Equation of Cone vs Elliptic Paraboloid I can't understand why $$\frac{z}{c} = \frac{x^2}{a^2} + \frac{y^2}{b^2} \tag{*}$$ corresponds to an elliptic paraboloid and $$\frac{z^2}{c^2} = \frac{x^2}{a^2} + \frac{y^2}{b^2} \tag{**}$$ to a cone, and not the other way around.
I tried to understand by looking at the traces of $z(x,y)$. For example, I did so for (**):
$\boxed{\text{When } x = k:}$ Then (**) becomes: $\displaystyle \frac{z^2}{c^2} - \frac{y^2}{b^2} =\underbrace{\frac{k^2}{a^2}}_{\text{a constant}} \color{green}{\text{: hyperbolas in the $yz$-plane.}}$
${\boxed{\text{When } y = k:}}$ Then (**) becomes: $\displaystyle \frac{z^2}{c^2} - \frac{x^2}{a^2} = {\underbrace{\frac{k^2}{b^2}}_{\text{a constant}}\color{red}{\text{: hyperbolas in the $xz$-plane.}}}$
$\boxed{\text{When } z = k:}$ Then (**) becomes: $\displaystyle \underbrace{\frac{k^2}{c^2}}_{\text{a constant}} = \frac{x^2}{a^2} + \frac{y^2}{b^2} \color{blue}{\text{:  ellipses in the $xy$-plane.}}$
I sketched the following shape based on this information, but it doesn't seem to tell me whether it's an elliptic paraboloid or a cone?


$\Large{\text{Supplementary Question:}}$ Thank you very much to all your answers, all of which helped! Based on them, the key step seems to be to look at the traces of $z(x,y)$ for $k = 0.$ I now understand that this answers my question, but why did my original work with $k \neq 0 $  fail to do so? My textbook doesn't mention the latter straightforward "trick." 

$\Large{\text{Question S.1:}}$ @bubba: Thank you very much for your answer to the Supplementary Question. To clarify your answer, did you mean: "but, in either case, these are hyperbolas curves} unless $k=1.$" As I wrote above, for $\text{(*)}$, the traces $x=k \text{ & }y = k$ do yield hyperbolas. But for $\text{(*)}$,
$ \boxed{\text{When } x = k:}$  Then (*) becomes $\displaystyle \frac{z}{c} - \frac{y^2}{b^2} =\underbrace{\frac{k^2}{a^2}}_{\text{a constant}} \text{: PARAbolas in the $yz$-plane.}$
${\boxed{\text{When } y = k:}}$ Then (*) becomes: $\displaystyle \frac{z}{c} - \frac{x^2}{a^2} = {\underbrace{\frac{k^2}{b^2}}_{\text{a constant}}{\text{: PARAbolas in the $xz$-plane.}}}$.
Of course, the fact the traces of (*) are parabolas and NOT hyperbolas still doesn't answer my original question. As you kindly explained and I now understand, it's necessary to consider $k = 0.$ 
 A: 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.
A: 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.
A: 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.
A: x^2/a^2 + y^2/b^2 = z/c^2 -> can not be a cone because general equation of cone is homogeneous 2nd degree equation passing through origin, since its not homogeneous so it is not cone.
for z= k^2 -> x^2/a^2 + y^2/b^2 = k^2/c^2 -> in xy plane it represent ellipse, as we increase k, size of ellipse also increase in xy plane.
in xz plane -> y=0 ->  x^2/a^2 = z/c^2 -> which is a equation of parabola for z= k^2,
and in yz plane -> x=0 -> y^2/b^2 = z/c^2 -> again equation of parabola for z= k^2.
so curve thus obtained is elliptical paraboloid.

x^2/a^2 + y^2/b^2 = z^2/c^2 -> homogeneous equation of 2nd degree passing through origin, so by definition of cone , the equation represent a cone. let us visualize further:
for z=k ->   x^2/a^2 + y^2/b^2 = k^2/c^2 -> represent ellipse in xy plane 
even power of x,y,z so symmetrical about each coordinate planes.
in xz plane -> y=0 -> x^2/a^2 = z^2/c^2 -> x/a = z/c and x/a = -z/c -> pair of lines which are the generator of cone.

