Hyperbolic Paraboloid: Understanding the Equation and Finding its Vertex

[jwxie]

Homework Statement

z = 2y^2 - x^2

Homework Equations

The Attempt at a Solution

I kind of know how to do it.

z = y^2/b^2 - x^2/a^2 the first power is the axis of paraboloid.


let x = k thus z = 2y^2 - k^2 and the vertex of this parabola (if x = 0 we see it is a parabola), v = (k, 0, -k^2)
if now x = k, the vertex becomes v = (0,0,0)

so far, do you guys agree?

similarly, for y = k v = (0,k, 2y^2) and when y = 0, we have v = (0,0,0)

and for z = k, we find z = k = 2y^2 - x^2 gives us a hyperbola.

 

[jwxie]

giving a bump :)
thank you!

 

[Mark44]

What do you need to do, draw a graph of the surface? The usual strategies are to get the traces in the three coordinate planes. To get the trace in the x-z plane, set y = 0.

Also helpful are cross-sections in planes parallel to the x-y plane. For each cross-section, set z to some value and you'll get some curve. Topographical maps are examples of this technique. All the points in a connected curve in a topo map are at the same elevation (z value).