How do I parameterize the intersection of $x^2-y^2=11$ and $z=xy$

Question

The minus sign is totally throwing me off. If it were $x^2 + y^2$ I know they would be functions of the sin and cos of the parameter, but I can find no trig identity to make this work.

Answer 1

Use $\sec$ and $\tan$, or $\cosh$ and $\sinh$ instead of $\cos$ and $\sin$. You just need to remember that $1+\tan^2t = \sec^2 t$, or that $\cosh^2t-\sinh^2t=1$. I'll go with the hyperbolic functions. Since $x^2-y^2 = 11$ is equivalent to $(x/\sqrt{11})^2 - (y/\sqrt{11})^2=1$, we may simply let $x = \sqrt{11}\cosh t$ and $y = \sqrt{11}\sinh t$. The second equation immediately gives $z$. One possible parametrization is $$t\mapsto (\sqrt{11}\cosh t, \sqrt{11}\sinh t, (11/2)\sinh(2t)),$$for example. Observe that the intersection is not connected, and the other half is parametrized by $$t\mapsto (-\sqrt{11}\cosh t, \sqrt{11}\sinh t, -(11/2)\sinh(2t)).$$