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Math Help - Analytic Geometry: Ellipse [2]

  1. #1
    Member looi76's Avatar
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    Analytic Geometry: Ellipse [2]

    Question:
    Find the coordinates of the foci, the end points of the major axes and minor axis, the length of latus rectum and the end points of latus rectum.

    \frac{(x-3)^2}{16} + \frac{(y-2)^2}{9} = 1

    Attempt:

    Center: (3,2) a=4, b=3 , c=\sqrt{16-9} = \sqrt{7}

    Foci: (3-\sqrt{7},2) and (3+\sqrt{7},2)

    End points of major axis: (-1,2) and (7,2)

    End points of minor axis: (3,-1) and (3,5)

    Length of latus rectum: 2\frac{b^2}{a} = 2\frac{(3)^2}{4} = \frac{9}{2}


    How can I find out the end points of latus rectum? and are my answers right tell now?
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Ellipse

    Hello looi76
    Quote Originally Posted by looi76 View Post
    Question:
    Find the coordinates of the foci, the end points of the major axes and minor axis, the length of latus rectum and the end points of latus rectum.

    \frac{(x-3)^2}{16} + \frac{(y-2)^2}{9} = 1

    Attempt:

    Center: (3,2) a=4, b=3 , c=\sqrt{16-9} = \sqrt{7}

    Foci: (3-\sqrt{7},2) and (3+\sqrt{7},2)

    End points of major axis: (-1,2) and (7,2)

    End points of minor axis: (3,-1) and (3,5)

    Length of latus rectum: 2\frac{b^2}{a} = 2\frac{(3)^2}{4} = \frac{9}{2}

    How can I find out the end points of latus rectum? and are my answers right tell now?
    You've got everything right so far. To get the end-points of the latus rectum is very easy. You know that the semi-latus-rectum has length \frac94, so just move up and down through this distance from each of the foci, and you get to each of the end points. So one of them will be (3 + \sqrt7, 2 + \frac94). I'll leave the rest to you.

    Grandad
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