Why aren't parametrizations equivalent to their "equation form"? Consider the parametrization $(\lambda,t)\mapsto (\lambda t,\lambda t^2,\lambda t^3)$. This is a union of lines (not sure how to visualize it precisely. I think it's a double cone). It doesn't appear that the $x$ or $z$ axis are in this parametrization (if $y$ and $z$ are both zero, so must $x$ be).
However, when you solve the parametrization in terms of $x,y,z$ you obtain the relationship $y^2=xz$. In which the $x$ and $z$ axes are solutions!
Why is this? Is this the closure? If so, how can we relate the closure to the original set?

When you say that you "solve" and obtain the relationship $y^2 = xz$ I image you just observed that $(\lambda t^2)^2 = (\lambda t)(\lambda t^3)$ correct?  In this case what you have shown is that the set for which you have a parameterization is a subset of the set of solutions to $y^2 = xz$.  What you have not shown is that every solution of $y^2 = xz$ is covered by the parameterization.  Indeed, as you pointed out, this is not true.
