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Math Help - problems with ellipse

  1. #1
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    problems with ellipse

    I was given this ellipse 4x^2 + 9y^2 - 8x + 36y + 4 = 0.
    Im really unsure how to put it in it's standar form but i have to i beleive to help me.
    I actually need to find the center and foci of the elliple, how would i go about doing that?
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  2. #2
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    Hello, 14041471!

    You've never done one of these before . . . ever?


    (a) Write in standard form: . 4x^2 + 9y^2 - 8x + 36y + 4 \:=\: 0

    (b) Find the center and foci of the ellipse.

    (a) We have: . 4x^2 - 8x + 9y^2 + 36y \:=\:-4

    Factor: . 4(x^2-2x \quad) + 9(y^2 - 4y \quad) \:=\:-4

    Complete the square: . 4(x^2-2x \:{\color{blue}+ 1}) + 9(y^2 + 4y \:{\color{red}+\: 4}) \;=\;-4 \:{\color{blue}+\:4} \:{\color{red}+\: 36}

    Factor: . 4(x-1)^2 + 9(y+2)^2 \:=\:36

    Divide by 36: . \frac{4(x-1)^2}{36} + \frac{9(y+2)^2}{36} \:=\:\frac{36}{36} \quad\Rightarrow\quad \boxed{\frac{(x-1)^2}{9} + \frac{(y+2)^2}{4} \:=\:1}


    (b) The center is: . \boxed{C(1,-2)}


    We have: . a = 3,\:b = 2
    This is a "horizontal" ellipse; the foci on are the horizontal axis,
    . . to the left and right of the center.

    Since c^2 \:=\:a^2-b^2, we have: . c^2 \:=\:9 - 4 \:=\:5\quad\Rightarrow\quad c \:=\:\pm\sqrt{5}

    The foci are: . \boxed{F\left(1\pm\sqrt{5},\:-2\right)}

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  3. #3
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    Cheers. I understand that concept alot better now, I also need to find the length of the major and minor axis of the same ellipsis "4x^2 + 9y^2 - 8x + 36y + 4 = 0" how would i do that?

    Edit: foget that i know now that
    2a = 6 = big axis
    and 2b = 4 = small axis
    Last edited by mr fantastic; January 8th 2009 at 12:22 PM. Reason: Merged posts
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