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Math Help - Reducing Quadratic Surface Equation, HELP!

  1. #1
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    Reducing Quadratic Surface Equation, HELP!

    Reduce the equation to one of the standard forms

    x2 y2 + z2 2x + 2y + 8z + 13 = 0

    And what kind of surface is it?

    I got (x-1)^2 - (y-1)^2 + (z+4)^2 = 5
    that answer was wrong, so then i switched the (y-1)^2 to (y+1)^2 and it was still wrong. I also used complete the square to solve it. So now I'm lost and not sure what to do, HELP!
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  2. #2
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    Quote Originally Posted by acg716 View Post
    Reduce the equation to one of the standard forms

    x^2 y^2 + z^2 2x + 2y + 8z + 13 = 0

    I got (x-1)^2 - (y-1)^2 + (z+4)^2 = 5
    that answer was wrong, so then i switched the (y-1)^2 to (y+1)^2 and it was still wrong. I also used complete the square to solve it. So now I'm lost and not sure what to do, HELP!
    (x^2-2x) - (y^2 -2y) + (z^2+8z) + 13 = 0
    ((x-1)^2 - 1) - ((y-1)^2 - 1) + ((z+4)^2 - 16) + 13 = 0
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  3. #3
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    Hello,
    Quote Originally Posted by acg716 View Post
    Reduce the equation to one of the standard forms

    x^2 y^2 + z^2 2x + 2y + 8z + 13 = 0

    I got (x-1)^2 - (y-1)^2 + (z+4)^2 = 5
    that answer was wrong, so then i switched the (y-1)^2 to (y+1)^2 and it was still wrong. I also used complete the square to solve it. So now I'm lost and not sure what to do, HELP!
    I get (x-1)-(y-1)+(z+4)=3

    We have (by ordering) :
    (x^2-2x)-(y^2-2y)+(z^2+8z)+13=0
    [(x-1)^2-1]-[(y-1)^2-1]+[(z+4)^2-16]+13=0
    (x-1)^2-1-(y-1)^2{\color{red}+}1+(z+4)^2-16+13=0

    which gives my solution
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