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Mathematics 6 Online
OpenStudy (lovelyharmonics):

hyperbolas

OpenStudy (lovelyharmonics):

Find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9).

OpenStudy (lovelyharmonics):

do you just square these to find the denominators?

OpenStudy (anonymous):

nope

OpenStudy (anonymous):

hmm you have to find b and a right so verices are (0,+-6) so b=6

OpenStudy (lovelyharmonics):

okay so then i square that and its 36

OpenStudy (anonymous):

your foci is (0,+-9) a will be \[\sqrt{6^2 - a^2} =9\]

OpenStudy (anonymous):

thats + underroot * typo

OpenStudy (lovelyharmonics):

\[\frac{ y^2 }{ 36 } - \frac{ x^2 }{ 81 }\]

OpenStudy (anonymous):

so a^2 = 81-36 a^2 = 45

OpenStudy (anonymous):

x^2/45

OpenStudy (anonymous):

so y^2/36 - x^2/45

OpenStudy (anonymous):

=1

OpenStudy (lovelyharmonics):

i have a few more questions im going to post if you would like to help c:

OpenStudy (anonymous):

yeah okay

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