Quadric Surfaces — Question 3

Question 3 – Solution

Strategy The sign of the $z^2$ coefficient controls the canonical type.

Case 1: $k<0$ Write $k=-a$ with $a>0$. Then $x^2+y^2+az^2=1$, an ellipsoid of revolution, bounded in every direction.

Case 2: $k=0$ The equation becomes $x^2+y^2=1$ with unrestricted $z$, so it is a circular cylinder of radius $1$ parallel to the $z$-axis, unbounded in $z$.

Case 3: $k>0$ The equation has two positive terms and one negative term, so it is a one-sheet hyperboloid about the $z$-axis. At $z=c$, the circle has radius $\sqrt{1+kc^2}$; the surface is unbounded in all three coordinates.

Verification In every case the $xy$-trace is the unit circle. Vertical traces change from ellipses, to parallel lines, to hyperbolas, confirming the transition.