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 $$2\sqrt{a}(x-\sqrt{a}) +0(x-0) -0(x-0)=0$$


I feel that something is wrong.Can you help me please?

  • $\begingroup$ The tangent plane is a plane in the space where your 2D manifold (i.e., surface) is imbedded. It is a property of the imbedding. The tangent space to any manifold is a construction (so you don't "calculate" it) on the manifold itself that has nothing to do with any imbedding. The tangent plane can be used to model the tangent space, but it is a different object. For example, in general, the tangent planes at two points will intersect. But the tangent spaces are always disjoint. $\endgroup$ Aug 29, 2015 at 20:18
  • $\begingroup$ Your procedure is right but I had chosen $f: \ \mathbb{R}^3\rightarrow \mathbb{R}\ : \ f(x,y,z)=x^2+y^2-z^2-a$. $\endgroup$
    – Willyf
    Aug 29, 2015 at 20:25
  • $\begingroup$ Thank you, I really appreciate your help :) $\endgroup$
    – letisya
    Aug 30, 2015 at 7:25

1 Answer 1


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}


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