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cutiecomittee123

  • one year ago

how do find the vertex and foci of this hyperbola (x-2)^2/36 - (y+1)^2/64 =1

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  1. anonymous
    • one year ago
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    you can simply guess the vertex of the hyperbola has (2,-1) the focii is (+ae,0) , (-ae,0) you can find e by b^2 = a^2(e^2-1) and just shift the vertex wrt the vertex

  2. cutiecomittee123
    • one year ago
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    Okay let me try that and see what I get

  3. anonymous
    • one year ago
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    sure :)

  4. cutiecomittee123
    • one year ago
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    wait so I plugged into my calculator "8=6(e^2-1) I got 16.3096

  5. cutiecomittee123
    • one year ago
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    I feel like I did something wrong

  6. anonymous
    • one year ago
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    dear a^2 =36 and b^2 = 64 you plugged a and b instead of a^2 and b^2

  7. cutiecomittee123
    • one year ago
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    oh wow thanks

  8. cutiecomittee123
    • one year ago
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    let me redo that using a^2 and b^2

  9. cutiecomittee123
    • one year ago
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    now I get 97

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