Prove that the paraboloid in $R^3$, defined by $x^2 + y^2 - z^2 =a$ is a manifold if $a>0$. why does not $x^2 + y^2 -z^2 =0$ define a manifold?

Question

Prove that the paraboloid in $R^3$, defined by $x^2 + y^2 - z^2 =a$ is a manifold if $a>0$. why does not $x^2 + y^2 -z^2 =0$ define a manifold?

Could anyone give me a hint of the solution of this question?

EDIT:

I found this solution:

But I need more details about why it is a manifold and why when $a = 0$ it is not, could anyone explain this to me in a better way?

Answer 1

This is $z = \pm \sqrt{x^2+y^2}$. For a fixed $z$, this is a circle, with radius $|z|$. So this is a two sided cone; any neighbourhood of the point $(0,0,0)\in M$ can be disconnected by removing the point $(0,0,0)$, and so isn't homeomorphic to an open subset of $\mathbb R^2$. For intuition about surfaces in $\mathbb R^3$, I reccomend spending some time with a plotter- observe the bad behavior at $0$.