.
For the values y = 0.1, 1, 2, 3 we have elliptic traces:
in the planes x k, y = k, z = k.
Then identify the surface and sketch it.
Solution: The surface is an elliptic parabolid.
Rewrite the equation as
.
For the values y = 0.1, 1, 2, 3 we have elliptic traces:
At z = 0,
, so we have the parabola:
At x = 0, 
,
resulting in the parabola:
Bringing these results together, we have an elliptic parabolid:
by: gm