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Math Help - Find Centre of symmetry of conic

  1. #1
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    Exclamation Find Centre of symmetry of conic

    I have the following conic and i want to find its centre of symmetry: 3y^2+2x+18y=-31

    I tried the following way but it didn't seem to work, could somebody tell me where i went wrong?

    (3y^2+18y)+2x=-31

    3(y^2+6y)+2x=-31

    completing the square:

    3(y^2+6y+9)+2x=-31+(3*9)

    3(y^2+6y+9)+2x=-4

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

    (3(y+3)^2)/(-4)+2x/(-4)=1

    so the centre is (0,-3)?

    Where am i going wrong?

    Thanks
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  2. #2
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    Re: Find Centre of symmetry of conic

    The conic you mentioned above is a shifted parabola .
    the full formula is (y+3)^2=4(-1/6)(x+2) and its centre lies at the point P(-2,-3).
    a shifted parabola of this type has the formula (y+b)^2=4p(x+a) with p the parameter of the parabola and centre at P(-a,-b)
    Minoas
    M.R
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  3. #3
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    Re: Find Centre of symmetry of conic

    Hey mcleja.

    Your working looks good but I don't think this is a normal conic since you will get many complex solutions.

    This also doesn't have the form of a circle, ellipse, or a hyperbola.
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  4. #4
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    Re: Find Centre of symmetry of conic

    As I said before this is a shifted Parabola. you may refere to any calculus book for further reference about shifted conics.
    As a general rule if a conic has the quadratic only in one variable like x or y only then it is a parabola.
    Minoas
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  5. #5
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    Re: Find Centre of symmetry of conic

    Hello, mcleja!

    I have the following conic and i want to find its centre of symmetry:
    . . 3y^2+2x+18y\:=\:-31

    This is conic is a parabola ... it does not have a center.
    It has a vertex.

    There are two basic forms for the parabola.

    \begin{Bmatrix}\text{Vertical }(\cup\text{ or }\cap): & (x-h)^2 \:=\:4p(y-k) \\ \text{Horizontal }(\subset\text{or}\supset): & (y-k)^2 \:=\:4p(x-h) \end{Bmatrix}\;\;\text{where }(h,k)\text{ is the vertex.}


    We have: . 3y^2 + 2x + 18y \:=\:\text{-}31 \quad\Rightarrow\quad 3y^2 + 18y \:=\:\text{-}2x - 31

    . . . . . . . . 3(y^2 + 6y) \:=\:\text{-}2x-31 \quad\Rightarrow\quad 3(y^2 + 6y + 9) \:=\:\text{-}2x - 31 + 27

    . . . . . . . . 3(y+3)^2 \:=\:\text{-}2x - 4 \quad\Rightarrow\quad 3(y+3)^2 \:=\:\text{-}2(x+2)

    . . . . . . . . (y+3)^2 \:=\:\text{-}\tfrac{2}{3}(x+2)


    The parabola opens to the left and has vertex (\text{-}2,\,\text{-}3)
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