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Question 422826: Find the Foci of the hyperbola defined by the equation.The equation is (x+5)^2/9 -(y+8)^2/64 = 1.Please help I have no Idea how to do this.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the Foci of the hyperbola defined by the equation.The equation is (x+5)^2/9 -(y+8)^2/64 = 1.Please help I have no Idea how to do this.
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(x+5)^2/9-(y+8)^2/64=1
Standard form of a hyperbola:
y=(x-h)^2/a^2-(y-k)^2/b^2=1, with (h,k) being the (x,y) coordinates of the
center. If the x^2 term is listed first, the hyperbola would have a horizontal transverse axis. If the y^2 term is listed first, the hyperbola would have a vertical transverse axis.
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Given hyperbola has a center at (-5,-8) and a horizontal transverse axis, that is, the hyperbola opens sideways.
a^2=9
a=3
b^2=64
b=8
c^2=a^2+b^2
c=sqrt(9+64)=sqrt(73)=8.54
The foci is on the transverse axis, y=-8,
Its coordinates are:(-5+-c,-8) or (13.54,-8) and (3.54,-8)
The graph below might help you understand the problem better.
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y=-8+-(64((x+5)^2/9-1))^.5
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