Equation of Cone vs Elliptic Paraboloid

Something to notice is that in the equation

$$\frac{z}{c}=\frac{x^2}{a^2}+\frac{y^2}{b^2}\tag{1}$$

$z$ will either be nonnegative or nonpositive, depending on the sign of $c$. This means that this conic section will never dip below the $xy$-plane if $c$ is positive, or never rise above the $xy$-plane if $c$ is negative. Since cones look like this:

$\hspace{2.25in}$

$(1)$ must not be a cone.

Alternatively, and perhaps more enlightening, is to notice that in the equation

$$\frac{z^2}{c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}\tag{2}$$

If we set $x=0$, that is, if we look at the intersection with the $yz$-plane, we get the two straight lines

$$y=\pm\frac{b}{c}z$$

which matches what we would expect from the picture. We get a similar result if we let $y=0$. So $(2)$ represents a cone.

Think about how in two dimensions, y^2 = x^2 represents a set of straight lines, while y = x^2 is a parabola.

In (*), fix x=0. Then it's easy to see that you get a parabola of the shape z = y^2 (constants ignored). Hence, the basis for an elliptic paraboloid.

In (**), fixing x=0 gives us something of the shape z^2 = y^2, or +-z = +-y. These are the straight lines needed for a cone.

Forget the "elliptical" part for a minute (in other words, suppose $a$ = $b$). This is just a scaling of the $x$ and $y$ axes, so it doesn't really change the problem. Then the second of your equations (**) is roughly equivalent to $\sqrt{x^2 + y^2} = mz$, where $m = 1/{c^2}$. This says the "radius" of the surface increases linearly as $z$ increases. So, it's a cone.