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animalsavior94

  • 4 years ago

find the standard form of the hyperbola with v( plus minus 1, 0) through point (2, 13)

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  1. amistre64
    • 4 years ago
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    vertices are (-1,0), and (1,0); its a convenient thing this is along the x axis i spose

  2. amistre64
    • 4 years ago
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    (2)^2 - (13)^2 = 1 is our goal i think

  3. amistre64
    • 4 years ago
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    4 196 --- - --- = 1 hmmm a^2 b^2

  4. amistre64
    • 4 years ago
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    a^2 = 1 since 1/1 = 1 for out vertice

  5. amistre64
    • 4 years ago
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    4 - 196/b^2 = 1 4b^2 -196 = b^2 3b^2 -196 = 0 b^2 = 196/3 x^2 - 3y^2/196 = 1 is te best i can do

  6. amistre64
    • 4 years ago
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    lol.... 13^2 = 169 ..... i always gets those confused

  7. amistre64
    • 4 years ago
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    \[x^2 - \frac{3y^2}{169}=1\] should be better

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