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Mathematics 12 Online
OpenStudy (anonymous):

Which points are the foci of this hyperbola?

OpenStudy (anonymous):

@Michele_Laino

OpenStudy (anonymous):

((x - 1)^2)/8^2 - ((y - 2)^2)/6^2 = 1

OpenStudy (anonymous):

(1, 2) and (11, 2) (1, 2) and (1, 10) (1, 12) and (1, –8) (11, 2) and (–9, 2)

OpenStudy (anonymous):

OpenStudy (michele_laino):

referring to the standard equation of the hyperbola: \[\frac{{{X^2}}}{{{a^2}}} - \frac{{{Y^2}}}{{{b^2}}} = 1\] we get: a=8, and b=6 Now the x-coordinate of the foci are : \[c = \pm \sqrt {{a^2} + {b^2}} = ...?\]

OpenStudy (michele_laino):

what are the values of c?

OpenStudy (anonymous):

I got a 10

OpenStudy (anonymous):

a ±10

OpenStudy (michele_laino):

ok! so our foci are the points: F1=(-c,0)=...? and F2=(c,0)=...?

OpenStudy (anonymous):

(-10,0) (10,0)?

OpenStudy (michele_laino):

ok! that's right!

OpenStudy (anonymous):

so what else I do?

OpenStudy (michele_laino):

we have finished!

OpenStudy (anonymous):

none of the answers LOL

OpenStudy (michele_laino):

sorry you are right, since we have performed this traslation: X=x-1 Y=y-2 so we have to solve this system: x-1=10--->x=..? and x-1=-10---> x=...? then we have to solve these equations: y-2=0---< y=...? y-2=0--->y=...?

OpenStudy (anonymous):

x=11 x=-9

OpenStudy (michele_laino):

yes! and y=...?

OpenStudy (anonymous):

y=2

OpenStudy (anonymous):

thanks bro

OpenStudy (michele_laino):

thanks!

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