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Math Help - Ellipse

  1. #1
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    Ellipse

    Find the coordinates of the vertices and foci and sketch the graph:

    4x^2 + 9y^2 - 32x - 36y + 64 = 0

    I tried completing the square and got:

    4x^2 - 32x + 256 + 9y^2 - 36y + 324 = 516

    but I cant factorise the x and y, since there seem to be no roots
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  2. #2
    Member rtblue's Avatar
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    Alright. The standard form of an ellipse is:

    \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1

    In your equation, we have:

    4x^2 + 9y^2 - 32x - 36y + 64 = 0

    4x^2 + 9y^2 - 32x - 36y = -64

    Now, group the x's and y's together:

    4x^2 - 32x + 9y^2 - 36y = -64

    Factor out a 4 for the x's, and a 9 for the y's:

    4(x^2-8x) + 9(y^2-4y) = -64

    Now complete the square, for the x's and y's.

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

    Since you are completing the square within the parenthesis, you have to multiply by the coefficient of the parenthesis, and that is how I get 36. ( -64 + 64 + 36 = 36)

    Next: Divide by 36

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

    So, now that we have the ellipse in the standard form, we can find the vertices and the foci. The foci of an ellipse =

    c^2 = a^2 - b^2 Where C is the length of the foci.

    Since we know that the foci are on the major axis of the ellipse, we can determine the placing of the foci:

    9 - 4 = 5

    \sqrt{5} is the length of the foci, so now we know that the two foci are at:

    (4\pm\sqrt{5},2)

    The using a and b, we have that the vertices are at:

    (1,2)  (7,2) along the major axis

    (4,4)  (4,0) along the minor axis.

    If you had some trouble understanding the terminology, or how I came to find the vertices and the foci, I believe this site will be of help to you:

    Focus of Ellipse. The formula for the focus and ...
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