anonymous
  • anonymous
Find the standard form of the equation of the hyperbola with the given characteristics. vertices: (2, - 6), (2, - 2) foci: (2, - 8), (2, 0) I know this much so far: (y+4)^2 (x-2)^2 ------ - ------ ? ?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
the y is the heavy side; so this opens up and down
amistre64
  • amistre64
teh distance between the foci is 8; so each foci is 4 from center; that 4 is either a or c :)
anonymous
  • anonymous
ok so a

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amistre64
  • amistre64
we have 4 between the vertices; so each vertex is 2 from center
anonymous
  • anonymous
right how do I find b^2 then?
amistre64
  • amistre64
under the y we have 2^2 right? let me look this up some more :)
anonymous
  • anonymous
ok I think actually I have to know what c^2 is cause C^2=a^2+b^2
anonymous
  • anonymous
I'm thinkin' b^2=16...?
amistre64
  • amistre64
tell me your understanding of a b and c; what do they measure?
anonymous
  • anonymous
the foci are c units from the center the vertices are a units from the center
amistre64
  • amistre64
so; a^2 + b^2 = c^2 then right?
amistre64
  • amistre64
2 + b^2 = 64 b^2 = 62 then
anonymous
  • anonymous
I think so....but that doesn't eem right maybe it's minus?
anonymous
  • anonymous
my options for b^2 are 12 or 16 here but I don't see how they get either of those
amistre64
  • amistre64
like this right?
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amistre64
  • amistre64
i used 8 instead of 4 :)
anonymous
  • anonymous
i think so...
amistre64
  • amistre64
16-4 = 12
amistre64
  • amistre64
b^2 = 12
anonymous
  • anonymous
oh ok I was gonna say 16 cause 4^2=16 but 12 works too
amistre64
  • amistre64
the minor axis is always smaller in number than the major axis
amistre64
  • amistre64
since the major is the y on this one; and it is 16; the other has to be less
amistre64
  • amistre64
or maybe i messed that up in thougth
anonymous
  • anonymous
ugh... idk this is all a review for my final and I am thinking that so far i'm not doing too hot. YIKES
amistre64
  • amistre64
y^2 x^2 --- - --- = 1 4 12 b^2 = focal distance(c)^2 - vertex distance(a)^2 in this case; the hyperbola; c is the greatest distance beacusae it is used as the hypotenust in the ellipse; a is the greatest distance becasue IT is used for ITS hypotenuse

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