anonymous
  • anonymous
Find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@phi
phi
  • phi
the center is the average of the vertices
anonymous
  • anonymous
So (0,0)

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phi
  • phi
the vertices are on the y-axis, so the hyperbola looks like a frown/smile combo that means the "y' goes first in the standard equation \[ \frac{(y-k)^2}{a^2}- \frac{(x-h)^2}{b^2}=1\] a is the distance from the center to the focus "c" is the distance from the center to the vertex we use a and c to find b a^2 + b^2 = c^2 we already know (h,k) is (0,0)
phi
  • phi
oops, got that swapped: a is the distance from the center to the vertex "c" is the distance from the center to the focus
anonymous
  • anonymous
alright.
anonymous
  • anonymous
so a is 6 and c is 9
phi
  • phi
yes, but in the equation you use a^2 = 36 and c^2=81 what is b^2 ?
anonymous
  • anonymous
\[3\sqrt{13}\]
anonymous
  • anonymous
b^2 is 117
phi
  • phi
a^2 + b^2 = c^2 36 + b^2 = 81
anonymous
  • anonymous
oh my bad
anonymous
  • anonymous
b^2 = 45 b=3sqrt5
phi
  • phi
now fill in the numbers to get the equation
anonymous
  • anonymous
thanks. I still have a few more

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