On the intersection of the hyperbolic paraboloid with a bundle of planes.

Question

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?

Answer 1

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.

Answer 2

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. $