Question 968003
For the following equation of a hyperbola determine the center, vertices, foci, and asymptotes:
36x^2−y^2−24x+6y−41=0
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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
{{{(x-1/3)^2/1-(y-3)^2/36=1}}}
given hyperbola has a horizontal transverse axis
Its standard form of equation: {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}
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)
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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)
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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
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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
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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