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Question 482543: 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
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12x^2-18y^2-18x-12y+12=0
divide by 3 to simplify
4x^2-6y^2-6x-4y+4=0
Complete the square
4(x^2-3x/2+9/16)-6(y^2+2y/3+1/9)=-12+9/4-2/3=-144/12+27/12-8/12=-163/12
4(x^2-3x/2+9/16)-6(y^2+2y/3+1/9)=-163/12
4(x-3/4)-6(y+1/3)=-163/12
multiply both sides by (-1)
6(y+1/3)^2-4(x-3/4)^2=163/12
divide by 163/12
(y+1/3)^2/163/72-(x-3/4)^/163/48=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, with (h,k) being the (x,y) coordinates of the center.
For given equation:
center: (3/4,-1/3)
a^2=163/72
a=√(163/72)
length of transverse axis=2a=2*√(163/72)
b^2=163/48
b=√(163/48)
length of conjugate axis=2b=2*√(163/48)
slope of asymptotes=± a/b=√(163/72)/√(163/48)
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