Question 634662
Find the equations of the asymptotes. 
y^2-9x^2-12y-36x-9=0
complete the squares
y^2-12y-9x^2-36x-9=0
(y^2-12y+36)-9(x^2+4x+4)=9+36-36
(y-6)^2-9(x+2)^2=9
{{{(y-6)^2/9-(x+2)^2/1=1}}}
This is an equation of a  hyperbola with vertical transverse axis:
Its standard form: {{{(y-k)^2/a^2-(x-h)^2/b^2=1}}}, (h,k)=(x,y) coordinates of center
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asymptotes are straight lines that intersect at the center.
form of  equation: y=mx+b, m=slope, b=y-intercept
slopes of asymptotes of hyperbolas with vertical transverse axis: ±a/b
For given hyperbola: {{{(y-6)^2/9-(x+2)^2/1=1}}}
center: (-2,6)
a^2=9
a=3
b^2=1
b=1
slope,m=±a/b=±3/1=±3
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equation of asymptote with slope=-3
y=mx+b
y=-3x+b
solve for b using coordinates of center
6=-3*-2+b
b=0
equation: y=-3x
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equation of asymptote with slope=3
y=mx+b
y=3x+b
solve for b using coordinates of center
6=3*-2+b
b=12
equation: y=3x+12
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equations of asymptotes: y=-3x and y=3x+12