anonymous
  • anonymous
I need help with hyperbolas please? Posting the hyperbola equations and questions in the comments.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
x^2 - y^2 = 1
anonymous
  • anonymous
\[\frac{ (x-1)^{2} }{ 2 }-\frac{ (y+3)^{2} }{ 2 }=1\]
anonymous
  • anonymous
The center of the hyperbola is (1, -3), right?

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anonymous
  • anonymous
yes
anonymous
  • anonymous
Also, I need to find the lengths of the transverse and conjugate axes, the slopes of the asymptotes, and the foci, but I'm really not quite sure how to do that...
anonymous
  • anonymous
the slope of the asymptotes is the easiest part if you have \[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\] the slopes are \(m=\pm\frac{a}{b}\) in this case \(\pm1\)
anonymous
  • anonymous
Oh my, I wrote the equation wrong, a^2 should be 64 and b^2 should be 81.
anonymous
  • anonymous
So, then the slopes of the asymptotes will be 64/81? or would that be vice versa, 81/64? is a always greater than b?
anonymous
  • anonymous
ok no problem in this case you have \[\frac{(x-1)^2}{8^2}-\frac{(y+3)^2}{9^2}=1\]
anonymous
  • anonymous
the slope is not \(\pm\frac{a^2}{b^2}\) it is \(\pm\frac{a}{b}\)
anonymous
  • anonymous
in your case \(\pm\frac{8}{9}\)
anonymous
  • anonymous
does that need to be simplified any further or can it stay as a fraction?
anonymous
  • anonymous
actually it is wrong
anonymous
  • anonymous
should be \(\pm\frac{9}{8}\) i had it upside down
anonymous
  • anonymous
oh, okey dokey. so does that need to be simplified or can it stay as a fraction?
anonymous
  • anonymous
not sure what you mean by "simplify" there is no common factors, so you cannot reduce it just leave it as it is
anonymous
  • anonymous
I mean't, like, turn it into a decimal, but I figured it should just be left as is, so thank you (:
anonymous
  • anonymous
Is there a formula I use to find the foci and the lengths of the axes?
anonymous
  • anonymous
i think the lengths are just 8 and 9
anonymous
  • anonymous
that is, semi major is 8 semi minor is 9 not much to that part
anonymous
  • anonymous
finding the foci starts with having an idea of what this looks like
anonymous
  • anonymous
|dw:1370481762189:dw|
anonymous
  • anonymous
there is my lousy picture, but i drew it to show that the foci are to the right and left of the center, which is \((1,-3)\)
anonymous
  • anonymous
Yes, I know where they're located, I'm just not sure how to find their coordinates using the equation.
anonymous
  • anonymous
since \(a^2+b^2=64+81=145\) you know \(c=\sqrt{145}\)
anonymous
  • anonymous
and since the center is \((1,-3)\) the foci are at \((1-\sqrt{145},-3)\) and \((1+\sqrt{145},-3)\)
anonymous
  • anonymous
Oh, okay, so the square root of 145 is 12.04159458, so that means the foci are (-11.04159458, -3) and (13.04159458, -3), right?
anonymous
  • anonymous
Is it okay to round those numbers? Will it still be accurate enough?
anonymous
  • anonymous
wait a second, is this hyperbola vertical? i remember reading something a while back saying that a (the major axis) is always longer (greater) than b (the minor axis), and if they're flipped, wouldn't that make it vertical?
anonymous
  • anonymous
yeah i drew it wrong answer was correct though you can check it here http://www.wolframalpha.com/input/?i=\frac{%28x-1%29^2}{8^2}-\frac{%28y%2B3%29^2}{9^2}%3D1
anonymous
  • anonymous
Okay! Thank you!

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