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Math Help - Problem representing a 2 degree equation in the form of a ellipse

  1. #1
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    Red face Problem representing a 2 degree equation in the form of a ellipse

    I am having problem with representing a 2 degree eq in the form of a ellipse
    2x^2 + 3y^2 - 4x - 12y +13 = 0

    please provide me with the starting few steps or clues on how to solve this .I have already done the tranformation procedure and the result is this equation .I am stuck at this step as i dont know what to do next ?
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  2. #2
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    Try completing the square for both x and y.
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  3. #3
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    Please can u elaborate

    Quote Originally Posted by MathTooHard View Post
    Try completing the square for both x and y.

    I didnt understand ur suggestion , could u please elaborate it
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by zorro View Post
    I am having problem with representing a 2 degree eq in the form of a ellipse
    2x^2 + 3y^2 - 4x - 12y +13 = 0

    please provide me with the starting few steps or clues on how to solve this .I have already done the tranformation procedure and the result is this equation .I am stuck at this step as i dont know what to do next ?
    Quote Originally Posted by MathTooHard View Post
    Try completing the square for both x and y.
    Quote Originally Posted by zorro View Post
    I didnt understand ur suggestion , could u please elaborate it
    If you rearrange the terms, we get

    2x^2-4x + 3y^2- 12y = -13\implies 2(x^2-2x)+3(y^2-4y)=-13.

    Now complete the square on x and y, just like MathTooHard suggested.

    Does this clarify things?
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  5. #5
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    Thanks for ur reply but u still not answered the question

    Quote Originally Posted by Chris L T521 View Post
    If you rearrange the terms, we get

    2x^2-4x + 3y^2- 12y = -13\implies 2(x^2-2x)+3(y^2-4y)=-13.

    Now complete the square on x and y, just like MathTooHard suggested.

    Does this clarify things?

    Thanks for ur reply but u still not answered the question, what do u mean by square x and y
    please can u provide me with the first few step ...I am not a so good with understanding the math verbiage
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    Is this correct

    Quote Originally Posted by MathTooHard View Post
    Try completing the square for both x and y.

    From the clue given by u i have formulated a solution but i dont know if its correct or no

    2(x^2 - 2x) + 3(y^2 - 4y) = -13
    2(x^2 - 2x + 1 - 1) + 3(y^2 - 4y + 4 - 4) = -13
    2[(x^2 - 2x + 1) - 1] + 3[(y^2 - 4y + 4) - 4] = -13
    2[(x^2 - 1)^2 - 1] + 3[(y^2 - 2)^2 - 4] = -13
    2(x^2 - 1)^2 - 2 + 3(y^2 - 2)^2 - 12 = -13
    2(x^2 - 1)^2 + 3(y^2 - 2)^2 - 14 = -13
    2(x^2 - 1)^2 + 3(y^2 - 2)^2 = 14 -13
    2(x^2 - 1)^2 + 3(y^2 - 2)^2 = 1

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

    Now shifting the center to (1,2) weget

    \frac{X^2}{\frac{1}{2}} + \frac{Y^2}{\frac{1}{3}} = 1
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  7. #7
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    Talking

    To learn how to complete the square to find the ellipse equation, try here.
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    Is the answer correct ?

    Quote Originally Posted by stapel View Post
    To learn how to complete the square to find the ellipse equation, try here.
    2(x^2 - 2x) + 3(y^2 - 4y) = -13
    2(x^2 - 2x) + 3(y^2 - 4y) = \frac{-13}{2} . 2
    4(x^2 - 2x) + 6(y^2 - 4y) = -36
    (x^2 - 2x) + \frac{3}{2}(y^2 - 4y) = -9
    \frac{1}{3}(x^2 - 2x) + \frac{1}{6}(y^2 - 4y) = -3

    \frac{1}{3}(x^2 - 2x + 4) + \frac{1}{6}(y^2 - 4y + 16) = -3 \frac{1}{3}(4) + \frac{1}{6}(16)

    \frac{1}{3}(x - 2)^2 + \frac{1}{6}(y - 4) = \frac{-9 + 4 + 8}{3}


    \frac{(x - 2)^2}{3} + \frac{(y - 4)^2 }{6}= 1
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