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Math Help - Putting the equation of a conic into standard form

  1. #1
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    Putting the equation of a conic into standard form

    Hi, I have the following exercise...

    By completing the square in x and y, and so by eliminating the linear terms, put the conic

    9x^2+4y^2-36x-24y+36=0

    into standard form

    ___

    Well, I have completed the square, but how do I get the standard form of this conic?? :-/

    (3x-6)^2+(2y-6)^2=36
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  2. #2
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    Quote Originally Posted by juanma101285 View Post
    Hi, I have the following exercise...

    By completing the square in x and y, and so by eliminating the linear terms, put the conic

    9x^2+4y^2-36x-24y+36=0

    into standard form

    ___

    Well, I have completed the square, but how do I get the standard form of this conic?? :-/

    (3x-6)^2+(2y-6)^2=36
    [3(x-2)]^2 + [2(y-3)]^2 = 36

    9(x-2)^2 + 4(y-3)^2 = 36

    \dfrac{9(x-2)^2}{36} + \dfrac{4(y-3)^2}{36} = 1

    \dfrac{(x-2)^2}{2^2} + \dfrac{(y-3)^2}{3^2} = 1
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  3. #3
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    Hello, juanma101285!

    \text{Write the conic: }\:9x^2+4y^2-36x-24y+36\:=\:0\:\text{ in standard form.}

    I recommend that you use this procedure . . .


    \text{Re-arrange terms: }\:9x^2 - 36x + 4y^2 - 24y \;=\;-36


    \text{Factor: }\quad 9(x^2 - 4x) + 4(y^2 - 6y) \;=\;-36


    \text{Complete the square:}

    . . . 9(x^2 - 4x {\bf+ 4 - 4}) + 4(y^2 - 6y {\bf\:+\: 9 - 9}) \;=\;-36

    . . 9(x^2-4x+4) - 36 + 4(y^2-6y+9) - 36 \;=\;-36

    . . . . . . . . . . . . . . . . . 9(x-2)^2 + 4(y-3)^2 \;=\;36


    \text{Divide by 36: }\;\;\dfrac{(x-2)^2}{4} + \dfrac{y-3)^2}{9} \;=\;1

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