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16 January, 08:58

What is the length of the focal width x=-1/8y^2

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  1. 16 January, 09:17
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    The standard equation is 4p (x-h) = (y-k) ^2 where |p| = focal distance |4p| = focal width.

    x=-1/8y^2

    -8x=y^2

    4 * (-2) (x-0) = (y-0) ^2

    p = - 2, h=0, k=0

    focal width = |4 * (-2) | = 8

    The answer to your question is that the length of the focal width is equal to 8.
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