Sketching Hyperbolas in Quadric Surfaces

[JeffNYC]

In general, say:

we have a surface: y^2/4 - x^2/3 - z^2 = 1

I know that this is a hyperboloid of 2 sheets, since the xz trace:

x^2/3 + z^2 =-1 doesn't exist,

But for the other traces:

xy trace: y^2/4 - x^2/3 = 1 and yz trace: y^2/4 - z^2 = 1

Which are both hyperbolas - how do I sketch these? What should I be looking at in the 2 equations:

xy trace: y^2/4 - x^2/3 = 1 and yz trace: y^2/4 - z^2 = 1

...to help me understand where they are positioned on the graph (intercepts, vertices, etc...)

Thanks,

Jeff

 

[HallsofIvy]

One method I like to use is to look at the trace in each coordinate plane. If z= 0, that becomes y^2/4 - x^2/3 = 1 in the xy plane. That is, of course, a hyperbola. If y= 0, that becomes - x^2/3 - z^2 = 1 which is impossible! The graph does not cross the xz-plane. Finally, if x= 0, this is y^2/4 - z^2 = 1, again a hyperbola. Draw those graphs on to sheets of paper (and the third is showing the xz coordinate system). Hold them so the x,y,z axes are orthogonal. That should give you an idea of what the surface looks like. It is, of course, a "hyperboloid of two sheets".