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December 9th, 2011, 09:15 AM   #1
 
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Sketching graphs in R3

Sketch the graph .

Let z = 0:



Now when drawing the graph on the y and x axis, I understand that there are hyperbolas which intersect y-axis (as letting you get ).

What I don't get is in my book their hyperbolas drawn, (on the same graph), which intersect the x-axis. This shouldn't be possible since letting you get , which can't be solved.
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December 9th, 2011, 01:47 PM   #2
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Re: Sketching graphs in R3

Your question is confusing. Where is z? Also k=0 is possible - that's how you get the lines crossing the origin.
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December 10th, 2011, 01:10 AM   #3
 
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Re: Sketching graphs in R3

What I meant to say is when you let (to get the view from above in R3) you get:





This would be graph of the hyperbolas intersecting the y-axis, but in the book they are hyperbolas which also intersect the x-axis. Don't understand why there are hyperbolas that intersect the x-axis (as letting , you get which can't be solved).
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December 10th, 2011, 01:29 AM   #4
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Re: Sketching graphs in R3

The hyperbolas intersecting the x-axis represent negative values of k.

When 0 < k they intersect the y-axis, when k < 0 they intersect the x-axis, and when k = 0, you get degenerate cases, the lines y = x.

This is the family of curves satisfying:

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December 16th, 2011, 02:47 AM   #5
 
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Re: Sketching graphs in R3

For k < 0, how can you draw , when the right-hand-side has a negative 1? (The standard equation of a hyperbola has a positive 1 on the right-hand-side).
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December 16th, 2011, 06:03 AM   #6
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Re: Sketching graphs in R3

The hyperbolas of the form:



have vertices at (-a,0) and (a,0), while the hyperbolas of the form:



have vertices at (0,-a) and (0,a).

In our case, we have , so we have:

or

When k < 0, we may choose the form:



so that now we have a positive value on the right and vertices on the x-axis.

When 0 < k, we choose the form:



so that we have a positive value on the right and vertices on the y-axis.

In both cases, the asymptotes are the lines .

When k = 0, we have:



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