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Math Help - Conic section

  1. #1
    Member OnMyWayToBeAMathProffesor's Avatar
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    Conic section

    I know this is a conic section and that it is a circle, I can get that much but I have forgotten how to simplify it down to a variation of the standard formula, x^2+y^2=r^2. I would love for someone to refresh my memory, Thank you.
    <br />
4x^2+y^2-8x+4y+4=0
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    I know this is a conic section and that it is a circle, I can get that much but I have forgotten how to simplify it down to a variation of the standard formula, x^2+y^2=r^2. I would love for someone to refresh my memory, Thank you.
    <br />
4x^2+y^2-8x+4y+4=0
    4x^2+y^2-8x+4y+4=0

    Group the x and y terms together:

    (4x^2-8x)+(y^2+4y+4)=0\implies 4(x^2-2x)+(y^2+4y+4)=0

    Complete the square for the x term. The equation then becomes:

    4(x^2-2x+1)+(y^2+4y+4)=4

    Now factor:

    4(x-1)^2+(y+2)^2=4

    The equation now becomes \color{red}\boxed{(x-1)^2+\frac{(y+2)^2}{4}=1}

    This is an ellipse with the major axis along the y-axis.

    I hope this makes sense!

    --Chris
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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    THANK YOU very much, I knew it was something simple like that. It has been so long since i did a problem like that.
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