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Math Help - Find the center the length of the axes and eccentricity of ellipse.

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    Find the center the length of the axes and eccentricity of ellipse.

    Find the center the length of the axes and eccentricity of ellipse.
    2x^2+3y^2-4x-12y+13=0
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    Quote Originally Posted by varunnayudu View Post
    Find the center the length of the axes and eccentricity of ellipse.
    2x^2+3y^2-4x-12y+13=0

    2x^2+3y^2-4x-12y+13= 0

    2(x^2 - 2x + 1)+3(y^2-4y+4) - 1 = 0

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

    \dfrac{(x-1)^2}{\frac12}+\dfrac{(y-2)^2}{\frac13} = 1

    Now its in standard form, can you continue?
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    Ellipse

    Hi -

    Quote Originally Posted by varunnayudu View Post
    Find the center the length of the axes and eccentricity of ellipse.
    2x^2+3y^2-4x-12y+13=0
    Complete the square for x:

    2x^2 - 4x = 2(x^2 - 2x)

    =2((x-1)^2-1)

    =2(x-1)^2-2

    and for y:

    3y^2-12y =3(y^2-4y)

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

    =3(y-2)^2-12

    So the equation can be re-written:

    2(x-1)^2-2+3(y-2)^2-12+13=0

    i.e. 2(x-1)^2 + 3(y-2)^2=1

    Substitute u=x-1 and v=y-2:

    2u^2+3v^2=1

    Re-arrange:

    \frac{u^2}{(\frac{1}{\sqrt{2}})^2} + \frac{v^2}{(\frac{1}{\sqrt{3}})^2}=1

    OK from here?

    Grandad
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