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briana.img

  • one year ago

How do you write an equation for a hyperbola that involves foci?

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  1. briana.img
    • one year ago
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  2. briana.img
    • one year ago
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    @jim_thompson5910 plz help you're the best

  3. jim_thompson5910
    • one year ago
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    |dw:1435360080154:dw|

  4. jim_thompson5910
    • one year ago
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    Foci |dw:1435360096843:dw|

  5. jim_thompson5910
    • one year ago
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    vertices |dw:1435360133480:dw|

  6. jim_thompson5910
    • one year ago
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    what is the midpoint of the vertices? |dw:1435360163097:dw|

  7. briana.img
    • one year ago
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    (4,0) ???

  8. jim_thompson5910
    • one year ago
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    4 is correct 0 is not

  9. briana.img
    • one year ago
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    (4,5)

  10. jim_thompson5910
    • one year ago
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    good

  11. jim_thompson5910
    • one year ago
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    |dw:1435360285728:dw|

  12. jim_thompson5910
    • one year ago
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    The distance from the center to either vertex is what numeric value?

  13. briana.img
    • one year ago
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    Ok I understand that the top half of the equation with be (x-4)^2 - (y-5)^2

  14. jim_thompson5910
    • one year ago
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    correct

  15. briana.img
    • one year ago
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    The distance from either vertex is two?

  16. jim_thompson5910
    • one year ago
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    from center to vertex (pick one) is 2 units, yes

  17. jim_thompson5910
    • one year ago
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    so a = 2 where \[\Large \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]

  18. jim_thompson5910
    • one year ago
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    'a' corresponds to the x portion because the hyperbola opens along the x axis direction (ie horizontally)

  19. jim_thompson5910
    • one year ago
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    c = 3 because it is 3 units from the center to either focus use a^2 + b^2 = c^2 to find b

  20. jim_thompson5910
    • one year ago
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    actually you don't need to solve for b you can stop at b^2 and that would be fine too

  21. briana.img
    • one year ago
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    @jim_thompson5910 oh okay! i clearly understand now!! thank you :)

  22. jim_thompson5910
    • one year ago
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    no problem

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