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powerangers69 Group Title

hyperbola help

  • 2 years ago
  • 2 years ago

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  1. powerangers69 Group Title
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    |dw:1351156478191:dw|

    • 2 years ago
  2. powerangers69 Group Title
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    how can you find b in \[\frac{x^2}{3^2} - \frac{y^2}{b^2} =1 \] v?

    • 2 years ago
  3. Jonask Group Title
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    intercept

    • 2 years ago
  4. soty2013 Group Title
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    @Jonask is correct

    • 2 years ago
  5. powerangers69 Group Title
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    do i sub (4,0) in ?

    • 2 years ago
  6. estudier Group Title
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    The foci are (plus/minus a*e,0) where the eccentricity e = sqrt(1+ (b^2/a^2))

    • 2 years ago
  7. powerangers69 Group Title
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    ? what do i do what tiwtha

    • 2 years ago
  8. powerangers69 Group Title
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    whats interecept??

    • 2 years ago
  9. estudier Group Title
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    foci, 4 = 4/e -< e = 1 = sqrt( 1 + b^2/3^2)) Solve for b^2

    • 2 years ago
  10. powerangers69 Group Title
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    can do intercept way??

    • 2 years ago
  11. estudier Group Title
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    What is "intercept way"? What intercept?

    • 2 years ago
  12. powerangers69 Group Title
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    @Jonask ??

    • 2 years ago
  13. Jonask Group Title
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    the intercept only works in an ellipse i thought you drew an ellipse

    • 2 years ago
  14. Jonask Group Title
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    @estudier is correct

    • 2 years ago
  15. estudier Group Title
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    S/B foci, 4 = 3/e -< e = 1 = sqrt( 1 + b^2/3^2)) Solve for b^2

    • 2 years ago
  16. estudier Group Title
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    Can u do this? It is not difficult.

    • 2 years ago
  17. powerangers69 Group Title
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    i think you can find by subbing in a point.. but i dunno which point

    • 2 years ago
  18. powerangers69 Group Title
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    yeah but i dont think i can remember ..

    • 2 years ago
  19. powerangers69 Group Title
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    can you find the corodinate of point directly above foci??

    • 2 years ago
  20. estudier Group Title
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    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

    • 2 years ago
  21. powerangers69 Group Title
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    yeah but i dont wanna remember that if in test -_-

    • 2 years ago
  22. powerangers69 Group Title
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    they are related by a^2+b^2=c^2 as well right

    • 2 years ago
  23. estudier Group Title
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    U can say c^2 = a^2 -b^2 (same thing)

    • 2 years ago
  24. estudier Group Title
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    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

    • 2 years ago
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is replying to Can someone tell me what button the professor is hitting...

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