anonymous
  • anonymous
HELP!!!! Find the foci of 16y^2-36x^2-576=0. Then graph the hyperbole.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Loser66
  • Loser66
move -576 to the right side, what do you get?
anonymous
  • anonymous
Positive 576
Loser66
  • Loser66
yes, now, divide both sides by 576, what do you get?

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More answers

anonymous
  • anonymous
1/36y^2-1/16x^2
Loser66
  • Loser66
=??
anonymous
  • anonymous
=1?
Loser66
  • Loser66
\[\dfrac{y^2}{6^2}-\dfrac{x^2}{4^2}=1\] right?
anonymous
  • anonymous
yeah
Loser66
  • Loser66
Hence, its b =6, a =4 ok?
anonymous
  • anonymous
yes
Loser66
  • Loser66
foci c, \(c=\sqrt{a^2+b^2}=??\)
anonymous
  • anonymous
thanks can you stay with me for just a second i need to write this down
anonymous
  • anonymous
ok is it square root of 52
Loser66
  • Loser66
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
Loser66
  • Loser66
|dw:1437522466412:dw|
anonymous
  • anonymous
do you put the foci into decimal form for the graph or do you just leave it as c=sqrt52
Loser66
  • Loser66
|dw:1437522532448:dw|
Loser66
  • Loser66
leave it as it is.
anonymous
  • anonymous
okay so I would just put the sqrt52 into the graph vertically and be done with the graphing
anonymous
  • anonymous
okay im gonna assume its done

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