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anonymous

  • 5 years ago

What are the retricemptotes of the hyperbola: (x-1)^(2)/(25)-(y-4)^(2)/(16)=1

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  1. amistre64
    • 5 years ago
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    bx/a where b and a are the sqruare roots of your denoms

  2. amistre64
    • 5 years ago
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    since the x spot is bigger; that isthe width of the rectangular asymptotes of the hyperbola

  3. amistre64
    • 5 years ago
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    -5,5 the height of this rectangular thing is -4,4

  4. amistre64
    • 5 years ago
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    i just noticed its off centered lol

  5. amistre64
    • 5 years ago
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    (x-1)^2 (y-4)^2 -------- - ------- = 1 a^2=25 b^2=16 if this thing were centered wed have the rectangle asymptote at: y = (4/5) x but lets put it back now

  6. amistre64
    • 5 years ago
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    (y-4) = (4/5)(x-1) y = (4/5)x - 4/5+20/5 y = (4/5)x + 16/5

  7. amistre64
    • 5 years ago
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    and the other is just the negative slope of that i believe

  8. anonymous
    • 5 years ago
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    so is (4/5)x+16/5 the retricepmtote?

  9. amistre64
    • 5 years ago
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    if i did it right, it should be one of them; th eother would be: y = -(4/5)x + 16/5

  10. amistre64
    • 5 years ago
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    y = (b/a)x is the normal asymptote; but since this is off centered, we gotta adjust for that discrepency

  11. amistre64
    • 5 years ago
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    gotta get to calc class and learn limits lol.... have fun ;)

  12. anonymous
    • 5 years ago
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    it can have multiple retricepmtotes, so would it be wrriten as (4/5)x, and 16/5?

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