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Math Help - Classify the graph of the equation

  1. #1
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    Classify the graph of the equation

    I have an equation that i think is an ellipse or hyperbola since looking at the equation it has a y^ and a x^2 .
    However, i am having problems finishing it since i think i have to do complete the square but cant seem to get it.

    Heres the equation.

     4x^2 + 4y^2 - 4x - 3 = 0

    Wouldn't i arrange like terms

     4x^2 -4x     + 4y^2 - 3 = 0

    then, completing the square has me baffled.
    First term i take half of -4 = 2^2 = 4

    so i now have
     (4x^2 -4x  + 4)   + (4y^2 - 3)   = 0 +  4

    would i then take half of 3 which is  (\frac {3}{2})^2 = \frac{9}{4}

    If true then:

     (4x^2 -4x  + 4)   + (4y^2 - 3 + \frac{9}{4})   = 0 +  4 + \frac{9}{4}

    pull out the 4 on the left side - but not ure how to handle the second side?

     4(x^2 -x  + 1)   + (4y^2 - 3 + \frac{9}{4})   = 0 +  4 + \frac{9}{4}

    simplifying the left side further

     4(x - 1)^2   + (4y^2 - 3 + \frac{9}{4})   = 0 +  4 + \frac{9}{4}


    Any help is much appreciated..
    Thank you
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  2. #2
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    Re: Classify the graph of the equation

    Quote Originally Posted by icelated View Post
    Heres the equation.
     4x^2 + 4y^2 - 4x - 3 = 0
     4x^2 + 4y^2 - 4x - 3 = 0

    x^2-x+y^2=\tfrac{3}{4}

    x^2-x+\tfrac{1}{4}+y^2=\tfrac{3}{4}+\tfrac{1}{4}

    \left( {x - \tfrac{1}{2}} \right)^2  + y^2  = 1
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