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4 May, 18:42

Which is the equation of an ellipse with directrices at x = ±4 and foci at (2, 0) and (-2, 0) ?

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  1. 4 May, 20:32
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    x^2/8 + y^2/4 = 1

    Step-by-step explanation:

    As the diretrix are vertical lines, we have a horizontal ellipse, which equation is:

    (x-h) ^2/a^2 + (y-k) ^2/b^2 = 1

    As the foci are at (2,0) and (-2,0), we have that k = 0, h = 2-2 = 0 and c = 2, where c^2 = a^2 - b^2

    As the diretrix are in x = ±4, we have that d = 4, where:

    c / a = a / d

    So now we can find a:

    2 / a = a / 4

    a^2 = 8

    a = 2.828

    And then we can find b:

    2^2 = 2.828^2 - b^2

    b^2 = 2.828^2 - 2^2 = 4

    b = 2

    So the ellipse equation is:

    x^2/8 + y^2/4 = 1
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