Say our question requires us to use traces to sketch the quadric surface with equation $x^2+y^2/9+z^2/4=1$. This is a worked example in our book, but I'm confused by their steps. So first we set $z=k$ to get $x^2+y^2/9=1-k^2/4$. First of all, is $k$ just a constant? And what is the difference between setting $z=k$ rather than $z=0$? Also, if $k$ is a constant (which I'm not sure if it is or not, or what it even represents for that matter), then how come when graphing using Wolfram, if I put in the equation with $k$, I get an ellipsoid, but if I just substitute say $0.5$ for $k^2/4$, I get a circle? And so then we repeat the same steps above with $x$ and $y$, and the same questions apply.
Yes, $k$ is a constant. The way you should think of this is that $z=k$ is the plane parallel to the $xy$-coordinate plane at "height" $k$. So $k=0$ corresponds to exactly the $xy$-plane, and as $k$ increases you are getting different planes higher above that one and $k<0$ gives you planes "below" the $xy$-coordinate plane.
Substituting in different numbers gives you a "slice" of the 3D picture, so it is just the curve that you get at that height. This is why when you substitute in a value for $k$ you get just a curve. When you typed it into Wolfram with the $k$, it interpreted $k$ as a variable instead of a constant so it gave you the full 3D ellipsoid. When you just look at a plane section of the ellipsoid you get ellipses (or circles).
The purpose of this method is that if you know what a bunch of the 2D slices of the 3D thing look like (which is much easier since you just keep getting curves that should be familiar), then you can piece them together to sketch the 3D thing.