Finding the equation of two lines lying on the surface of a hyperbloid

Question

Trying to solve this question:
Surface S is obtained by revolving the line $x^2-z^2=4$ around the "z" axis.

1. Write the equation for S.
2. Show that exactly two lines pass through M=(2,0,0) which also lies exactly on the surface of S. Obtain the equation for the lines.

For part 1, we use $x=\sqrt (x^2+y^2)$.
If I'm correct; $S=x^2+y^2-z^2=4$ which makes a hyperboloid and hyperboloids are doubly ruled. However, I have no idea how to obtain the equation for the lines.

If it's possible I don't want to use trigonometry(sin, cos, θ) or integration.

I've read this post but couldn't really understand it.

Answer 1

Your hyperboloid can be represented as

$$ S=\left(\matrix{x\\ y\\ z\\}\right)\left(\matrix{1&0&0\\ 0&1&0\\ 0&0&-1\\}\right)\left(\matrix{x\\ y\\ z\\}\right)-4=0 $$

or $p^TMp-4=0$

now given a line $p = p_0 + \lambda \vec v$ with $\vec v = (v_x,v_y,v_z)$, if the line is contained in $S$ we should have

$$ (p_0+\lambda\vec v)^tM(p_0+\lambda\vec v)-4=p_0^TMp_0+2\lambda p_0 M \vec v+\lambda^2\vec v^TM\vec v-4=0 $$

and this should be true for all $\lambda$ hence

$$ \cases{ p_0^TMp_0-4=0\\ p_0 M \vec v=0\\ \vec v^TM\vec v=0\\ \|\vec v\|=1 } $$

as an example, given $p_0 = (0,2\sqrt{2},2)'\in S$ we need

$$ \cases{ 2\sqrt{2}v_y-2v_z=0\\ v_x^2+v_y^2-v_z^2=0\\ v_x^2+v_y^2+v_z^2=1 } $$

giving

$$ \vec v = \cases{(\frac 12,\frac 12,\frac {1}{\sqrt{2}})\\ (\frac 12,-\frac 12,-\frac {1}{\sqrt{2}}) } $$