Given the vertex of a parabola, finding the self-polar triangle Given the point $R=[1:-4:0]$ of the parabola $\mathscr{P}: x^2+y^2-2xy+4x+4y=0$, find a self-polar triangle for $\mathscr{P}$ with vertex in $R$.
Now, a self-polar triangle is composed by a point $P \in r_{\infty} \setminus \mathscr{P}$, then we find the polar line $Pol_{\mathscr{P}}(P) = P^tCX=0$, where $C$ is the matrix which represents the conic section and $X$ the general point $[x_0:x_1:x_2]$. Then we calculate $Pol_{\mathscr{P}}(P) \setminus r_{\infty}$ finding a point $Q$, if $P$ isn't orthogonal to Q, we must change $P$ with a $P'\in r_{\infty} \setminus \mathscr{P}$ s.t. $P'Q=0$ and so $Pol_{\mathscr{P}}(P')$ will be the axis of symmetry of $\mathscr{P}$. Finally $Pol_{\mathscr{P}}(Q) \cap \mathscr{P}$ gives the vertex $R$.
How to do this process having $R$?
 A: $R$ lies on $\mathscr P$, so its polar line is the tangent to $\mathscr P$ at $R$. Another vertex of the triangle must lie on this line, but no other point on this line lies on its own polar, which means that the third vertex of the triangle must do so. Therefore, the third vertex, too, lies on $\mathscr P$. So, choose any other point $Q$ on $\mathscr P$ as a vertex of the self-polar triangle that you are constructing. The remaining vertex is the pole of the line through $P$ and $Q$, which is the intersection of the tangents to $\mathscr P$ at those points.  
From the comments to your question, this is a step in constructing an affinity that converts $\mathscr P$ into standard form. Assuming, then, that this standard form is $y^2=4px$, you can also construct such an affinity from three known lines: the line at infinity, which is fixed by any affinity, the tangent at $R$, which will be mapped to the $y$-axis, and the line through $R$ parallel to $\mathscr P$’s axis, which will be mapped to the $x$-axis. The rows of the matrix of this affinity are just the homogeneous representations of these lines. If you adjust their signs and scale factors appropriately, you can always arrange for the coefficient of $y^2$ in the transformed equation to be $1$ and the coefficient of $x$ to be negative.
