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Question 968003: Hello, I'm looking for some guidance with this question, can anyone help?
For the following equation of a hyperbola determine the center, vertices, foci, and asymptotes:
36x^2−y^2−24x+6y−41=0
Found 2 solutions by stanbon, lwsshak3:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! For the following equation of a hyperbola determine the center, vertices, foci, and asymptotes:
36x^2−y^2−24x+6y−41=0
36(x^2-(24/36)x) - (y^2-6y) = 41
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Complete the square::
36(x^2 - (2/3)x + (1/3)^2) - (y^2-6y+9) = 41 + 36(1/9) - 9
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36(x-(1/3)^2 - (y-3)^2 = 36
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Rewrite:
(x-(1/3))^2 - (y-3)^2/36 = 1
a = 6 ; b = 1 ; c = sqrt(36+1) = sqrt(37)
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center: (1/3, 3)
vertices:: (1/3,3-6);(1/3,3+6)
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foci:: ((1/3),-3-sqrt(37)),((1/3),9+sqrt(37))
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asymptotes: I'll leave that to you
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Cheers,
Stan H.
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! For the following equation of a hyperbola determine the center, vertices, foci, and asymptotes:
36x^2−y^2−24x+6y−41=0
***
36x^2−24x−y^2+6y=41
36(x^2−(24/36)x)−(y^2-6y)=41
36(x^2−(2/3)x)−(y^2-6y)=41
complete the square:
36(x^2−(2/3)x+1/9)−(y^2-6y+9)=41+4-9
36(x-1/3)^2-(y-3)^2=36
given hyperbola has a horizontal transverse axis
Its standard form of equation: 
For given hyperbola:
center: (1/3, 3)
a^2=1
a=1
vertices: (1/3±a,3)=(1/3±1,3)=(-2/3, 3) and (4/3, 3)
..
foci:
b^2=36
b=6
c^2=a^2+b^2=1+36=37
c=√37≈6.1
foci: (1/3±c,3)=(1/3±6.1,3)=(-5.8, 3) and (6.4, 3)
..
Asymptotes: (two straight-line equations(mx+b) that intersect at center
For hyperbolas with horizontal transfer axis, slopes of asymptotes=±b/a=±6/1=±6
..
Equation of asymptote with positive slope:
y=6x+b
solve for b using coordinates of center
3=6*1/3+b
b=1
equation: y=6x+1
..
Equation of asymptote with negative slope:
y=-6x+b
solve for b using coordinates of center
3=-6*1/3+b
b=5
equation: y=-6x+5
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