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Math Help - Finding equation of ellipse

  1. #1
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    Finding equation of ellipse

    The major axis of ellipse has end points (8,4) and (-4,4). If the ellipse passes through the origin, what is the equation of the ellipse?

    My solution:

    End points (8,4) and (-4,4)
    so 2a = 12, a =6

    therefore:
    x/a + y/b =1

    x/6 + y/b =1

    Passes through origin, put (0,0) in x and y:

    0/6 + 0/b =1
    b = 0
    b = 0

    So the equation for the ellipse is:

    x/6 + y =1

    Did I do this right? Is it really x/6 + y =1 ?

    Thanks.
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  2. #2
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    Quote Originally Posted by shenton View Post
    The major axis of ellipse has end points (8,4) and (-4,4). If the ellipse passes through the origin, what is the equation of the ellipse?

    My solution:

    End points (8,4) and (-4,4)
    so 2a = 12, a =6

    therefore:
    x/a + y/b =1

    x/6 + y/b =1

    Passes through origin, put (0,0) in x and y:

    0/6 + 0/b =1
    b = 0
    b = 0

    So the equation for the ellipse is:

    x/6 + y =1

    Did I do this right? Is it really x/6 + y =1 ?

    Thanks.
    Since when does 0/b^2 = 1 imply b = 0 and at what point does 1/0 = 1?

    Let's try this again...

    Yes, a = 6.

    But the equation for the ellipse would be:
    (x - h)^2/36 + (y - k)^2/b^2 = 1

    Now, we know the center of the ellipse is the point (2, 4) since this is the midpoint of the line segment containing the two vertices. Thus h = 2 and k = 4:
    (x - 2)^2/36 + (y - 4)^2/b^2 = 1

    Now put in the origin:
    4/36 + 16/b^2 = 1

    1/9 + 16/b^2 = 1

    (1/9)b^2 + 16 = b^2

    (1/9 - 1)b^2 = -16

    -(8/9)b^2 = -16

    b^2 = 18

    b = sqrt{18}

    So:
    (x - 2)^2/36 + (y - 4)^2/18 = 1

    -Dan
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  3. #3
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    Thanks. I knew my solution looks too bad to be right.
    Last edited by ThePerfectHacker; March 11th 2007 at 04:49 PM.
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