anonymous
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hyperbola help
Mathematics
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anonymous
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hyperbola help
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
|dw:1351156478191:dw|
anonymous
  • anonymous
how can you find b in \[\frac{x^2}{3^2} - \frac{y^2}{b^2} =1 \] v?
anonymous
  • anonymous
intercept

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anonymous
  • anonymous
@Jonask is correct
anonymous
  • anonymous
do i sub (4,0) in ?
anonymous
  • anonymous
The foci are (plus/minus a*e,0) where the eccentricity e = sqrt(1+ (b^2/a^2))
anonymous
  • anonymous
? what do i do what tiwtha
anonymous
  • anonymous
whats interecept??
anonymous
  • anonymous
foci, 4 = 4/e -< e = 1 = sqrt( 1 + b^2/3^2)) Solve for b^2
anonymous
  • anonymous
can do intercept way??
anonymous
  • anonymous
What is "intercept way"? What intercept?
anonymous
  • anonymous
anonymous
  • anonymous
the intercept only works in an ellipse i thought you drew an ellipse
anonymous
  • anonymous
@estudier is correct
anonymous
  • anonymous
S/B foci, 4 = 3/e -< e = 1 = sqrt( 1 + b^2/3^2)) Solve for b^2
anonymous
  • anonymous
Can u do this? It is not difficult.
anonymous
  • anonymous
i think you can find by subbing in a point.. but i dunno which point
anonymous
  • anonymous
yeah but i dont think i can remember ..
anonymous
  • anonymous
can you find the corodinate of point directly above foci??
anonymous
  • anonymous
foci, 4 = 3/e -> e = 4/3 = sqrt( 1 + b^2/3^2)) Solve for b^2 So 4/3 = sqt(1 + b^2/3^2) -> b^2 = 12/9
anonymous
  • anonymous
yeah but i dont wanna remember that if in test -_-
anonymous
  • anonymous
they are related by a^2+b^2=c^2 as well right
anonymous
  • anonymous
U can say c^2 = a^2 -b^2 (same thing)
anonymous
  • anonymous
Actually, I made mistake in figures: 4/3 = sqrt(1 +b^2/3^2) -> 16/9 = 1 + b^2/3^2 -> 3^2 * 16/9 = 3^2 + b^2 -> b^2 = 3^2 (16/9 -1) = 3^2 *7/9 = 7 You can check it here http://www.wolframalpha.com/input/?i=%28x^2%2F3^2%29++++-+y^2%2F7+%3D+1

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