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Write the equation of the hyperbola whose vertices are at (0,±4) and whose foci are at (0,±11)
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the y axis is the parallel; and in hyper that gets the lower number
the trick is finding the missing 'b^2'
the center is the origin;
y^2 x^2 --- - ---- = 1 16 b^2
5.5^2 - 4^2 = b^2; if I recall it correctly
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14.25 = b^2 then; but we can prolly work it up better
57/4 hmmm
So that is the equation of the hyperbola?
no.... still trying to get it in my head right; i halved the 4 and 11 and wasnt sposed to
b^2 = 105 ..... 11^2 = b^2 + 4^2 11^2 - 4^2 = b^2 = 105
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y^2 x^2 ---- - ---- = 1 ; is my best shot at it 16 105
Okay, thanks!
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