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9 March, 08:23

Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1.

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Answers (2)
  1. 9 March, 08:32
    0
    The vertices are at (3, 0) and (-3, 0).

    The foci are at (5, 0) and (-5, 0).

    Step-by-step explanation:

    Compare given equation x^2/9 - y^2/16 = 1

    against standard hyperbola form (x - h) ^2 / a^2 - (y - k) ^2 / b^2 = 1,

    the hyperbola opens left and right with its center (h, k) at (0, 0).

    a^2 = 9 so a = + / -3

    The vertices are at (3, 0) and (-3, 0).

    a^2 + b^2 = c^2 where c is 2x distance between foci.

    9 + 16 = c^2

    c = + / -5

    The foci are at (5, 0) and (-5, 0).
  2. 9 March, 08:34
    0
    Step-by-step explanation:

    equation x squared over nine minus y squared over sixteen = 1 is the same as

    x^2/9 - y^2/16 = 1

    x is at its min/max when y^2=0, y=0

    x^2/9 = 1

    x = + 3

    vertices (+3, 0)

    c^2 = 9 + 16 = 25

    c = + 5

    foci (+5, 0)
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