I'm attempting to graph the function: $$\frac{x^2}{4} + \frac{y^2}{9} = 1$$
I can tell that at constant $z$ the graph is elliptical, but I found, at constant $x$ and $y$, let's say $x=0$, $y=0$:
Let $y=0$ $$\frac{x^2}{4} -1 = z$$
This is parabolic.
For $x=0$:
$$\frac{y^2}{9}-1=z$$
This is also parabolic. However, according to the answer for this exercise this object is an elliptical cylinder, yet I can't understand how would that be possible given the fact that it's quadratic in the $xz$ and $yz$ planes. Can someone tell me where I'm going wrong?
If you intersect with plane $y=0$ you get $\frac{x^2}{4}=1$ which is a pair of straight lines in the plane $y=0$ equations $x=2;\;x=-2$. And for $x=0$ a similar thing happens $\frac{y^2}{9}=1$ pair of lines in the plane $x=0$, namely $y=3;\;y=-3$
The elliptic thing is that in any plane $z=k$ the intersection is $\frac{x^2}{4}+\frac{y^2}{9}=1$, an ellipse as you can see in the picture below