Show that Lorentzian metric restricts to Riemannian metric on hyperbolic space 
Define the Lorentzian metric $\langle \ ,\ \rangle$ on $\mathbb{R}^{n+1}$ by $\langle x, x \rangle = - x_0^2 + x_1^2 + \cdots + x_n^2$.  Let 
  $$\mathbb{H}^n = \{ x \in \mathbb{R}^{n+1} | \langle x , x \rangle = -1, \ x_0 > 0 \}$$
  Prove that $\langle \ , \ \rangle$ pulls back to a Riemannian metric on $\mathbb{H}^n$

Preliminary question / remark: Does specifying a quadratic form as above uniquely specify a metric?  Although I am familiar with the usual Lorentzian metric where $g_{11} = -1$, $g_{ii} = 1$ for $i \neq 1$, and $g_{ij} = 0$ for $i \neq j$, I do not see why specifying the quadratic form given above would uniquely determine the metric's value for all pairs: $\langle x , y \rangle$. 
Primary Question:  How do I prove the desired result?  My first strategy was to try to write down explicitly the tangent space for any point in $\mathbb{H}^n$, but I have not been able to do so.  Any help would be appreciated.
 A: Geometrically, this is the upper sheet of the two-sheeted hyperboloid $x_0^2-x_1^2-\dots -x_n^2=1$.
Consider the smooth function $f\colon \mathbb{R}^{n+1}\to\mathbb{R}$ defined by $f(X)=g(X,X)$. Because of
\begin{align*}
T_Xf(Y_X)(h)=Y_X(h\circ f)&=\left. \frac{d}{dt}\right|_{t=0} (h\circ f)(X+tY)\\
&=\left. \frac{d}{dt}\right|_{t=0} h(f(X)+2tg(X,Y)+t^2g(Y,Y))=2g(X,Y)\left(\frac{d}{dt}\right)_{f(X)}h,
\end{align*}
we have $df_X(Y)=2g(X,Y)$, which yields $df_X=2X^{\flat}$.
Thus, $df_X$ has rank $1$ for any $X\neq 0$, and $-1$ is a regular value of $f$.
Then a regular level set $f^{-1}(\{-1\})$ is a smooth manifold, which also holds for $\mathbb{H}^n$.
Moreover, we have $T_X \mathbb{H}^n=\ker T_Xf=X^{\perp}$ for each $X\in \mathbb{H}^n$.
Because $g(X,X)=-1<0$, the restriction of the ambient metric to $X^{\perp}$ is positive definite,
so $g$ induces a Riemannian metric $\imath^* g$ on $\mathbb{H}^n$.
A: As I noted in the comments, given any quadratic form on a real vector space (such as an indefinite metric), one has
$$ Q(x+y)-Q(x)-Q(y) = 2B(x,y) \\
Q(x-y)-Q(x)-Q(y) = -2B(x,y) $$
by the definition. Subtracting and dividing gives
$$ B(x,y) = \frac{1}{4}(Q(x+y)-Q(x-y)), $$
or in this case,
$$ \langle x,y \rangle = \frac{1}{4}( \langle x+y,x+y \rangle - \langle x-y,x-y \rangle). $$
So given $\langle x,x \rangle $ for every $x$, we can find $\langle x,y \rangle$ for any $x$ and $y$.

We can parametrise the surface as
$$ \phi(y) = (\sqrt{1+\lVert \mathbf{y} \rVert^2},\mathbf{y}), $$
where $\mathbf{y} = (y_1,y_2,\dotsc,y_n)$. 
The pullback metric is then given by $ h(X,Y) = (\phi^*g)(X,Y) = g(d\phi(X),d\phi(Y)) $. In coordinates,
$$ h_{ij} = g_{kl}\frac{\partial x^i}{\partial y_k} \frac{\partial x^j}{\partial y_l} = \delta_{ij} + \frac{y_i y_j}{1+\lVert \mathbf{y} \rVert^2}, $$
so
$$h_{ij} dy^i dy^j = \lVert dy \rVert^2 + \frac{(y \cdot dy)^2 }{1+\lVert y \rVert^2}, $$
where $\cdot$ is the ordinary Euclidean scalar product metric. This is bounded below by 
$$ \lVert dy \rVert^2 \left( 1-\frac{\lVert y \rVert^2}{1+\lVert y \rVert^2} \right) = \frac{\lVert{dy}\rVert^2}{1+\lVert y \rVert^2} > 0 $$
for $dy \neq 0$, and so is positive-definite.
