Q:

Smooth quadrics as quotient of $SO(2n+1)$

I've just started studying parabolic subgroup, and I read I can obtain smooth quadrics of the form
$$x_0x_{n+1}+\ldots+x_{n-1}x_{2n}+x_n^2=0$$
in $\mathbb{P}^{2n}$ as quotients of $SO(2n+1)$ by a appropriate parabolic subgroup, that is a subgroup containing a Borel subgroup. 
I know $SO(2n+1)=\{A\in M_{2n+1}(\mathbb{C})\mid A^t Q A=Q \text{ and } \det(A)=1\}$, where
$$Q=\begin{pmatrix} 0_{n,n} & 0 & I_n \\ 0 & 1 & 0 \\ I_n & 0 & 0_{n,n}
\end{pmatrix},$$
but to be honest I cannot go much further, I always struggle to translate the theory regarding algebraic groups in a concrete example.  Since this is very new to me I'd like to understand why this statement is true and how to find such a subgroup: also a proper reference may be fine. 

A:

Partial Answer:
To reiterate the ideas from the comments: we note that the set $C = \{p \in \Bbb P^{2n} : p^TQp = 0\}$ is invariant under the action of $SO(2n + 1)$. If we can show that $C$ is the orbit of the point $p = [1:0:\cdots:0]$ under $SO(2n+1)$, then it follows that our desired quadric is homeomorphic to the quotient $SO(2n+1)/SO(2n+1)_p$.
Equivalently, we want to show the following: for any $x \in \Bbb P^{2n}$ with $x^TQx = 0$, there exists an $A \in SO(2n+1)$ with $Ap = x$.  That is, we need an $A$ whose first column is $x$.
An idea for an indirect approach: I claim that there exists a size $2n+1$ square matrix $F$ such that $A \in SO(2n + 1)$ iff $B = FAF^{-1}$ satisfies $B^TB = I, \det(B) = 1$. With that established, it suffices to show that there is a matrix $B$ with $B(Fp) = B(Fx)$ and $B^TB = I$.

