Question 968003
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.