cutiecomittee123
  • cutiecomittee123
how do find the vertex and foci of this hyperbola (x-2)^2/36 - (y+1)^2/64 =1
Mathematics
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cutiecomittee123
  • cutiecomittee123
how do find the vertex and foci of this hyperbola (x-2)^2/36 - (y+1)^2/64 =1
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
cutiecomittee123
  • cutiecomittee123
Okay let me try that and see what I get
anonymous
  • anonymous
sure :)

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cutiecomittee123
  • cutiecomittee123
wait so I plugged into my calculator "8=6(e^2-1) I got 16.3096
cutiecomittee123
  • cutiecomittee123
I feel like I did something wrong
anonymous
  • anonymous
dear a^2 =36 and b^2 = 64 you plugged a and b instead of a^2 and b^2
cutiecomittee123
  • cutiecomittee123
oh wow thanks
cutiecomittee123
  • cutiecomittee123
let me redo that using a^2 and b^2
cutiecomittee123
  • cutiecomittee123
now I get 97

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