Polarity on a Hyperboloid of one sheet 
Given a quadric $Q = \{v \in \mathbb{R}^n \mid \alpha(v,v) = 1\} \subset \mathbb{R}^n$, defined by a bilinear form $\alpha: \mathbb{R}^n\times\mathbb{R}^n \to \mathbb{R}^n$,  and an affine subspace $L \subset \mathbb{R}^n$.
  We define $L$'s polar subspace with respect to $Q$ as $L^\circ = \{ q \in \mathbb{R}^n \mid \alpha(p, q) = 1 \forall p \in L \}$. If $L$ contains just a single point $p$, then $L^\circ = \{p\}^\circ = \{q \in \mathbb{R} \mid \alpha(p,q) = 1\}$.

Now let $Q$ be the Hyperboloid of one sheet centered around the origin in $\mathbb{R}^3$, that is $Q = \{ x \in \mathbb{R}^3 \mid x_1^2 + x_2^2 - x_3^2 = 1\}$. The corresponding bilinear form is $\alpha(p,q) = p_1q_1 + p_2q_2 - p_3q_3$.

Choose a line $l \subset Q$ on the hyperboloid and a point $p \in l$. Prove, that the polar of $p$ contains $l$, e.g. $l \subset p^\circ$.

A well-known fact is that this Hyperboloid of one sheet can be generated either by rotating the hyperbola $x_1^2 - x_2^2 = 1$ around the $x_3$-axis or aquivalently by selecting a line on the hyperboloid and rotate this line around the $x_3$-axis. There are essentially two families of lines on the Hyperboloid of one sheet; my guess is that the polar of a point on $Q$ are the two lines that pass through it. However, i'm a at a loss on how to tackle this proof. Any help is appreciated!
 A: Call ${\bf p}(t) = {\bf p}+t{\bf v}$ a line in the hyperboloid. We then have $$\alpha({\bf p},{\bf p}) = \alpha({\bf p}(t),{\bf p}(t)) = 1$$ for all $t$. We want to check that $\alpha({\bf p},{\bf p}(t))=1$ for all $t$. But: $$\alpha({\bf p},{\bf p}(t)) = 1 + t\, \alpha({\bf p},{\bf v}),$$as you can readily see. 
Now I claim that $\alpha({\bf p},{\bf v}) = 0$. The condition $ \alpha({\bf p}(t),{\bf p}(t)) = 1$ for all $t$ gives $2t \alpha({\bf p},{\bf v}) + t^2\alpha({\bf v},{\bf v}) = 0$ for all $t$. Assume $t \neq 0$, divide everything by $t$, then plug $t = 1$, $t=2$ to get: $$\begin{cases} 2\alpha({\bf p},{\bf v}) + \alpha({\bf v},{\bf v}) = 0 \\  2\alpha({\bf p},{\bf v}) + 2\alpha({\bf v},{\bf v}) = 0  \end{cases} \implies \alpha({\bf v},{\bf v}) = 0 \implies \alpha({\bf p},{\bf v}) = 0,$$by subtracting both equations.
The space $\Bbb R^3$ furnished with that bilinear form $\alpha$ is actually a model used in special relativity for a flat spacetime, and that one-sheeted hyperboloid is the "unit sphere" of that ambient. One can see that $\alpha({\bf p},{\bf v}) = 0$ by noticing that $\bf v$ is tangent to the hyperboloid and $\bf p$ is on the hyperboloid - and a radius of a sphere is orthogonal to any tangent line to it at the point, for example.
