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anonymous
 5 years ago
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 (x2)^2
  
? ?
anonymous
 5 years ago
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 (x2)^2    ? ?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the y is the heavy side; so this opens up and down

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0teh distance between the foci is 8; so each foci is 4 from center; that 4 is either a or c :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we have 4 between the vertices; so each vertex is 2 from center

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right how do I find b^2 then?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0under the y we have 2^2 right? let me look this up some more :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok I think actually I have to know what c^2 is cause C^2=a^2+b^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm thinkin' b^2=16...?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0tell me your understanding of a b and c; what do they measure?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the foci are c units from the center the vertices are a units from the center

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so; a^2 + b^2 = c^2 then right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02 + b^2 = 64 b^2 = 62 then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think so....but that doesn't eem right maybe it's minus?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my options for b^2 are 12 or 16 here but I don't see how they get either of those

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i used 8 instead of 4 :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok I was gonna say 16 cause 4^2=16 but 12 works too

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the minor axis is always smaller in number than the major axis

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0since the major is the y on this one; and it is 16; the other has to be less

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or maybe i messed that up in thougth

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ugh... idk this is all a review for my final and I am thinking that so far i'm not doing too hot. YIKES

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y^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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