Find a parameterisation and compute $r_{\alpha},r_{\beta},r_{\alpha}$ x $r_{\beta}$ and the tangent plane at the point mentioned of the surface $$x^2+y^2-z^2=2y+2z$$ where $-1\leq z \leq 0$ and the point is $(1,1,0)$.
is there a technique to parametrise this? no one teaches properly how to parametrise...
Honestly, I have no idea how to "properly" parametrize this. However, you can take $$ x^2+y^2-z^2=2y+2z $$ and turn this into $$ x^2=z^2-y^2+2y+2z. $$ Then you have an $$ x=f^+(y,z)=\sqrt{z^2-y^2+2y+2z} $$ and an $$ x=f^-(y,z)=-\sqrt{z^2-y^2+2y+2z} $$ functions.
In the half-space of the positive $x$ axis, you can have $$ \psi^+(y,z)=(f^+(y,z),y,z), $$ and in the half-space of the negative $x$ axis, you can have $$ \psi^-(y,z)=(f^-(y,z),y,z) $$ which are parametrized surfaces. You need to check for which $y$ and $z$ do we define this though, which I am too tired to right now, so I'll assume everything works fine.
Then the tangent basis vectors are $$ \left.\frac{\partial\psi^+}{\partial y}\right|_{(1,1,0)},\ \ \left.\frac{\partial\psi^+}{\partial z}\right|_{(1,1,0)} $$ and the tangent plane is, I have no idea in what form do they ask it, but it is basically the $$ \mathrm{span}\left(\left.\frac{\partial\psi^+}{\partial y}\right|_{(1,1,0)},\left.\frac{\partial\psi^+}{\partial z}\right|_{(1,1,0)}\right) $$ subspace.