How to find the equations of the lines that make up the ruled surface 2xy

Question

How to find the equations of the lines that make up the ruled surface z= 2xy? There are a lot of lines and solving this seems non-obvious. I imagine the first thing to know is the equation of a line in 3d. What is the equation for a 3D line?

Answer 1

The points $\;M=\Bigl(x,\dfrac 2x, 4\Bigr)$ and $\;N=\Bigl(x,-\dfrac 2x, -4\Bigr)$ are on the quadric. One easily check that any point $P=tM+(1-t)N$ also lies on the quadric. Indeed the coordinates of such a point are $$\Bigl(tx+(1-t)x, \dfrac{2t}x-\dfrac{(1-t)2}x, 4t-4(1-t)\Bigr)=\Bigl(x, \dfrac{2(2t-1)}x, 4(2t-1)\Bigr).$$

Answer 2

The lines are given by the planes intersection

• $z=xy$
• $x=k$

and

• $z=xy$
• $y=k$

In parametric form for

• $z=xy$
• $x=k$

we have that a generic point is

$$P=(k,t,tk)=(k,0,0)+t(0,1,k)$$

that is the parametric equation for this family of lines and similarly we can find that for the other family of lines.

Answer 3

A systematic way to find these lines uses the following facts:

1. If the surface is given by the implicit equation $f(\mathbf x)=0$, then the line $\mathbf p+t\mathbf v$ is a generator iff $f(\mathbf p+t\mathbf v)=0$ for all $t$. For an $n$th-degree surface, this equation is an $n$th-degree polynomial in $t$ and the coordinates of $\mathbf v$.

2. A generator of the surface lies in a tangent plane to the surface. Therefore, if $\mathbf p$ lies on the surface, then $\nabla f(\mathbf p)\cdot(\mathbf p+t\mathbf v)=0$. If $\nabla f(\mathbf p)\ne0$, this is a linear equation in the coordinates of $\mathbf v$ that can be used to eliminate one of the variables.

3. The length of $\mathbf v$ is unimportant, but it must be non-zero, so one of its coordinates can be set to $1$. You might have to consider several cases here.

Facts 2 and 3 eliminate two of the variables, leaving you with an equation in $t$ and the remaining variable. Viewed as a polynomial in $t$, all of the coefficients must vanish, which gives you a system of single-variable polynomial equations to solve. You will end up with one or two solutions, depending on whether the surface is singly or doubly ruled.

So, starting with $f(x,y,z)=2xy-z$ we have $$2(p_x+t v_x)(p_y+t v_y)-p_z+t v_z = 0. \tag 1$$ The orthogonality condition of fact 2 gives us $2(p_x v_y+p_yv_x)-v_z=0$, which we can solve for $v_z$ and substitute into (1) to obtain, after simplifying, $$2p_x p_y-p_z+2t^2 v_x v_y = 0. \tag 2$$ Setting $v_x=1$ results in $v_y=0$ and vice-versa, therefore there are two generators through each nonsingular point $\mathbf p$ on the surface: $\mathbf p+t(1,0,2p_y)$ and $\mathbf p+t(0,1,2p_x)$. ${\partial f\over\partial z}=-1$ everywhere, so the gradient never vanishes and there are no singular points to consider.

If you need a directrix curve for this surface, find a plane that’s not parallel to any of these generators and intersect it with the surface. For this one, the plane $x=y$ works, producing a curve that can be parameterized as $(s,s,2s^2)$, which then generates the parameterization $(s,s+t,2s(s+t))$ of the surface.