# Conic Envelope of a Double Line Conic

In the real projective plane $$\mathbb{R}\mathbb{P}^2$$ where points are represented by homogenous coordinates $$[x,y,z]$$, there is a degenerate conic consisting of a real double line. This conic is of equation $$x^2=0$$. I am looking to determine its corresponding conic envelope.

In Jurgen Richter-Gebert's Perspectives on Projective Geometry, he says there are two such envelopes, one consiting of all lines passing through one point, and the other consisting of all lines passing through two points. Is this correct, or is it a misunderstanding?

• What do you mean in this context by "envelope of $x^2=0$", (which is nothing else that the $y$ axis) ? Usualy this term is used as "envelope of a family of curves..." – Jean Marie Aug 21 at 12:20
• @JeanMarie Jurgen Richter-Gebert defines it as the set of lines $\left\{ l \in \mathcal{L}_{\mathbb{R}\mathbb{P}^2} \mid l^T A^\triangle l = 0 \right\}$, where $A^\triangle$ is the transposed comatrix of $A$, and $A$ is the matrix of the original conic. – Shurik Goyal Aug 21 at 12:54
• I understand : the conic curve is the envelope of these lines. – Jean Marie Aug 21 at 13:28
• You will agree with me that the set of tangent lines to a line $L$ is reduced to this unique line. – Jean Marie Aug 21 at 13:30
• I have no idea... In such degenerate cases, one can as well stick to definitions... sometimes better than intuition... – Jean Marie Aug 21 at 14:53

Definition 9.5. A primal/dual pair of conics is given by a pair $$(A,B)$$ of real symmetric nonzero $$3\times3$$ matrices such that there exists a factor $$\lambda\in\mathbb R$$ with $$AB=\lambda E$$.
Note that $$\lambda$$ can be zero, which is in fact the case for a double line. For your canonical double line $$x^2=0$$, then, we have $$A=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}$$ and any nonzero matrix of the form $$B=\begin{bmatrix}0&0&0\\0&a&b\\0&b&c\end{bmatrix}$$ is dual to it. A rank-one $$B$$ (which occurs when $$b^2=ac$$) represents a double point, while a rank-two $$B$$ represents two distinct points. In the latter case, you can recover the two points by splitting the conic, using the line $$(1,0,0)^T$$ as the “intersection point” for the algorithm.