How can I prove that a point on a surface can be parametrized by a certain parametrization?

Question

So the specific question is as follows: how can I show that the surface $y(x-a)+zx=0$ can be parametrized by $\alpha(u,t)=(au,ut,t-ut)$? Or equivalently, the set of points defined by the first equation is a subset of the set of points defined by the second?

Answer 1

Assume first that $a \neq 0$. Given a point $(x,y,z)$ that satisfies $y(x-a) + zx = 0$, note that if $x = 0$ then $y = 0$. Define $u = \frac{x}{a}$ and

$$ t = \begin{cases} \frac{ay}{x} & x \neq 0 \\ z & x = 0 \end{cases}. $$

Then if $x \neq 0$, we have

$$ \alpha(u, t) = (au, ut, t - ut) = \left(x, y, \frac{ay}{x} - y \right) = \left( x, y, \frac{ay - yx}{x} \right) = \left( x, y, \frac{y}{x}(a - x) \right) = \left( x, y, \frac{zx}{x} \right) = \left( x, y, z \right) $$

while if $x = 0$, then $y = 0$ and we have

$$ \alpha(u, t) = (au, ut, t - ut) = (0, 0, z - 0) = (x,y,z). $$

If $a = 0$, the equation becomes $x(y+z) = 0$, so for example $(1,1,-1)$ lies on the surface $x(y+z) = 0$ (which is the union of two planes). However, $(1,1,-1)$ is not in the image of $\alpha(u,t)$ as the first coordinate of $\alpha(u,t)$ must be zero.