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anonymous

  • one year ago

Find the equation of the tangent line to the hyperbola at the given point: (x^2/a^2) - (y^2/b^2) = 1 (m, n) and write your answer in the form y=f(x) for some function f(x).

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  1. anonymous
    • one year ago
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    the point \((m,n)\)?

  2. anonymous
    • one year ago
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    I used implicit diff. to get y' = (2a^2y)/(2b^2x) but then what's next?

  3. anonymous
    • one year ago
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    Yes the point (m, n)

  4. anonymous
    • one year ago
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    cancel the twos

  5. anonymous
    • one year ago
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    Yup

  6. anonymous
    • one year ago
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    replace \(x\) by \(m\) and \(y\) my \(n\) for your slope

  7. anonymous
    • one year ago
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    so a^2y / b^2x becomes a^2n / b^2m ?

  8. anonymous
    • one year ago
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    yeah

  9. anonymous
    • one year ago
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    then use the point - slope formula

  10. anonymous
    • one year ago
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    So, y = (a^2n / b^2m)(x-m) + n ?

  11. anonymous
    • one year ago
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    yeah, nice and ugly

  12. anonymous
    • one year ago
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    i actually didn't check your derivative but i assume it is right

  13. anonymous
    • one year ago
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    hmm maybe it is not

  14. anonymous
    • one year ago
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    Yeah its not... >.<

  15. anonymous
    • one year ago
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    shouldn't it be \[\frac{b^2x}{a^2y}\]?

  16. anonymous
    • one year ago
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    I still get the wrong answer with that

  17. anonymous
    • one year ago
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  18. anonymous
    • one year ago
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    you still need to replace m and n for x and y

  19. anonymous
    • one year ago
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    \[\frac{b^2m}{b^2n}(x-m)+n\] try that

  20. freckles
    • one year ago
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    put an a^2 on bottom there instead of that b^2 :p

  21. anonymous
    • one year ago
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    it is late

  22. freckles
    • one year ago
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    yes I know it is time for the old folks to retire

  23. freckles
    • one year ago
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    that includes me of course

  24. anonymous
    • one year ago
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    ((b^2m) / (a^2n))(x-m) + n thats right

  25. anonymous
    • one year ago
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    :D Thanks so much!! @satellite73 @freckles :D

  26. anonymous
    • one year ago
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    yw

  27. freckles
    • one year ago
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    I'm just satellite's cheerleader on this one! Go satellite!!!

  28. freckles
    • one year ago
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    And Tracy! :)

  29. anonymous
    • one year ago
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    Ahaha XD thats an important role @freckles !!

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