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anonymous

  • one year ago

Anyone good with hyperbolas? Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1

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  1. anonymous
    • one year ago
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    \[\frac{ x^2 }{ 3^2 } -\frac{ y^2 }{ 4^2 } = 1\]

  2. anonymous
    • one year ago
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    Yes this was the problem we were working on last night. (: @saseal

  3. anonymous
    • one year ago
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    |dw:1439046272960:dw|

  4. anonymous
    • one year ago
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    \[c=\sqrt{a^2+b^2}\]

  5. anonymous
    • one year ago
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    Ok, so would it be the SqRt of 9+16? so the SqRt of 27?

  6. anonymous
    • one year ago
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    yea

  7. anonymous
    • one year ago
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    um no

  8. anonymous
    • one year ago
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    how did you get 9+16=27?

  9. anonymous
    • one year ago
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    Isn't a^2 = 9 and b^2 = 16?

  10. anonymous
    • one year ago
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    9+16=25

  11. anonymous
    • one year ago
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    Oh my bad. Stupid arithmetic problem. Ok, so c = 5?

  12. anonymous
    • one year ago
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    yea

  13. anonymous
    • one year ago
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    its +/- 5 actually because hyperbola goes both ways, just like vertices

  14. anonymous
    • one year ago
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    Oh ok!

  15. anonymous
    • one year ago
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    So is that for foci or vertices? @saseal

  16. anonymous
    • one year ago
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    c is for foci and a is your vertices

  17. anonymous
    • one year ago
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    Ok...so would that make the foci: (+/-5, 0) And vertices: (+/-3, 0) @saseal

  18. anonymous
    • one year ago
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    @saseal Could you help me with a couple more?

  19. anonymous
    • one year ago
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    yea

  20. anonymous
    • one year ago
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    and ok ill help you with the questions

  21. anonymous
    • one year ago
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    So the answer is correct with the foci and vertices? How do I know where to put the 0? Meaning, the zero always goes in the y values?

  22. anonymous
    • one year ago
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    @saseal

  23. anonymous
    • one year ago
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    |dw:1439048329264:dw|this is for \[\frac{ x^2 }{ a^2 } - \frac{ y^2 }{ b^2 } = 1\]

  24. anonymous
    • one year ago
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    |dw:1439048416352:dw|this is for \[\frac{ y^2 }{ a^2 } - \frac{ x^2 }{ b^2 } = 1\]

  25. anonymous
    • one year ago
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    Ok... @saseal

  26. anonymous
    • one year ago
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    the center of your hyperbola is deternined by the x^2 like \[\frac{ x^2 }{ a^2 } - \frac{ y^2 }{ b^2 } = 1\] has center at origin while\[\frac{ (x-1)^2 }{ a^2 } - \frac{( y+1)^2 }{ b^2 } =1\] has a center of (1,-1)

  27. anonymous
    • one year ago
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    Ok I get that

  28. anonymous
    • one year ago
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    its actually determined by x^2 and y^2 not just x^2, got distacted dem parents

  29. anonymous
    • one year ago
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    Oh ok, still makes sense. (:

  30. anonymous
    • one year ago
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    So my answer would be correct? @saseal

  31. anonymous
    • one year ago
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    yea

  32. anonymous
    • one year ago
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    Ok so my next question is: Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y plus five squared divided by nine = 1

  33. anonymous
    • one year ago
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    \[\frac{ (x+1)^2 }{ 4^2 } + \frac{ (y+5)^2 }{ 3^2 } = 1\] where do you think is the center of the hyperbola?

  34. anonymous
    • one year ago
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    I graphed it and got the center to be ( -1, -5 ) @saseal

  35. anonymous
    • one year ago
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    you should be able to get the center without even graphing it

  36. anonymous
    • one year ago
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    you see the pattern there with the (x+1)^2 and the (y+5)^2 and the center?

  37. anonymous
    • one year ago
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    Oh ok. Yes. -1, -5

  38. anonymous
    • one year ago
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    I get the center is (-1, -5), so now what? @saseal

  39. anonymous
    • one year ago
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    you know wheres your a and b?

  40. anonymous
    • one year ago
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    My a is 16, b is 9. So would that make c = 5?

  41. anonymous
    • one year ago
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    a^2 is 16 not a. a is 4, dont be mistaken.

  42. anonymous
    • one year ago
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    Oh ok! So would the c still be 5?

  43. anonymous
    • one year ago
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    yes

  44. anonymous
    • one year ago
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    now where do you think the hyperbola will open out to? left/right or up/down

  45. anonymous
    • one year ago
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    left/right

  46. anonymous
    • one year ago
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    correct

  47. anonymous
    • one year ago
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    |dw:1439050835472:dw|

  48. anonymous
    • one year ago
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    where do you think the vertices are?

  49. anonymous
    • one year ago
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    (-5, -5) and (3, 5) ?

  50. anonymous
    • one year ago
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    there you have it

  51. anonymous
    • one year ago
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    now we solve for c to get your foci

  52. anonymous
    • one year ago
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    \[c=\sqrt{a^2+b^2}\]

  53. anonymous
    • one year ago
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    c = +/- 5

  54. anonymous
    • one year ago
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    correct, now put that into the coords just like you did for vertices

  55. anonymous
    • one year ago
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    (-6, -5) and (4, -5) ?

  56. anonymous
    • one year ago
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    yup

  57. anonymous
    • one year ago
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    Ok, can you help me with one more? And check one?

  58. anonymous
    • one year ago
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    sure

  59. anonymous
    • one year ago
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    For: Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±7). Would the answer be: y^2/4 - x^2/45 = 1

  60. anonymous
    • one year ago
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    the x^2/45 part is wrong

  61. anonymous
    • one year ago
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    you need to meddle a bit with the foci formula, its not so straightforward

  62. anonymous
    • one year ago
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    Ok so how do I do that?

  63. anonymous
    • one year ago
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    ok if c^2 = a^2 + b^2 then c^2 - a^2 = b^2

  64. anonymous
    • one year ago
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    so for the x^2 would we subtract x^2/49?

  65. anonymous
    • one year ago
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    no

  66. anonymous
    • one year ago
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    foci is c which is 7 and your 4 is a^2

  67. anonymous
    • one year ago
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    so 33?

  68. anonymous
    • one year ago
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    \[7^2-2^2=?\]

  69. anonymous
    • one year ago
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    so yea its 45 lol

  70. anonymous
    • one year ago
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    i didnt square the stuff i got which is 6.708 thats why lol

  71. anonymous
    • one year ago
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    Oh ok!

  72. anonymous
    • one year ago
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    And the last problem on this lesson is: Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at y = ±five divided by six times x.

  73. anonymous
    • one year ago
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    im used to putting em with the ^2 on

  74. anonymous
    • one year ago
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    Oh ok

  75. anonymous
    • one year ago
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    this is definitely a hyperbola opening up/down

  76. anonymous
    • one year ago
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    |dw:1439051936537:dw|

  77. anonymous
    • one year ago
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    I agree

  78. anonymous
    • one year ago
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    \[\frac{ y^2 }{ 10^2 } - \frac{ x^2 }{ b^2 } = 1\]\[y=\pm \frac{ 5 }{ 6 }x\]

  79. anonymous
    • one year ago
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    So...how could I get it into standard form equation?

  80. anonymous
    • one year ago
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    and gradient is \[y=\frac{ b }{ a } x \]

  81. anonymous
    • one year ago
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    \[\frac{ b }{ a } = \frac{ 5 }{ 6 } = \frac{ b }{ 10 }\]

  82. anonymous
    • one year ago
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    Ok. So how can I solve for the b and a terms?

  83. anonymous
    • one year ago
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    multiply through\[\frac{ 5 }{ 6 } = \frac{ 50 }{ 60 } = \frac{ 6b }{ 60 }\]\[6b=50\]\[b=8.333333333333\]

  84. anonymous
    • one year ago
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    Ok that's what I thought.

  85. anonymous
    • one year ago
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    lemme recheck

  86. anonymous
    • one year ago
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    So would my answer be: y^2/64 - x^2/100 ? Because I just checked and b should equal 8...

  87. anonymous
    • one year ago
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    yup i made a mistake at the slope part b/a is for hyperbola opening left/right b/a is for opening up/down

  88. anonymous
    • one year ago
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    Ok so my final answer is y^2/64 - x^2/100 = 1?

  89. anonymous
    • one year ago
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    no lemme redo the last part

  90. anonymous
    • one year ago
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    cause isn't a^2 = 10 and b^2 = 8 ?

  91. anonymous
    • one year ago
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    i used the wrong formula for the asymtote lol

  92. anonymous
    • one year ago
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    lemme explain to ya

  93. anonymous
    • one year ago
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    |dw:1439052898040:dw|

  94. anonymous
    • one year ago
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    |dw:1439052978322:dw|

  95. anonymous
    • one year ago
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    see the difference?

  96. anonymous
    • one year ago
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    Yes

  97. anonymous
    • one year ago
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    One is going up/down, the other left/right

  98. anonymous
    • one year ago
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    \[\frac{ a }{ b } = \frac{ 5 }{ 6 } = \frac{ 10 }{ b }\] so b is 12

  99. anonymous
    • one year ago
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    Oh ok. Well I was just looking on open study, to see what others did, and someone else listed the answer as: y^2/64 - x^2/100 = 1 But I understand also what you're saying.

  100. anonymous
    • one year ago
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    same question?

  101. anonymous
    • one year ago
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    yes

  102. anonymous
    • one year ago
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    he prolly flipped the gradient like i did

  103. anonymous
    • one year ago
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    Oh ok. So would the answer be: y^2/144 - x^2/100 = 1?

  104. anonymous
    • one year ago
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    \[\frac{ y^2 }{ 10^2 }-\frac{ x^2 }{ 12^2 } = 1\]

  105. anonymous
    • one year ago
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    Oh ok! So simplified: y^2/100 - x^2/144 = 1

  106. anonymous
    • one year ago
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    Correct? @saseal

  107. anonymous
    • one year ago
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    yea

  108. anonymous
    • one year ago
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    Ok thanks!

  109. anonymous
    • one year ago
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    @saseal could you help me again? so sorry!!!

  110. anonymous
    • one year ago
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    ok

  111. anonymous
    • one year ago
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    maybe just open a new one lol this one is getting really long

  112. anonymous
    • one year ago
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    Ok lol

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