Naming quadric surfaces

Non-degenerate real quadric surfaces
$\frac{x^2}{a^2}+\frac{y^2}{b^2}-z=0\\ z=\frac{x^2}{a^2}+\frac{y^2}{b^2}$
Consider the horizontal cross section (level curve) of z by setting $z=c$.
We have $\frac{x^2}{a^2}+\frac{y^2}{b^2}=c$, which is the equation for an ellipse.
Now consider the vertical cross section of z at $x=c$.
We get $z=\frac{c^2}{a^2}+\frac{y^2}{b^2}$ -- a constant plus $\frac{y^2}{b^2}$, which is the equation of a parabola. Similarly, the vertical cross section of z at $y=c$ is also a parabola.
We know that the figure is a elliptical paraboloid.

When we change b (denominator) to a, we have a circular paraboloid.

$\frac{x^2}{a^2}-\frac{y^2}{b^2}-z=0\\ z=\frac{x^2}{a^2}-\frac{y^2}{b^2}$
Horizontal cross section: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=c\:\:[z=c]\Rightarrow$ hyperbola
Vertical cross section: $\frac{x^2}{a^2}- \text{constant}$ or $\text{constant} - \frac{y^2}{b^2} \Rightarrow$ parabola
Surface: Hyperbolic paraboloid

$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$
Horizontal cross section: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{k^2}{c^2}+1\:\:[z=k] \Rightarrow$ ellipse
Vertical cross section: $\frac{x^2}{a^2}-\frac{z^2}{c^2}=1-\frac{m^2}{b^2}\:\:[y=m]$ or $\frac{y^2}{b^2}-\frac{z^2}{c^2}=1-\frac{n^2}{a^2}\:\:[x=n] \Rightarrow$ hyperbola
Surface: Elliptical hyperboloid of one sheet

$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=-1$
Horizontal cross section: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{k^2}{c^2}-1\:\:[z=k] \Rightarrow$ generally ellipse
There are no horizontal cross sections when z is between -1 and 1. For simplicity, we can look at the equation $-x^2-y^2+z^2=1$. Suppose $z=0$. Then the left hand side is negative, but the right hand side is positive. There's no way to fix this, so the cross section simply doesn't exist!
Vertical cross section: $\frac{z^2}{c^2}-\frac{x^2}{a^2}=1+\frac{m^2}{b^2}\:\:[y=m]$ or $\frac{z^2}{c^2}-\frac{y^2}{b^2}=1+\frac{n^2}{a^2}\:\:[x=n] \Rightarrow$ hyperbola
Surface: Elliptical hyperboloid of two sheets

Here's a hint about telling the two kinds of hyperboloids apart: look at the cross sections x=0, y=0, and z=0 (respectively but not simultaneously). If they exist, then it's a hyperboloid of one sheet. If you end up with something negative equal to something positive, it's a hyperboloid of two sheets.

Degenerate quadric surfaces
$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=0$
Horizontal cross section: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{k^2}{c^2}\:\:[z=k] \Rightarrow$ ellipse
Vertical cross section:
(1) $\frac{z^2}{c^2}-\frac{x^2}{a^2}=\frac{m^2}{b^2}\:\:[y=m]$ or $\frac{z^2}{c^2}-\frac{y^2}{b^2}=\frac{n^2}{a^2}\:\:[x=n] \Rightarrow$ hyperbola
(2) $\frac{x^2}{a^2}-\frac{z^2}{c^2}=0\:\:[y=0]$ or $\frac{y^2}{b^2}-\frac{z^2}{c^2}=0\:\:[x=0]$ $\iff (\frac{x}{a}-\frac{z}{c})(\frac{x}{a}+\frac{z}{c})=0$ or $(\frac{y}{b}-\frac{z}{c})(\frac{y}{b}+\frac{z}{c})=0 \Rightarrow$ pair of straight lines
Surface: Elliptic Cone

$\frac{x^2}{a^2}+\frac{y^2}{a^2}-\frac{z^2}{b^2}=0$ -- Special case of cone (Circular cone)
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ -- Elliptic cylinder
$\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$ -- Circular cylinder
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ -- Hyperbolic cylinder

Useful website:
Quadric surfaces