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21 July, 17:38

Y^2 - 4x² - 4y-8x-16=0

transverse axis

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Answers (1)
  1. 21 July, 18:02
    0
    line segment of length 8 between the vertices at (-1, - 2) and (-1, 6)

    Explanation:

    The equation can be rearranged to standard form.

    (y^2 - 4y) - 4 (x^2 + 2x) = 16

    (y^2 - 4y + 4) - 4 (x^2 + 2x + 1) = 16 + 4 - 4

    (y - 2) ^2 - 4 (x + 1) ^2 = 16

    (y - 2) ^2 / 16 - (x + 1) ^2/4 = 1

    This is of the form ...

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

    where the transverse axis is 2a and the center is (h, k). Here, a=4, so 2a = 8.

    The transverse axis is a vertical line segment of length 8, centered on (-1, 2).
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