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anonymous
 one year ago
Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y equals inverse of plus minus 1 divided by 2x.
anonymous
 one year ago
Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y equals inverse of plus minus 1 divided by 2x.

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1what is the center of this hyperbola?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1so (h,k) = (0,0) h = 0 k = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hey that makes a cool face 0,0

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1lol yes it does how far do go from (0,0) to either vertex?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1so a = 8 a = half of the length of the transverse axis the transverse axis connects the two vertices and goes through the center

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0SO what is the formula that we need to plug these into?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1how is the hyperbola oriented? is it opening left/right? or up/down?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1according to this http://www.purplemath.com/modules/hyperbola.htm it says that if the hyperbola opens up/down, then the basic form is \[\Large \frac{(yk)^2}{a^2} + \frac{(xh)^2}{b^2} = 1\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1on that same page it says that the asymptotes have a slope of +a/b focus on say the positive slope slope = 1/2 a/b = 1/2 8/b = 1/2 b = ??

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \frac{(yk)^2}{a^2}  \frac{(xh)^2}{b^2} = 1\] \[\Large \frac{(y0)^2}{8^2}  \frac{(x0)^2}{16^2} = 1\] \[\Large \frac{y^2}{64}  \frac{x^2}{256} = 1\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I had a + in there for some reason. It should have been 

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok thanks! I have 4 more questions of similar nature if you dont't mind :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll make new questions

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1try the next one out and I'll take a look at the work you have so far on it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not the same type of question but the same chapter in my book
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