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Let $M$ be a hyperboloid of one sheet satisfying $x^2+y^2-z^2=1$. Show that $x(u,v)=(\frac{uv+1}{uv-1},\frac{u-v}{uv-1},\frac{u+v}{uv-1})$ gives a parametrization of $M$ where both sets of parameter curves are rulings.

I have been working on this for quite a while but have not been able to write $x(u,v)$ with the above parametrization as a ruled surface in $u$. I tried letting $u=\tan(\varphi)$ and $v=\tan(\psi)$ but it did not help.

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HINTS: First you should check that the points given by the parametrization do in fact lie on the hyperboloid. Next, you want to see that the $u$- and $v$-curves are in fact lines. You need to do some algebraic manipulations with the rational functions (like "long division"). It will help to fix $v=v_0$, say.

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  • $\begingroup$ That was what confused me the most. Say a u-curve is $x(u,v_0)$. How do I long divide $x(u,v)$ coordinate by coordinate? For instance, I cannot find a real-valued function $f(u)$ and a real number $A$ such that $$\frac{uv+1}{uv-1}=f(u)+A\frac{uv_0+1}{uv_0-1}.$$ $\endgroup$ – user269715 Feb 10 '16 at 10:42
  • $\begingroup$ I'm not quite sure where that equation came from. You should be considering $\dfrac{uv_0+1}{uv_0-1} = 1 + \dfrac 2{uv_0-1}$. $\endgroup$ – Ted Shifrin Feb 10 '16 at 15:43
  • $\begingroup$ Isn't a ruled surface $x(u,v)=\alpha(u)+v\beta(u)$ where $v\in\mathbb{R}$ with $\beta(u)$ being a ruling by definition? So if we want to show a $u$-curve is a ruling, then we should find the corresponding $\alpha(u)$ and constant $v$. Did I misunderstand the question or the definition? $\endgroup$ – user269715 Feb 10 '16 at 18:55
  • $\begingroup$ You're correct, as far as the standard parametrization goes. But there's nothing that says that you have to move along the rulings with constant speed. Any reasonable function will do in the place of $v$ in that formula. $\endgroup$ – Ted Shifrin Feb 10 '16 at 18:59
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    $\begingroup$ @WillJagy: I worked it out a number of years ago, never having seen it anywhere. I put it as an exercise in my differential geometry notes (which the OP may well be doing). I have recently run across it elsewhere. It basically comes from the fact that the hyperboloid of one sheet is projectively equivalent to the standard saddle surface. $\endgroup$ – Ted Shifrin Feb 10 '16 at 21:49

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