Finding tangent planes using differential geometry "language" I'm studying differential geometry and I have a couple of questions (which are related).
I'm supposed to find the tangent planes to the surface given by $x^2+y^2-z^2=1$ at points $(x,y,0)$.

*

*Is it enough to consider only the "upper" part of this hyperboloid, by using $\varphi(x,y)=(x,y,\sqrt{x^2+y^2-1})$ and work it as I would with the graph of functions?

If so, why? My reasoning was that even if I had to "complete" the surface with the lower part of the hyperboloid,  $\tilde{\varphi}(x,y)=(x,y,-\sqrt{x^2+y^2-1})$, the tangent planes would have to agree at $(x,y,0)$ so I can choose either one. I guess this would imply that it is sufficient to investigate tangent planes in any local parametrization that contains the points of interest.
Supose I can proceed as stated. Then I know how to find the tangent plane as I used to do in multivariable calculus, but I was also given the following deffinition:
Given a local parametrization $\varphi: U \subset \mathbb{R}^2 \to S$ where $S$ is a regular surface and a point $p \in \varphi(U)$. The tangent plane to this point is :
$$T_pS=D\varphi(\varphi^{-1}(p))(\mathbb{R}^2)$$
Which leads me to my next problem:


*I'm having a hard time understanding the differential notation.

I tried to apply this to $\varphi(x,y)=(x,y,\sqrt{x^2+y^2-1})$ at points $p=(x_0,y_0,0)$.
As far as I'm concerned, $\varphi^{-1}(p)=(x_0,y_0)$ and
$$
D\varphi((x_0,y_0))=\begin{bmatrix}
1 &0 \\ 0& 1 \\ \dfrac{x_0}{\sqrt{x_0^2+y_0^2-1}} &\dfrac{y_0}{\sqrt{x_0^2+y_0^2-1}}\\
\end{bmatrix}
$$
I don't know what to do next. What does the $\mathbb{R^2}$ has to do in the definition of the tangent plane, and how do I get to the tangent plane expression from here?
This was a long question but I would really appreciate anyone taking the time to help me here :´)
 A: 1- I don't think its a good idea to use that parametrization. In differential geometry, parametrizations of regular surfaces are defined at open sets, so all the properties are well defined. That parametrization would not work for that, at least not in $z=0$
2-The two vectors on the columns are the basis of the tangent plane at that point. So any vector on the plane can be espressed by a linear combination of those vectors. Let's say you wanna find the tangent of a curve $\alpha (u(t), v(t))$ at a point $t=0$ (This curve is at your surface parametrized by $u$ and $v$) If $D_u$ and $D_v$ are the column vectors, so the tangent would be $D_u u'(0)+ D_v v'(0)$. In other words, it would be the matrix multiplication $D * (u'(0), v'(0))^T$
I'm kind new to differential geometry, but I have a test about it after tomorrow, so I've been studying. Hope this helps!
A: I agree with Trandus's answer, but here is a slightly different version.
Question 1: Your $\varphi$ is differentiable only when $x^2 + y^2 > 1$; this shows up when you examine the denominators in your calculation of $D\varphi$. It would be fine to use it, were the question not asking for the tangent plane at points $(x,y,0)$ which exactly have $x^2 + y^2 =1$. In other words, the question is asking for tangent planes at points where that plane is vertical, so it won't be okay to use a function of $(x,y)$ to describe them.
You could find another parametrization by describing your surface as a graph over, say, the $(y,z)$ plane: solving the given equation for $x$ gives $x = \pm\sqrt{1 + z^2 - y^2}$, and then defining $\psi(y,z) = (\sqrt{1 + z^2 -y^2}, y, z)$ is a parametrization you can use to find tangent planes at points $(x,y,0)$ with $x >0$. For $x < 0$ choose the other sign for the square root, and if $x= 0$ do a similar construction solving for $y$ instead.
Question 2. As Trandus says, the columns in the differential of $\psi$ will give a basis for the tangent plane. If you prefer to find an equation of the plane rather than a basis, another method is to remember that the gradient of a function is always normal to the level sets. Looking at the function $f(x,y,z) = x^2 + y^2 - z^2$, your surface is the level set $f(x,y,z) = 1$. So compute that the gradient is $\nabla f = \langle 2x, 2y, -2z\rangle$, which gives you a normal vector to the desired plane. At the point $(a,b,0)$ this becomes $\langle 2a, 2b, 0\rangle$, and the plane with this normal vector is given by the equation
$$
2a(x-a) + 2b(y-b) = 0.
$$
Note that in ``differential geometry language'', the fact that $\nabla f$ is a normal vector for the tangent to the level set is expressed by the fact that the tangent space of the level set of $f$ at a point $p$ is equal to the kernel of the differential of $f$ at $p$.
