Rectilinear generators of hyperbolic paraboloid.

Question

Given a point on a hyperbolic paraboloid, prove that there exist exactly 2 lines that pass through that point and lie on the surface of that paraboloid.

Basically it's asked to prove that there are exactly 2 rectilinear generators passing through any given point of such paraboloid (it's also necessary to prove their existence). I know that hyperbolic paraboloid is a ruled surface and know the equations of its rectilinear generators passing through a given point, but I have no idea how to prove the claim.

Answer 1

Let

$$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=z$$

be the equation of the Hyperbolic Paraboloid (HP in short).

Consider the following family of lines with non zero parameter $c$

$$(L_c) \ \begin{cases}\dfrac{x}{a}-\dfrac{y}{b}&=&c\\ \dfrac{x}{a}+\dfrac{y}{b}&=&\dfrac{z}{c} \end{cases} \implies \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=z$$

(the implication is obtained by multiplying the equations)

But implication means for corresponding sets, inclusion. In this way, we have proven that $(L_c) \subset (HP).$

Same reasoning for the other (distinct) family:

$$(L'_d) \ \begin{cases}\dfrac{x}{a}+\dfrac{y}{b}&=&d\\ \dfrac{x}{a}-\dfrac{y}{b}&=&\dfrac{z}{d} \end{cases}$$

Answer 2

comment

BTW, The 2 lines are geodesics, unique euclidean straight lines for

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 2z.$$