How do you determine a system of Cartesian equations?

Answer

The surface z = 2xy is a quadric. There are different kinds of quadric, but some have the property that through every point of the surface at least one straight line can be found that lies completely in the surface. A quadric with this property is called a ruled surface.
The equation you give is such a regulated surface, more specifically a parabolic hyperboloid, also called saddle surface. A saddle surface is therefore a control surface. Two straight lines that lie completely in the surface pass through each point.

Now for those rights. In geometry it is almost always much cheaper to write a line in parameter form, so in the form :

x = x0 + at
y = xy + bt
z = z0 + ct

where (x0,y0,z0) is a point on the line, (a,b,c) is a direction vector. t is the parameter that takes all values ​​from minus to plus infinity. For each t you get exactly one point on the line, and each point on the line corresponds to exactly one value of t. A parametric form can be set up “in two lines” in a Cartesian form.

An arbitrary line through the point p from your question is thus of the form:

x = 1 + at
y = -1 + bt
z = -2 + ct

with a,b,c yet to be determined so that they yield a line that lies in M, and with the ever-present condition that they may not be zero together (otherwise you have no direction left). On the other hand, a,b,c are determined except for a mutual multiple: after all, the direction (2, 4,-1) is parallel to, for example (20, 40, -10)

We now want that line to lie completely in M. So the coordinates of the line must satisfy the equation of M:

-2 + ct = 2 ( 1 + at ) ( -1 + bt )

This must hold for all values ​​of t, because the complete line must lie in M.

So, after rearranging some terms:

2 abbot² + ( -2a + 2b – c ) t + 0 = 0

To be sure that this is zero for all possible real values ​​of t, we require that the coefficients of t-squared and of t are both zero:

ab = 0

-2a + 2b – c = 0

Obviously a or b must be zero because of this first equation. However, both at the same time cannot be zero, because then c would also be zero, and we know that a,b and c cannot be zero at the same time.

(1) First possibility : Choose a = 0,
then b =1 or any other non-zero number and further => c = 2

This gives a first line :

x = 1
y = -1 + t
z = -2 + 2t

(you see that these indeed satisfy M) To convert to Cartesian form you have to eliminate t. This gives the cartesian equation

x = 1 and z – 2y = 0

The line cannot be written Cartesian in the form (x-x0) / a = (y-y0) / b = (z-z0) / c
because one of the directional numbers is zero. This shows once again that the parametric form of a line is preferable to a Cartesian equation.

(2) Second possibility : Choose : b = 0,
then a =1 or any other non-zero branch and further => c = -2

This gives a second straight:

x = 1 + t
y = -1
z = -2 – 2t

(you see that these indeed satisfy M. To convert to Cartesian form you have to eliminate t. This gives the Cartesian equation

y = 1 and z + 2x = 0

Also here: the straight line cannot be written here in Cartesian in the form
(x-x0) / a = (y-y0) / b = (z-z0) / c because one of the directional numbers is zero.