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Mathematics 7 Online
OpenStudy (anonymous):

Write the equation of the hyperbola whose vertices are at (0,±4) and whose foci are at (0,±11)

OpenStudy (amistre64):

the y axis is the parallel; and in hyper that gets the lower number

OpenStudy (amistre64):

the trick is finding the missing 'b^2'

OpenStudy (amistre64):

the center is the origin;

OpenStudy (amistre64):

y^2 x^2 --- - ---- = 1 16 b^2

OpenStudy (amistre64):

5.5^2 - 4^2 = b^2; if I recall it correctly

OpenStudy (amistre64):

14.25 = b^2 then; but we can prolly work it up better

OpenStudy (amistre64):

57/4 hmmm

OpenStudy (anonymous):

So that is the equation of the hyperbola?

OpenStudy (amistre64):

no.... still trying to get it in my head right; i halved the 4 and 11 and wasnt sposed to

OpenStudy (amistre64):

b^2 = 105 ..... 11^2 = b^2 + 4^2 11^2 - 4^2 = b^2 = 105

OpenStudy (amistre64):

y^2 x^2 ---- - ---- = 1 ; is my best shot at it 16 105

OpenStudy (anonymous):

Okay, thanks!

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