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cutiecomittee123

  • one year ago

how do find the vertex and foci of this hyperbola (x-2)^2/36 - (y+1)^2/64 =1

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  1. misty1212
    • one year ago
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    HI!!!

  2. misty1212
    • one year ago
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    standard form is \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\] you get the center with your eyeballs right?

  3. misty1212
    • one year ago
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    this is where you say "yes i see it" or "no i have no clue"

  4. cutiecomittee123
    • one year ago
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    yes:)

  5. misty1212
    • one year ago
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    ok what is the center?

  6. cutiecomittee123
    • one year ago
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    (2,-1)

  7. misty1212
    • one year ago
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    ok good now we need to know one other thing what does it look like left or right?|dw:1433186027332:dw|

  8. cutiecomittee123
    • one year ago
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    second, its horizontal

  9. misty1212
    • one year ago
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    hmmm no

  10. misty1212
    • one year ago
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    the \(x\) term comes first, vertical

  11. misty1212
    • one year ago
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    now that we (presumably) have these two facts, the orientation and the center, the rest is easy

  12. cutiecomittee123
    • one year ago
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    oh gotcha

  13. misty1212
    • one year ago
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    the center is \((2,-1)\) and it is vertical that means the vertices are to the LEFT and RIGHT of the center (that is why we need the orientation first)

  14. misty1212
    • one year ago
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    if it was horizontal they would be up and down

  15. cutiecomittee123
    • one year ago
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    okay makes sense.

  16. misty1212
    • one year ago
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    you have \[\frac{(x-2)^2}{36}-\frac{(y+1)^2}{64}=1\] which looks like \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\] what is \(a\)?

  17. cutiecomittee123
    • one year ago
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    a=6

  18. misty1212
    • one year ago
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    right so how easy is this? the vertices are 6 units to the right and left of \((2,-1)\)

  19. misty1212
    • one year ago
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    got that ?

  20. cutiecomittee123
    • one year ago
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    okay so (-8, -1) and (5, -1)

  21. misty1212
    • one year ago
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    hmm no to the second one

  22. misty1212
    • one year ago
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    oh no no to both of them

  23. misty1212
    • one year ago
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    what is 6 units to the left of \((2,-1)\)?

  24. misty1212
    • one year ago
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    or is simple english , what is \(2-6\) and \(2+6\)?

  25. cutiecomittee123
    • one year ago
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    oh lol. (-4,-1) and (8,-1)

  26. misty1212
    • one year ago
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    whew!

  27. cutiecomittee123
    • one year ago
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    better?

  28. misty1212
    • one year ago
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    ok that takes care of the vertices now for the foci

  29. misty1212
    • one year ago
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    much yes

  30. cutiecomittee123
    • one year ago
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    alright so how do we find the foci?

  31. misty1212
    • one year ago
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    we need one number number , namely \(c\) and we find it like pythagoras \[a^2+b^2=c^2\]

  32. misty1212
    • one year ago
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    you got \(a^2=36,b^2=64\) what is \(c\)?

  33. cutiecomittee123
    • one year ago
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    so 36+64=100

  34. misty1212
    • one year ago
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    right and so \(c^2=100\) making \(c=?\)

  35. cutiecomittee123
    • one year ago
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    c=10

  36. cutiecomittee123
    • one year ago
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    well + or - 10

  37. misty1212
    • one year ago
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    just plus

  38. misty1212
    • one year ago
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    so now (don't mess this up!) what is ten units to the right and left of \((2,-1)\)?

  39. cutiecomittee123
    • one year ago
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    (12,-1) and (8,-1)

  40. misty1212
    • one year ago
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    lol

  41. misty1212
    • one year ago
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    i know what you meant (i hope)

  42. misty1212
    • one year ago
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    \[(12,-1)\] is ten units the the right, \[(\color{red}-8,1)\] is ten units to the left

  43. cutiecomittee123
    • one year ago
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    yeah :)

  44. misty1212
    • one year ago
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    ok that is it

  45. cutiecomittee123
    • one year ago
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    thank you :)

  46. misty1212
    • one year ago
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    it is really not that hard people here don't like it and math teachers make it seem way harder than it is you see we really only needed two things the center and the orientation

  47. cutiecomittee123
    • one year ago
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    yeah it seems simple

  48. misty1212
    • one year ago
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    \[\color\magenta\heartsuit\]

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