Cecilia Wilson

Answered

2022-11-25

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

Answer & Explanation

artyleriaCuy

Expert

2022-11-26Added 10 answers

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

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