On the intersection of the hyperbolic paraboloid with a bundle of planes. Suppose we are in Euclidian space in 3 dimensions.
I intersect a bundle of planes $\alpha(x-y) + z = 0$ with a hyperbolic paraboloid $x^2 - y^2 = 2z$
\begin{cases} \alpha(x-y) + z = 0 \\ x^2 - y^2 = 2z \end{cases}
\begin{cases} \alpha(x-y) + z = 0 \\ (x-y)(x+y+2 \alpha) = 0 \end{cases}
I don't understand what we obtain from this. I wrote that I obtain a family of lines but can't see it and would not be sure how to solve a system of this kind.
How do I see it and what is the way to solve these kind of systems?
 A: The equation $(x-y)(x+y+2 \alpha) = 0$ is equivalent to either $(x-y)= 0$ or $(x+y+2 \alpha) = 0$, which are the equations of two planes. 
Combining the first one with $\alpha(x-y)+z = 0$, one gets $x=y$ and $z=0$,
the equation of a fixed line in the $(x,y)$-plane.
The other solutions of your system are then given by
$$
\begin{cases} 
\alpha(x-y)+z = 0 \\ 
x+y+2 \alpha = 0 
\end{cases}
$$
These are the equations of two planes, whose intersection is a line varying with $\alpha$, that is a family of lines.
A: HINT: 
Not a proper answer. I suppose you can study projections of intersections of various planes to the hyperbolic paraboloid. If that is the aim, however it is better to begin with:
\begin{cases} y = x \,\alpha \\ x^2 - y^2 = 2z \end{cases}
to get intersections by varying the inclination of cutting plane rotated around z-axis through  slope $\alpha$  { asymptotic lines , parabolas drooping down , parabolas opening up },
and then proceed with inclined planes as given in the problem.
$ z=0 $ gives a pair of asymptote straight lines of intersection, $ x = y, x = -y. $
