SOLUTION: Find the equations of the asymptotes. y^2-9x^2-12y-36x-9=0

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Question 634662: Find the equations of the asymptotes.
y^2-9x^2-12y-36x-9=0
Found 2 solutions by richwmiller, lwsshak3:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
asymptotes | y = -3 x | y = 3 x+12

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
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
%28y-6%29%5E2%2F9-%28x%2B2%29%5E2%2F1=1
This is an equation of a hyperbola with vertical transverse axis:
Its standard form: %28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1, (h,k)=(x,y) coordinates of center
..
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: %28y-6%29%5E2%2F9-%28x%2B2%29%5E2%2F1=1
center: (-2,6)
a^2=9
a=3
b^2=1
b=1
slope,m=±a/b=±3/1=±3
..
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
..
equations of asymptotes: y=-3x and y=3x+12