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Question 734909: Find an asymptote of this conic section..
9x^2-36x-4y^2+24y-36=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find an asymptote of this conic section..
9x^2-36x-4y^2+24y-36=0
complete the square:
9(x^2-4x+4)-4(y^2-6y+9)=36+36-36
9(x-2)^2-4(y-3)^2=36
This is an equation of a hyperbola with horizontal transverse axis:
Its standard form of equation:  , (h,k)=(x,y) coordinates of center
For given equation:
center: (2,3)
a^2=4
a=2
b^2=9
b=3
Slopes of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±3/2
Asymptotes are equations of straight lines that go thru the center of the hyperbola: y=mx+b, m=slope, b=y-intercept
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Equation of asymptote with negative slope, m=-3/2
y=-3x/2+b
solve for b using coordinates of the center (2,3)
3=-3*2/2+b
b=6
equation:y=-3x/2+6
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Equation of asymptote with positive slope, m=3/2
y=3x/2+b
solve for b using coordinates of the center (2,3)
3=3*2/2+b
b=0
equation:y=3x/2
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Equations of asymptotes: y=-3x/2+6 and y=3x/2
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