Showing a property with regards to a vector tangent to a point on the hyperboloid! Considering the positive half of the standard hyperboloid $$S^+ = \{(x,y,z) : -x^2 - y^2 + z^2 = 1, z > 0\},$$ any point $(x,y,y) \in S^+,$ and the vector $<a,b,c>$ tangent to that point $(x,y,z),$ I must show that $-ax - by + cz = 0.$ Then, I must use this fact to show that $a^2 + b^2 - c^2 > 0.$ 
I have been having a lot of difficulty. I am not very comfortable with vectors.
(1) I have recognized that $-ax - by + cz = 0$ implies that the vectors $<-a,-b,c>$ and $<x,y,z>$ are perpendicular. However, I do not know how to leverage the fact that $\langle a,b,c \rangle$ is tangent to the point $(x,y,z)$ in order to show that $\langle -a,-b,c \rangle,$ or $\langle a,b,-c \rangle.$ Is perpendicular to the vector $\rangle x,y,z \rangle.$ The angle between $\langle a,b,-c \rangle$ and $\langle a,b,c \rangle,$ as well as the angle between $\langle a,b,c \rangle$ and $ \langle x,y,z\rangle $ varies according to the values of $x,y,z,a,b,c.$ 
May I have some guidance or some direction? Maybe a hint as to how to employ the fact that $\langle a,b,c \rangle$ is tangent to $(x,y,z).$
(2) Moreover, That $a^2 + b^2 - c^2 > 0$ implies that the angles between $\langle a,b,c \rangle$ and $\langle a,b,-c \rangle$ is between $0$ and $90$ degrees, right? Is there anything that suggests that angle between the two vectors is so?
Thank you very much! :)
 A: (1) A tangent vector must be the velocity vector of some path $(x(t),y(t),z(t))$ on the hyperboloid. That is, we get the vector $(x'(0),y'(0),z'(0))=(a,b,c)$. If we write out the equation
$$ -x(t)^2-y(t)^2+z(t)^2=1, $$
implicitly differentiate both sides and plug in $t=0$, what do we get?
(2) The hyperboloid is asymptotic to the cone $x^2+y^2=z^2$. A vector on the top half of the cone, and a vector on the bottom half with the same $x,y$ coordinates (and opposite $z$ coordinate) are exactly a right angle apart from each other. Tangent vectors to the hyperboloid are on the other side of this cone (pointing more horizontally than vertically) so $(a,b,c)$ and $(a,b,-c)$ are $<90^{\circ}$ apart.

Here is a way to show $-a^2-b^2+c^2<0$. Assume $(x,y,z)$ and $(a,b,c)$ are nonzero.
Define the "pseudo-Euclidean" dot product $(x,y,z)\cdot(a,b,c)=-ax-by+cz$.
The hyperboloid is then "pseudosphere" $\vec{x}\cdot\vec{x}=1$. Define the quadratic function
$$ f(t)= (\vec{x}+t\vec{v})\cdot(\vec{x}+t\vec{v})=\vec{x}\cdot \vec{x}+2t(\vec{x}\cdot \vec{v})+t^2(\vec{v}\cdot \vec{v})=\vec{x}\cdot \vec{x}+t^2(\vec{v}\cdot \vec{v}) $$
where $\vec{v}=(a,b,c)$. The third components $z$ and $c$ are both nonzero, so there exists a $t$ for which the third component of $\vec{x}+t\vec{v}$ is zero, in which case $f(t)\le 0$, impossible unless $\vec{v}\cdot\vec{v}<0$.
