Question 927550: Given the hyperbola with the equation x^2−y^2−6x−2y+7=0, find the vertices, the foci, and the equations of the asymptotes.
I can't seem to get the right answer, I get it to this equation(which is the standard equation but the denominators don't look right to me):
(((x-3)^2)/1)-(((y-1)^2)/1)=1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Given the hyperbola with the equation x^2−y^2−6x−2y+7=0, find the vertices, the foci, and the equations of the asymptotes.
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x^2−6x−y^2−2y+7=0
complete the square:
(x^2−6x+9)−(y^2+2y+1)=-7+9-1
(x-3)^2-(y+1)^2=1
(x-3)^2/2-(y+1)^2/2=1
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form of equation:  , (h,k)=coordinates of center.
center:(3,-1)
a^2=1
a=1
vertices:(3±a,-1)=(3±1,-1)=(4,-1) and (2,-1)
b^2=1
b=1
c^2=a^2+b^2=1+1=2
c=√2
foci:(3±c,-1)=(3±√2,-1)=(3+√2,-1) and (3-√2,-1)
..
Asymptotes:Straight line equations of the form: y=mx+b, m=slope, b=y-intercept
Asymptotes go thru center
slope of asymptotes: ±b/a=±1
..
equation of asymptote with positive slope:
y=x+b
solve for b with coordinates of center
-1=3+b
b=-4
equation: y=x-4
..
equation of asymptote with negative slope:
y=-x+b
solve for b with coordinates of center
-1=-3+b
b=2
equation: y=-x+2
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