anonymous
  • anonymous
Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±7).
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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bahrom7893
  • bahrom7893
http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx Look at the standard form of a hyperbola.
anonymous
  • anonymous
How do you get a^2 and b^2?
bahrom7893
  • bahrom7893
let me refresh this stuff real quick.

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anonymous
  • anonymous
Thank you so much
bahrom7893
  • bahrom7893
http://www.purplemath.com/modules/hyperbola.htm The vertices are some fixed distance a from the center.
bahrom7893
  • bahrom7893
|dw:1370613894271:dw|
anonymous
  • anonymous
So the points on the graph would be at 2 and -2?
bahrom7893
  • bahrom7893
yea so a = 2 because both points are two away from the center.
anonymous
  • anonymous
And b would equal 7?
bahrom7893
  • bahrom7893
@amistre64 im getting a bit confused, can you help us out here?
bahrom7893
  • bahrom7893
nevermind, i got this.
bahrom7893
  • bahrom7893
http://home.windstream.net/okrebs/page63.html Finally, decent website: So basically, Vertices: (a,0); (-a,0), so we know that a=2 Foci: (c,0); (-c,0), so we know that c = 7
bahrom7893
  • bahrom7893
c^2 = a^2+b^2 7^2 = 2^2+b^2 49 = 4 + b^2 b^2 = 49-4 = 45, b=sqrt(45) = 3sqrt(5)
amistre64
  • amistre64
hyperbolas are nice enough to use the pythag setup: c^2 = a^2 + b^2 it the ellipses that are mean and dirty and by convention decide to twist things about: a^2 = b^2 + c^2
bahrom7893
  • bahrom7893
hahah im sure she can get the equation from here. Gonna go have my breakfast. Thanks for visiting us amistre :)
amistre64
  • amistre64
youre welcome, and the ab parts tend to be a little more strict in hyperbolas. if you consider the asymptotes as b/a slope equations, then b is always under y and a is always under x trying to place them depending on x-y or y-x formats is just futile and pointless since the construct of abc aligns to the pythag thrm

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