Question 765498: Find the vertices, asymptotes, and foci of the hyperbola.
y^2-9x^2-10y-54x-65=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the vertices, asymptotes, and foci of the hyperbola.
y^2-9x^2-10y-54x-65=0
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complete the square:
y^2-10y-9x^2-54x-65=0
(y^2-10y+25)-9(x^2+6x+9)=65+25-81
(y-5)^2-9(x+3)^2=9
(y-5)^2/9-(x+3)^2=1
This is an equation of a hyperbola with vertical transverse axis.
Its standard form:  , (h,k)=(x,y) coordinates of the center.
For given hyperbola:
center: (-3,5)
a^2=9
a=√9=3
vertices: (-3,5±a)=(-3,5±3)=(-3,2) and (-3,8)
..
b^2=1
b=1
..
c^2=a^2+b^2=10
c≈√10≈3.16
foci: (-3,5±c)=(-3,5±3.16)=(-3,1.84) and (-3,8.16)
..
slopes of asymptotes with vertical transverse=±a/b=±3/1=±3
asymptotes are straight line equations that go thru center of hyperbola
For asymptote with positive slope:
y=3x+b
solve for b using coordinates of center
5=3*-3+b
b=14
equation: y=3x+14
..
For asymptote with negative slope:
y=-3x+b
solve for b using coordinates of center
5=-3*-3+b
b=-4
equation: y=-3x-4
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