Consider the conic bundle of $\mathbb{P}^2(\mathbb R)$with matrix $$A(\lambda,\mu)=\begin{pmatrix} 0 & \mu & \mu \\ \mu & 0 & \lambda \\ \mu & \lambda & 0 \end{pmatrix}$$ This is a tangent conic bundle with base cycle $2A,B,C$ with $$A= \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} B= \begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix} C= \begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}$$ The fixed tangent in $A$ is $r:x_1+x_2=0$. I want to understand which points $P \in \mathbb{P}^2(\mathbb R)$ have the same polar for every conic in the bundle, i.e. for which $p_0,p_1,p_2 \in \mathbb R$ the plucker coordinates $$p=P^TA(\lambda,\mu)=\begin{pmatrix}p_0 &p_1 & p_2 \end{pmatrix} \begin{pmatrix} 0 & \mu & \mu \\ \mu & 0 & \lambda \\ \mu & \lambda & 0 \end{pmatrix}=\begin{pmatrix}\mu(p_1+p_2) & \mu p_0 +\lambda p_2 & \mu p_0 +\lambda p_1\end{pmatrix}$$ of the polar line with pole $P$ do not depend on $\lambda, \mu$. Obviously, one of these points is $A$: I would like to verify if there are other points of this kind.
For the polar to be defined, the conic must be non-degenerate i.e. $\lambda \neq 0$ and $\mu \neq 0$, hence
$$\begin{pmatrix}\mu(p_1+p_2) & \mu p_0 +\lambda p_2 & \mu p_0 +\lambda p_1\end{pmatrix}\stackrel{\text{x}}{=}\begin{pmatrix}p_1+p_2 & p_0 +\frac{\lambda}{\mu} p_2 & p_0 +\frac{\lambda}{\mu} p_1\end{pmatrix}$$ Let $k:=\frac{\lambda}{\mu}$. The polar of $P$ is the same for every conic in the bundle if and only if $$\begin{pmatrix}p_1+p_2 & p_0 +k_1 p_2 & p_0 +k_1 p_1\end{pmatrix}\stackrel{\text{x}}{=}\begin{pmatrix}p_1+p_2 & p_0 +k_2 p_2 & p_0 +k_2 p_1\end{pmatrix}$$ for every $k_1,k_2 \in \mathbb{R}\setminus \{0\}$. If $p_1+p_2 \neq 0$, the two remaining coordinates must be constant, implying $p_1=p_2=0$, contradiction. Such equation is true if and only if $$\begin{cases}p_1+p_2=0 \\(p_0+k_1p_2)(p_0+k_2p_1)=(p_0+k_1p_1)(p_0+k_2p_2) \end{cases}\iff \begin{cases}p_1+p_2=0 \\ p_0(p_2-p_1)(k_1-k_2)=0\end{cases}$$
The two projective points for which these equality hold for every $k_1,k_2$ are
$$A=\begin{pmatrix} 1 \\ 0 \\ 0\end{pmatrix},Q=\begin{pmatrix} 0 \\ 1 \\ -1\end{pmatrix}$$
I cannot figure out geometrically why the point $Q$ has a fixed polar. Can somebody give an intuition as to why this happens?
Further thinking about the problem brought me to this conclusion.
Let $\mathcal C$ be a tangent bundle of conics in $\mathbb{P}^2(\mathbb{R})$ with base cycle $2A,B,C$.Let $a$ be the fixed tangent through $A$. Let $D:=a~ \land(B~ \lor C)$. Since $D\in a$, its polar line $d$ passes through $A$ by reciprocity. Moreover, $X\in d$ because the cross-ratio $(BCDX)$ is harmonic. This implies that the polar $d$ is fixed, because it passes through $X$ and $A$ for every conic in the bundle.
In my problem, $D$ is exactly $\begin{pmatrix} 0 \\ 1 \\ -1\end{pmatrix}.$