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Given that the ellipse x^2/a^2 + y^2/b^2 =1 has foci at(+/-ae,0)where e is the eccentricity and b^2=a^2(1-e^2),find the coordinates of the foci and the value of e^2 for the ellipse X^2+4y^2 =16

asked by muzungu
  1. Well you subtract the 16 to get your equation equal to one.
    Do x^2/16 + y^2/4=1

    Your center is 0,0
    Your eclipse would be horizontal.
    Do when graphing put a point at (0,0) then go 4 spaces in the negative and positive direction on the x-axis. Then go up 2 spaces from (0,0) in the positive and negative directions on the y-axis. Put your points on the graph. Draw your ellipse.
    Then you know to find your c, you do c^2=a^2 -b^2. So that's c^2= 16-4. Which makes c=2radical3

    Eccentricity= c/a
    So 2radical3/ 4.......e= radical 3 / 2

    Foci= (+/- 2radical3, 0)

    posted by L.Bianchessi

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