\varphi:\mathbb{R}P^2\rightarrow(\mathbb{R}P^2)^\ast$ by $\varphi(q)=q^tA$ and $\varphi^*:(\mathbb{R}P^2)^\ast\rightarrow\mathbb{R}P^2$ by $\varphi(p)=A^{-1}p^t$.
If $x$ is a point on $C$, we would like to show that $\varphi(x)$ is a line tangent to $C$ at $x$. Observe that $\varphi(x)x=x^tAx=0$. Geiges says, "[therefore] $x$ lies on its polar $\varphi(x)$." This is the point I am confused about. Here's why:
One can imagine $\mathbb{R}P^2$ as a northern hemisphere in $\mathbb{R}^3$ (with antipodal boundary points identified). Given a point $p$ on this hemisphere, consider the tangent plane to the sphere at $p$, and now parallel transport it to intersect the origin of $\mathbb{R}^3$. The points of intersection of this plane with the hemisphere should be exactly the set $p^\perp=\{q\in\mathbb{R}P^2\mid q^tp=0\}$. My reasoning for this is that the lines through the origin of the tangent plane are exactly the lines perpendicular to the line through $p$ and the origin of $\mathbb{R}^3$, so moving the tangent plane to the origin of $\mathbb{R}^3$ does not change the vectors, and we projectivize them by looking at the intersections with the hemisphere.
The problem is that, by this construction, it's obvious that $p\not\in p^\perp$, so how could $x$ lie on the polar $\varphi(x)$ if $\varphi(x)x=0$?
If I misinterpreted something, let me know: the presentation above is a little foreign to me.
It seems to me that we're discussing the conic $C=\{q\mid q^tAq=0\}$, and that the pole-polar correlation in question is the one where $q\mapsto (q^tA)^\perp$. That is, the right hand side is a plane in $\Bbb R^3$ that determines a projective line in $\Bbb R P^2$.
If $q\in C$, then $q^tAq=(q^tA)q=0$, but that is saying that $q\in (q^tA)^\perp$, so that $q$ lies on its own polar.