HELP!!!! Find the foci of 16y^2-36x^2-576=0. Then graph the hyperbole.

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HELP!!!! Find the foci of 16y^2-36x^2-576=0. Then graph the hyperbole.

Mathematics
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move -576 to the right side, what do you get?
Positive 576
yes, now, divide both sides by 576, what do you get?

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Other answers:

1/36y^2-1/16x^2
=??
=1?
\[\dfrac{y^2}{6^2}-\dfrac{x^2}{4^2}=1\] right?
yeah
Hence, its b =6, a =4 ok?
yes
foci c, \(c=\sqrt{a^2+b^2}=??\)
thanks can you stay with me for just a second i need to write this down
ok is it square root of 52
You need know where c is. Your equation has y first, hence its graph is vertical, right? Hence foci will be (0, -c) and (0,c) . Yes for c =sqrt 52
|dw:1437522466412:dw|
do you put the foci into decimal form for the graph or do you just leave it as c=sqrt52
|dw:1437522532448:dw|
leave it as it is.
okay so I would just put the sqrt52 into the graph vertically and be done with the graphing
okay im gonna assume its done

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