multivariable calculus
posted by cheuk on .
Consider the surface given by the equation:
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.

2x^2  3z^2  4x  4y  12z + 2 = 0
2(x1)^22  3(z+2)^2+12  4y + 2 = 0
2(x1)^2  3(x+2)^2  4y + 12 = 0
4(y3) = 2(x1)^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.