Prove that for every point in one-sheeted hyperboloid, there exists at least one line which is full contained in it

Question

Please help me with the task:

Prove that for every point in one-sheeted hyperboloid, there exist at least one line, which is full contained in it.

Firstly, I've noticed that I can transform the hyperboloid to a isomorphic one without loss, so I've choosed generic one-sheeted hyperboloid.

I've tried to put the equation of this hyperboloid ($H$):

$x_1^2 - x_2^2 - x_3^2 + 1 = 0$

and line ($L$) in $\mathbb{R}^3$

$x_1 + ax_2 + bx_3 + c = 0.$

Then to show that if we choose any point on the $H$ (by fixing $k=x_2$ and $l=x_3$) we get

$x_1^2 = k^2 + l^2 - 1$.

So, we are looking for a line satisfing both equations, described:

$k^2 + l^2 - 1 + ak^2 + bl^2 + c = 0$

$(\sqrt{a+1}k+\sqrt{b+1}l)^2 - \sqrt{(a+1)(b+1)} kl + c - 1 =0$

Now I can't finish this from this point...

All hints appreciated!

Answer 1

Using rotational invariance, you need to show it only for the points of the form $(r, 0,\sqrt{r^2-1}$) for $r \geq 1$. Then $x=r+ \sqrt{r^2 -1}t$, $y= \pm t$, $z=\sqrt{r^2-1} +rt$ give two lines that go through the point and are contained in the surface.