How to calculate the tangent space? first of all, what is the difference between the tangent space and the tangent plane?
I tried to find the tangent space of the hyberpoloid $$x^2 +y^2 -z^2 =a$$ , $$a>0
$$ at the point $$(\sqrt{a},0,0)$$ in this way:
$$f(x,y,z)=x^2 +y^2 -z^2$$
$$f_{x}=2x$$, $$f_{y}=2y$$, $$f_{z}=-2z$$
then, at the point $$(2\sqrt{a},0,0)$$ we get $$(2\sqrt{a},0,0)$$ hence the tangent space is 
$$2\sqrt{a}(x-\sqrt{a}) +0(x-0) -0(x-0)=0$$
$$x=\sqrt{a}$$
I feel that something is wrong.Can you help me please?
 A: I just saw that this post was linked to a post (Calculating the tangent space to a hyperboloid) I just answered. As this post here does not have an answer anymore, I'll post my answer here again:
You can calculate the tangent space via the inverse of a function that defines your hyperboloid.
Let $f:\mathbb{R}^3\rightarrow \mathbb{R}$ be defined by $f(x,y,z):=x^2+y^2-z^2-a$. Then
    \begin{equation*}
  \begin{split}
   & df = (2x,2y,-2z),
  \end{split}
\end{equation*}
    which is isomorphic to a basis vector of $T_{f(x)}(\mathbb{R})\simeq \mathbb{R}$
    if any of the $x,y,z$ are non-zero. This must be the case for
    $f(x,y,z)=0$, so by the preimage theorem
    $H:=f^{-1}(0)$ defines the manifold of the
    hyperboloid. 
Now because $df_x^{-1}(0)=T_x(f^{-1}(0))$, we can obtain the tangent space $T_x(H)$ by looking for all vectors $v$ for which $df_x(v)=0$. On $f^{-1}(0)$, we have the condition
    $z=\sqrt{x^2+y^2-a}$, so
    \begin{equation*}
  df_x(v)=
  \begin{pmatrix}
    2x\\
    2y\\
   - 2\sqrt{x^2+y^2-a}
  \end{pmatrix}
  \begin{pmatrix}
    v_1\\
    v_2\\
    v_3
  \end{pmatrix} =
  2(xv_1+yv_2-\sqrt{x^2+y^2-a}v_3)
\end{equation*}
    This is zero if $v_1 = (\sqrt{x^2+y^2-a}~v_3-yv_2)/x$ (and if one wants the tangent space at $x=0$, then one must choose other combinations like $v_3=(xv_1+yv_2)/\sqrt{x^2+y^2-a}$ etc), so the tangent space is two-dimensional, depends
    on $x,y,z$ and is given (except at $x=0$) by
\begin{equation}
   T_{x,y,z}(H)=\left\{(v_1,v_2,v_3)~|~v_2,v_3\in\mathbb{R},~v_1=(\sqrt{x^2+y^2-a}~v_3-yv_2)/x\right\}.
\end{equation}
    In particular, like the whole space, it is only defined for $x^2+y^2\ge a$. 
At the point $x=(\sqrt{a},0,0)$, the tangent space is thus
\begin{equation}
   T_{\sqrt{a},0,0}(H)=\left\{(v_1,v_2,v_3)~|~v_2,v_3\in\mathbb{R},~v_1=0\right\}\simeq \mathbb{R}^2.
\end{equation}
