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I am looking at the following exercise:

The hyperboloid of one sheet is $$S=\{(x,y,z)\in \mathbb{R}^3 \mid x^2+y^2-z^2=1\}$$ Show that, for every $\theta$, the straight line $$(x − z) \cos \theta = (1 − y) \sin \theta, \ \ (x + z) \sin \theta = (1 + y) \cos \theta \tag 1$$ is contained in $S$, and that every point of the hyperboloid lies on one of these lines.

Deduce that $S$ can be covered by a single surface patch, and hence is a surface.

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I have done the following:

Multiplying the two equations of $(1)$ we get $$(x^2+y^2-z^2-1)\cos \theta \sin \theta=0 \\ \Rightarrow x^2+y^2-z^2=1 \text{ unless } \cos \theta =0 \text{ or } \sin \theta =0$$

If $\cos \theta=0$ then $y=1, x=-z$ $\rightarrow (-z,1,z)\in S$.

If $\sin \theta=0$ then $x=z, y=-1$ $\rightarrow (z,-1,z)\in S$.

So, for each $\theta$, the line $(1)$ is contained in $S$.

Is this correct?

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Then we take a point of the hyperboloid, $(x,y,z)$.

How can we show that this lies on the one of these lines?

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EDIT:

How could we deduce that $S$ can be covered by a single surface patch, and hence is a surface?

Also how could we find a second family of straight lines on $S$, and show that no two lines of the same family intersect, while every line of the first family intersects every line of the second family with one exception?

Mary Star
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1 Answers1

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Take a point on $S$, $p = (x,y,z)$. We know that

$x^2 + y^2 - z^2 - 1 = 0$.

Thus $(x+z)(x-z) = (1+y)(1-y)$. Suppose $(x-z) \not = 0$ and $(1+y) \not = 0$ (you can do these two cases by hands). Then of course

$\frac{x+z}{1+y} = \frac{1-y}{x-z} = a$ for some $a \in \mathbb{R}$.

Since $ctg(x)$ is onto, there is $\theta$ such that $a = ctg(\theta) = \frac{cos(\theta)}{sin(\theta)}$, so you get

$(x+z)sin(\theta) = (1+y)cos(\theta)$

and

$(x-z)cos(\theta) = (1-y)sin(\theta)$.

Maffred
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    I forgot to say that what you have done is actually correct! – Maffred Jan 08 '16 at 19:45
  • We have to suppose that $x-z\neq$ and $1+y\neq 0$, right? What can we do at the cases where $x-z=0$ or $1+y=0$ ? – Mary Star Jan 08 '16 at 19:52
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    The case $x-z = 0$ and $1+y =0$ is easy. These two equations define a line, that you can rewrite making multiplication with $sin(0)$ and $cos(0)$ as $(x-z)cos(0) = (1-y)sen(0)$ and $(x+z)sin(0) = (1+y)cos(0)$. When just one of them is verified, one equation is trivially true $0=0$, you can just solve the other in order to find $\theta$. For example when $x-z=0$ and $y-1=0$ then just take $\theta = arcctg(x)$ you get $x = cos/sin$ thus $2xsin(\theta) = 2cos(\theta)$ and so using the two equations you get $(x+z)sin(\theta) = (1+y)cos(\theta)$. – Maffred Jan 09 '16 at 05:05
  • I got it!! Thank you for your answer!! :-) $$$$ I added two more questions in my inital post... Could you take a look at it? Do you have an idea? – Mary Star Jan 09 '16 at 10:10
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    This is your surface https://en.wikipedia.org/wiki/Hyperboloid – Maffred Jan 09 '16 at 17:16
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    The Jacobian of $F(x,y,z):=x^2 + y^2 - z^2 - 1$ is $(0,0)$ only in $(0,0)$ so it is a smooth surface. You can move the lines continuously to go from a point $p$ to any point $q$ so it is a single surface patch (ie connected) – Maffred Jan 09 '16 at 17:18
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    Probably good lines are the one you get dividing for the other factors, ie $\frac{x+z}{1-y} = \frac{1+y}{x-z} = a$ but I'm not sure about that and I'm not an expert! Check if they work! – Maffred Jan 09 '16 at 17:22
  • Why can we move the lines continuously to go from a point $p$ to any point $q$ ? To conclude to that do we use the fact that the line is contained in $S$ for each $\theta$ and that each point of $S$ lies on the line for a specific $\theta$ ? – Mary Star Jan 09 '16 at 22:10
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    Exactly, you can change $\theta$ from $\theta_1$ to $\theta_2$ continuously remaining in $S$! – Maffred Jan 10 '16 at 01:54
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    It's like you start on $p$, then you move thanks to $\theta$ in a circular path and arrive on a line containing $q$, then you move on that line to reach $q$. – Maffred Jan 10 '16 at 01:56
  • The single surface patch that covers the surface is the line $(1)$, right? Which is a parametrization of this surface patch? – Mary Star Jan 10 '16 at 10:03
  • I didn't know you were looking for 1 chart! I don't know differential geometry to be honest, I just liked your question. By the way as far as I understood, you can't use a single chart in something like this that is similar to a cylinder, unless you allow charts from a generic open set, not only from simply connected one. If you allow punctured unit disk to be a chart you can use just 1 chart. It's like you stretch the punctured disk in the vertical direction while you enlarge the missing point. Try read here http://math.stackexchange.com/questions/243257/looking-for-an-atlas-with-1-chart?rq=1 – Maffred Jan 10 '16 at 17:03