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Babynini

  • one year ago

Identity and graph, Hyperbola?

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  1. Babynini
    • one year ago
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  2. Babynini
    • one year ago
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    @jim_thompson5910

  3. jim_thompson5910
    • one year ago
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    it's not a hyperbola you can use something like geogebra (which is what I just used) or desmos to check https://www.geogebra.org/ https://www.desmos.com/calculator

  4. Babynini
    • one year ago
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    oo thanks.

  5. Babynini
    • one year ago
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    ellipse?

  6. jim_thompson5910
    • one year ago
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    yep it's an ellipse

  7. Babynini
    • one year ago
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    fabulous, so...what next? I grouped stuff and factored out but i'm not sure how to finish the squares.

  8. Babynini
    • one year ago
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    4(x^2-8x ) 25(y^2+6y )=-189

  9. jim_thompson5910
    • one year ago
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    the x coordinate is -8 cut that in half to get -4 then square it to get 16 you will add and subtract 16 inside the parenthesis \[\Large 4(x^2-8x \ \ \ \ \ \ \ \ ) + 25(y^2+6y \ \ \ \ \ \ \ \ )=-189\] \[\Large 4(x^2-8x {\color{red}{+16-16}} ) + 25(y^2+6y \ \ \ \ \ \ \ \ )=-189\] the +16-16 is to make sure that the expression doesn't change (since we're effectively adding 0). Then notice how x^2-8x+16 factors to (x-4)^2

  10. Babynini
    • one year ago
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    hmm ok yeah

  11. Babynini
    • one year ago
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    so now we've got 4(x-4)^2 ?

  12. Babynini
    • one year ago
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    and is the right side 25(x-3)^2

  13. Babynini
    • one year ago
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    (sorry, by right I meant the y stuff)

  14. jim_thompson5910
    • one year ago
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    you're forgetting about the -16 in the parenthesis

  15. Babynini
    • one year ago
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    how do I write that?

  16. jim_thompson5910
    • one year ago
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    \[\Large 4(x^2-8x + \underline{ \ \ \ \ \ \ \ \ } \ ) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\] \[\Large 4(x^2-8x+16-16) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\] \[\Large 4((x^2-8x+16)-16) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\] \[\Large 4((x-4)^2-16) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\] \[\Large 4(x-4)^2-64 + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\]

  17. jim_thompson5910
    • one year ago
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    what goes in the blank for the y terms?

  18. Babynini
    • one year ago
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    ahh ok. y^2-6y+9

  19. Babynini
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    25((x-3)^2-9)

  20. Babynini
    • one year ago
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    25(x-3)^2-225 ?

  21. jim_thompson5910
    • one year ago
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    yeah +9-9 goes in the blank on line 1 for the y terms

  22. jim_thompson5910
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    be careful it is NOT y^2 - 6y it's y^2 + 6y

  23. Babynini
    • one year ago
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    sorry, my bad!

  24. jim_thompson5910
    • one year ago
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    so you should have this \[\Large 4(x-4)^2-64 + 25(y+3)^2 - 225=-189\]

  25. Babynini
    • one year ago
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    yeah. ok so since we're adding and doing stuff to the left side we need to do the same on the right, yeah?

  26. jim_thompson5910
    • one year ago
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    no need

  27. jim_thompson5910
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    that's why I had +16 and -16 to balance things out

  28. jim_thompson5910
    • one year ago
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    same for +9-9

  29. Babynini
    • one year ago
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    oh! that's fabulous :) I was wondering because in class I remember the +9 but and doing stuff to the left side also but I like this better haha

  30. jim_thompson5910
    • one year ago
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    well you could add things to both sides, but you'd have to be careful

  31. Babynini
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    Yep yep. So now what do we do?

  32. Babynini
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    add 64 and 225 to both sides?

  33. jim_thompson5910
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    yep

  34. Babynini
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    so 100 on the right side :P

  35. Babynini
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    then divide the whole thing by 100 so we get 0 on the right again?

  36. jim_thompson5910
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    you mean 1 on the right side

  37. jim_thompson5910
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    but yeah

  38. Babynini
    • one year ago
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    \[\frac{ 4(x-4)^2 }{ 100 }+\frac{ 25(y+3)^2 }{ 100 }=1\]

  39. Babynini
    • one year ago
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    haha yeah sorry.

  40. jim_thompson5910
    • one year ago
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    and you can rewrite that into \[\Large \frac{(x-4)^2}{25} + \frac{(y+3)^2}{4} = 1\]

  41. Babynini
    • one year ago
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    perfect :)

  42. Babynini
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    so now we get all the other stuff from there?

  43. jim_thompson5910
    • one year ago
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    yeah you can find the center, foci, vertices, co-vertices, etc

  44. Babynini
    • one year ago
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    a = 5 b = 4 c= sq29

  45. jim_thompson5910
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    how did you get c?

  46. Babynini
    • one year ago
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    oops b = 2

  47. Babynini
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    a^2+b^2=c^2

  48. Babynini
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    25+4=c^2

  49. jim_thompson5910
    • one year ago
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    http://www.mathwarehouse.com/ellipse/images/formul-focus.gif

  50. Babynini
    • one year ago
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    ou. then c = sqrt 21

  51. jim_thompson5910
    • one year ago
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    yep

  52. Babynini
    • one year ago
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    kks, and then the ellipse will be sideways yeah?

  53. Babynini
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    so foci: \[(0,\pm \sqrt{21})\]

  54. jim_thompson5910
    • one year ago
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    not quite

  55. jim_thompson5910
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    the foci will shift along with the center

  56. jim_thompson5910
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    |dw:1433386754426:dw|

  57. jim_thompson5910
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    |dw:1433386763450:dw|

  58. jim_thompson5910
    • one year ago
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    |dw:1433386776494:dw|

  59. Babynini
    • one year ago
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    so we've got to find the center first.

  60. jim_thompson5910
    • one year ago
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    yes

  61. jim_thompson5910
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    in this case, the major axis is horizontal, so the foci + center all lay on the same horizontal level. Only the x values will change

  62. jim_thompson5910
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    oh nearly forgot the vertices |dw:1433387027602:dw|

  63. Babynini
    • one year ago
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    wouldn't want to forget those! :P

  64. Babynini
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    ok. that makes sense.

  65. jim_thompson5910
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    for vertical ellipses (where b > a) \[\Large \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] we have this schematic |dw:1433387213931:dw|

  66. Babynini
    • one year ago
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    this one isn't horizontal? o.0

  67. jim_thompson5910
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    no I'm saying IF it were vertical, then you'd have what you see above

  68. Babynini
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    ooh right right

  69. jim_thompson5910
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    the ellipse you have is horizontal, like this |dw:1433387651593:dw|

  70. Babynini
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    so how do we find the center? :)

  71. jim_thompson5910
    • one year ago
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    Compare \[\Large \frac{(x-4)^2}{25} + \frac{(y+3)^2}{4} = 1\] with \[\Large \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]

  72. jim_thompson5910
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    The center of \[\Large \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] is (h,k)

  73. Babynini
    • one year ago
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    so (4,-3)

  74. jim_thompson5910
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    yes

  75. Babynini
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    so foci : (4- sq21, -3) , (4sq21, -3)

  76. Babynini
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    ?

  77. jim_thompson5910
    • one year ago
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    I think you meant to say \[\Large \left(4 {\color{red}{ \textbf{+}}}\sqrt{21},-3\right)\] for the second focus

  78. Babynini
    • one year ago
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    yeah

  79. jim_thompson5910
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    you have the correct foci

  80. Babynini
    • one year ago
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    haha yay! Ok, vertex: (+/-5, -3) ?

  81. jim_thompson5910
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    no

  82. jim_thompson5910
    • one year ago
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    did you mean \[\Large \left(4 \pm 5, -3\right)\]??

  83. Babynini
    • one year ago
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    ai ai ai yes sorry!

  84. Babynini
    • one year ago
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    (9,-3) (-1,-3)

  85. jim_thompson5910
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    both correct

  86. Babynini
    • one year ago
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    phew.

  87. Babynini
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    Length of major axis: 2a? minor axis 2b?

  88. jim_thompson5910
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    yes correct

  89. jim_thompson5910
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    only because a > b

  90. Babynini
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    so major: 50 minor: 8

  91. jim_thompson5910
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    50?

  92. Babynini
    • one year ago
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    lol I did (a^2)2 major = 10

  93. jim_thompson5910
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    major is 10, yep

  94. jim_thompson5910
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    minor isn't 8

  95. Babynini
    • one year ago
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    4

  96. jim_thompson5910
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    yep minor = 4

  97. Babynini
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  98. Babynini
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    all good? :)

  99. jim_thompson5910
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    everything is perfect

  100. Babynini
    • one year ago
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    thanks so much! a thousand medals for your efforts :)

  101. jim_thompson5910
    • one year ago
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    you're welcome

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