SOLUTION: How do you find the center, vertices, foci, and asymptotes of the hyperbola? x^2-9y^2+36y-72=0

Question 618494: How do you find the center, vertices, foci, and asymptotes of the hyperbola?
x^2-9y^2+36y-72=0
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
How do you find the center, vertices, foci, and asymptotes of the hyperbola?
x^2-9y^2+36y-72=0
complete the square
x^2-9(y^2-4y+4)=72-36
x^2-9(y-2)^2=36
x^2/36-(y-2)^2/4=1
This is an equation of a hyperbola with horizontal transverse axis.
Form of equation:(x-h)^2/a^2-(y-k)^2/b^2=1
For given equation:
center: (0,2)
a^2=36
a=√36=6
vertices: (0�a,2)=(0�6,2)=(-6,2) and (6,2)
..
b^2=4
b=√4=2
..
c^2=a^2+b^2=36+4=40
c=√40≈6.3
foci: (0�c,2)=(0�6.3,2)=(-6.3,2) and (6.3,2)
..
slopes of asymptotes with horizontal transverse axis=�b/a=2=�2/6=�1/3
Asymptotes are straight lines that intersect at center. Equation: y=mx+b, m=slope, b=y-intercept
..
Equation of asymptote with negative slope.
y=-x/3+b
solving for b using coordinates of center(0,2)
2=0+b
b=2
Equation: y=-x/3+2
..
Equation of asymptote with positive slope.
y=x/3+b
solving for b using coordinates of center(0,2)
2=0+b
b=2
Equation: y=x/3+2

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