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Converting from General to Standard

  1. Mar 14, 2005 #1
    I have the answer to this equation in standard form but I don't know how I they got it,


    [tex] 2x^2 - y^2 - 4x - 8 = 0 [/tex]

    so far I get this:

    [tex] 2x^2 - 4x - y^2 - 8 = 0 [/tex]

    [tex] 2 (x^2 - 2x +1) - y^2 -1 - 8 = 0 [/tex]

    [tex] 2 (x-1)^2 - y^2 -1 - 8 = 0 [/tex]

    the equation in standard form is...

    [tex] \frac {(x-1)^2}{4}+ \frac {y^2}{10} =1[/tex]
     
    Last edited: Mar 14, 2005
  2. jcsd
  3. Mar 14, 2005 #2

    dextercioby

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    It should be

    [tex] \frac{(x-1)^{2}}{\left (\frac{\sqrt{10}}{\sqrt{2}}\right)^{2}}-\frac{y^{2}}{(\sqrt{10})^{2}}=1 [/tex]

    A hyperbola...

    Daniel.
     
    Last edited: Mar 14, 2005
  4. Mar 14, 2005 #3
    But why? How did you get that answer. The answer I got was from my teacher so I'm surprised he'd be wrong.


    So where did I go wrong in my calculations?
     
  5. Mar 14, 2005 #4

    dextercioby

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    When adding & subtracting 2.

    Daniel.
     
  6. Mar 14, 2005 #5
    :confused:
     
  7. Mar 14, 2005 #6

    dextercioby

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    [tex] 2x^{2}-4x+2-2-y^{2}-8=0 [/tex]

    [tex] 2(x-1)^{2}-y^{2}=10 [/tex]

    Then it's simple to reach to my formula.

    Daniel.
     
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