Find line that is contained in the set I want to find a line that is contained in the set that is defined from the equation $x^2+y^2-z^2=1$.
So we are looking for a line of the form $at+b$, right?
Then $x=a_1 t+b_1, y=a_2t+b_2, z=a_3 t+b_3$.
How can we find some restrictions? 
 A: Plugging in we get
$$ (a_1t+b_1)^2+(a_2t+b_2)^2-(a_3t+b_3)^2=1$$
or
$$(a_1^2+a_2^2-a_3^2)t^2+(2a_1b_1+2a_2b_2-2a_3b_3)t+(b_1^2+b_2^2-b_3^2-1)=0 $$
and this shall hold for all $t$. Thus we need
$$ a_1^2+a_2^2-a_3^2=0$$
$$ 2a_1b_1+2a_2b_2-2a_3b_3=0$$
$$ b_1^2+b_2^2-b_3^2=1$$
To simplify the search we can try to consider the special case $a_1=0$, $b_1=1$. Then the first equation suggests $a_3=\pm a_2$, the third suggests $b_3=\pm b_2$ and then the second holds only if we take opposite signs.
In its simplest case this leads to $x=1$, $z=-y$ or $x=1$, $z=y$.
A: Extending the remark of @Étienne Bézout, there is a double generation  of the hyperboloid $H_1$of one sheet by the following straight lines:
$(D_{\lambda})\begin{vmatrix}(x-z)=\lambda(1-y) \\ (x+z)=\dfrac{1}{\lambda}(1+y) \end{vmatrix}
\Rightarrow (x-z)(x+z)=(1-y)(1+y) \Rightarrow x^2+y^2-z^2=1  $
$(D_{\mu})\begin{vmatrix}(x+z)=\mu(1-y) \\ (x-z)=\dfrac{1}{\mu}(1+y) \end{vmatrix}
\Rightarrow (x-z)(x+z)=(1-y)(1+y) \Rightarrow x^2+y^2-z^2=1.$
One knows that implication corresponds to set inclusion ; thus each straight line $(D_{\lambda})$ is included in $H_1$. The same for straight lines $(D_{\mu}).$
In fact, it can be easily proven that each point $(x,y,z)$ of $H_1$ belongs exactly to one straight line of type $(D_{\lambda})$ and one straight line of type $(D_{\mu}).$

Figure: 100 straight lines of type $(D_{\lambda})$ and 10 of type $(D_{\mu}).$
A: The line 
$$l: \begin{cases} x=1 \\ y=z \end{cases}$$
satisfies the equation of your curve. Other similar ones work: invert $x$ and $y$, or use $y=-z$.
A: this is similar to hagen von eitzens idea. take a point $(\cos s, \sin s, 0), 0 \le s < 2\pi, $ on the $xy$-plane. now consider the lines $$x = \cos s - t\sin s, y = \sin s + t\cos s, z = t, -\infty < t < \infty\tag1 $$ parametrized by $s, t.$ 
we can verify that 
$$\begin{align}x^2 + y^2 - z^2 &=( \cos s - t\sin s)^2+(\sin s + t\cos s)^2 -t^2\\&= \cos^2 s + \sin^2 s+t^2(\sin^2 s + \cos^2 s )-t^2\\&=1\end{align}$$
as required.
