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Find the coordinates of the center, the foci, and the vertices of an ellipse

Find the coordinates of the center, the foci, and the vertices, sketch the curve, and determine the eccentricity of the ellipse given by the equation

    \[ \frac{x^2}{100} + \frac{y^2}{36} = 1. \]


This is an equation for an ellipse in standard form with center at (0,0).

    \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

with a = 10 and b = 6. We then know its foci are located (c,0) and (-c,0) where

    \[ c = \sqrt{a^2 - b^2} = \sqrt{100 - 36} = \sqrt{64} = 8. \]

Therefore, the foci of this ellipse are (8,0) and (-8,0). Further, the vertices are given by \pm a N (page 505 of Apostol) so they are at the points (10,0) and (-10,0). Finally, we can compute the eccentricity

    \[ c = |a| e \quad \implies \quad 8 = 10e \quad \implies \quad e = \frac{4}{5}. \]

4 comments

  1. yacine says:

    from the book the vertices should be at x=a/e and x=-a/e , in this case it should be at x=10*5/4=10.5 and x=-10*5/4=-12.5

    • William C says:

      This makes sense to me, but the vertirx doesnt seem to be equidistant to the line and focus (ie (-10,0) with x = -12.5 and (-8,0). Could someone explain what’s happening here please?

      • William C says:

        Nevermind, I got it. The distances shouldn’t be equal, but should have a constant ratio represented by the eccentricity (in this case 4/5, or 2/2.5 in what I was observing)

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