Consider the surface given by the equation:

Question

2x2 - 3z2 - 4x -4y - 12z + 2 = 0
(a) Write in standard form )
(b) Determine and classify the traces with planes parallel to the xy-, xz- and yz- planes.
(c) Hence classify the surface.

Answer 1

2x^2 - 3z^2 - 4x - 4y - 12z + 2 = 0
2(x-1)^2-2 - 3(z+2)^2+12 - 4y + 2 = 0
2(x-1)^2 - 3(x+2)^2 - 4y + 12 = 0
4(y-3) = 2(x-1)^2 - 3(z+2)^2

Looks like an hyperbolic paraboloid with vertex at (1,3,-2)

The traces are an hyperbola in the xz plane, and parabolas in the others.