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Mathematics 19 Online
OpenStudy (cutiecomittee123):

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

OpenStudy (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

OpenStudy (cutiecomittee123):

Okay let me try that and see what I get

OpenStudy (anonymous):

sure :)

OpenStudy (cutiecomittee123):

wait so I plugged into my calculator "8=6(e^2-1) I got 16.3096

OpenStudy (cutiecomittee123):

I feel like I did something wrong

OpenStudy (anonymous):

dear a^2 =36 and b^2 = 64 you plugged a and b instead of a^2 and b^2

OpenStudy (cutiecomittee123):

oh wow thanks

OpenStudy (cutiecomittee123):

let me redo that using a^2 and b^2

OpenStudy (cutiecomittee123):

now I get 97

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