Intersection of a polar hyper plane and a quadric is a conic Let $\mathbb{R} P^3$ be the real projective space of $\mathbb{R}^4$ and Q be a quadric defined as
$$ 
Q = \left\{\,\left[\,\left(\begin{smallmatrix} x_1 \\ x_2 \\ x_3 \\ x_4\end{smallmatrix}\right)\,\right] \in \mathbb{R}P^3 \mid x_1^2 + x_2^2 -x_3^2 -x_4^2 = 0 \,\right\}
$$
The associated symetric bilinear form of the the quadratic form mentioned in the quadric above is non-degenerate. By using the polarisations theorem I arrive at the result
$$B(x,y) = x_1y_1 + x_2y_2 - x_3y_3 - x_4y_4 \qquad \forall x,y \in \mathbb{R}^4$$
The polar hyperplane $E \subset \mathbb{R} P^3$ of a point $[\, p\,] \in \mathbb{R} P^3$ is defined as
$$ E = \{\,\left[\, x \,\right] \in \mathbb{R}P^3 \mid B(p,x) = 0 \,\} $$
Show that  


*

*$\mathcal{C} = Q \cap E \neq \emptyset$  

*$\mathcal{C}$ is a non-degenerate conic (conic section: circle, ellipse, parabola, hyperbola)


I have tried a couple of things - here is a GeoGebra plot of mine 
I'm not allowed to post a picture (<>_<>)
With an image, it is somehow clear that the intersection is surely not empty. Here are the problems 


*

*The quadric (even from the projective point of view) is a manifold of codimension 2 and the plane is a 2-dimensional subspace but this is a not enough to conclude the first question. 

*I have tried to construct a point that lies in the intersection. I could not find anything

*To prove that the intersection is a conic, I can equivalently show that $|C| \geq 5$. Since $\mathcal{C}$ is not empty, there is at least one point $[\, r\,] \in \mathcal{C}$ so do $-r$. I failed to find another 3 points.

*My idea of how to show, that the conic is non-degenerate is to show, that it does not contain a line. This can be achieved by showing, that no 3 points in $\mathcal{C}$ lie on the same line.


Since I failed at showing 1, 3, I am just out of ideas. Here is the honor mention
For any $[\,x\,] \in \mathcal{C}$ - the line through $[\,p\,],  [\,x\,]$ is a tangent of the quadric $Q$ at $[\, x \,]$ and only $[\, x\,]$. What worries me the most is, that none of these ideas involves the bilinear form (except for the last one). What am I missing?  
 A: Thank you @amd ! $B$ can be expressed as
\begin{align*}
    B\left(\left[\begin{matrix}
        x_1 \\
        x_2 \\
        x_3 \\
        x_4  \\
    \end{matrix}\right],
    \left[\begin{matrix}
        y_1 \\
        y_2 \\
        y_3 \\
        y_4   \\
    \end{matrix}\right]\right) &= 
    \left[\begin{matrix}
        y_1 & y_2 & y_3 & y_4 \\
    \end{matrix}\right]
    \left[\begin{matrix}
        1 & 0 & 0 & 0\\
        0 & 1 & 0 & 0\\
        0 & 0 & -1& 0\\
        0 & 0 & 0 & -1\\
    \end{matrix}\right]
    \left[\begin{matrix}
        x_1 \\
        x_2 \\
        x_3 \\
        x_4  \\
    \end{matrix}\right]
    = y^T A x
\end{align*}
Let $[\,p\,] \in \mathbb{R} P^3$ and denote the set
    $$E = \{\, [\,x\,] \in \mathbb{R} P^3 \mid B(x,p) =  0 \,\} $$
    as the polar hyper plane w.r.t $P$. Let $[\,v^{(1)}\,], [\,v^{(2)}\,], [\,v^{(3)}\,] \in E$  be any arbitrary linear independent points. For some $r_1, r_2, r_3 \in \mathbb{R}\setminus\{\,0\,\}$ the representative $x$ of any point $[\,x\,] \in E$ can be characterized as
    $$ x =  r_1 v^{(1)} + r_2 v^{(2)} + r_3v^{(3)} = \left[\begin{matrix} v^{(1)} &  v^{(2)} & v^{(3)} \end{matrix}\right] \left[\begin{matrix}  r_1 \\  r_2  \\ r_3\end{matrix}\right] = Mr $$
    Where $M$ denote the matrix on the left and $r$ the  vector on the right.
    We can write
    $$x^T A \,x = (Mr)^T A\, (Mr) = r^T (M^T A \,M) \,r$$
    where
    \begin{align*}
    B^\ast = M^T A \,M = 
    \left[\begin{matrix}
    B(v^{(1)}, v^{(1)}) &  B(v^{(1)}, v^{(2)}) &  B(v^{(1)}, v^{(3)})   \\
    B(v^{(2)}, v^{(1)}) &  B(v^{(2)}, v^{(2)}) &  B(v^{(2)}, v^{(3)})   \\
    B(v^{(3)}, v^{(1)}) &  B(v^{(3)}, v^{(2)}) &  B(v^{(3)}, v^{(3)})   \\
    \end{matrix}\right]
\end{align*}
    Since $B$ is non-degenerate and $v^{(1,2,3)}$ build a linear independent system, the matrix $B^\ast$ is symmetric with
    $$\textrm{det}(B^\ast) = \textrm{det}(M^T) \textrm{det}(A) \textrm{det}(M)   \neq 0$$
    representing a non-degenerate bilinear form in $\mathbb{R}^3$. $r$ can be seen as a representative of a point $[\,r\,] \in \mathbb{R} P^2$, therefore the set
    $$\mathcal{C} = \{\,[\,r\,] \in  \mathbb{R} P^2 \mid r^T B^\ast r = 0 \,\}$$
    defines a non-degenerate quadric in $\mathbb{R} P^2$ - a conic, especially $\mathcal{C} \neq \emptyset $
