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