SOLUTION: Explain which conic section this equation and explain how to solve it: 12x^2-18y^2-18x-12y+12=0

Question 482542: Explain which conic section this equation and explain how to solve it:
12x^2-18y^2-18x-12y+12=0
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Explain which conic section this equation and explain how to solve it:
12x^2-18y^2-18x-12y+12=0
***
12x^2-18y^2-18x-12y+12=0
dividing by 3 to simplify
3x^2-6y^2-6x-4y+4=0
completing the square
3(x^2-2x+1)-6(y^2+(2/3)y+1/9)=-4+3-2/3=-1-2/3=-5/3
3(x-1)^2-6(y-1/3)^2=-5/3
multiply both sides by (-1)
6(y-1/3)^2-3(x-1)^2=5/3
divide both sides by 5/3
(y-1/3)^2/(5/18)-(x-1)^2/(5/9)=1
This is an equation of a hyperbola with vertical transverse axis of the standard form:
(y-k)^2/a^2-(x-h)^2/b^2=1
For given equation:
Center:(1,1/3)
a^2=5/18
a=√(5/18)
Length of transverse axis=2a=2*√(5/18)
Vertices: (1,1/3�a), (1,1/3�√(5/18))
b^2=5/9
b=√(5/9)
Length of conjugate axis=2b=2*√(5/9)
c^2=a^2+b^2=5/18+5/9=15/18
c=√(15/18)
Foci: (1,1/3�c)=(1,1/3�√(15/18))
Slope of asymptotes=�a/b=�√(5/18)/√(5/9)=�√(1/2)=�√2/2
Equation of asymptotes:
Asymptotes are straight lines of the standard form: y=mx+b, m=slope, b=y-intercept
Asymptote with positive slope:
y=(√2/2)x+b
solving for b, using center coordinates thru which asymptotes pass
1/3=(√2/2)*1+b
b=-.3737
equation: y=.707x-.3737
..
Asymptote with negative slope:
Equation: y=-.707x+1.0403
..
Comment: As you can see, there is a whole bunch of things that can be asked about a hyperbola.
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