Question

Consider the level surface given by

Answer 1

$x^2-y^2+z^2=2$

If $y=2$, then you get $x^2-2^2+z^2=2\iff x^2+z^2=6$ which is a circle in the x-z plane with radius $\sqrt6$.

If $x=1$, then you get $1^2-y^2+z^2=2\iff z^2-y^2=1$ which is a hyperbola in the y-z plane.

If $y=0$, you get another circle in the x-z plane defined by $x^2+z^2=2$, this time with radius $\sqrt2$.

If $x=2$, then $2^2-y^2+z^2=2\iffy^2-z^2=2$, another hyperbola in the y-z plane with branches perpendicular to the case where $x=1$.