Question 407063: The equation of the hyperbola is x^2-6x-9y^2=0
show the equation in standard form
find the coordinates for the vertices and the foci and give the equations of the asymptotes
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! The equation of the hyperbola is x^2-6x-9y^2=0
show the equation in standard form
find the coordinates for the vertices and the foci and give the equations of the asymptotes
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standard forms of the hyperbola:
(x-h)^2/a^2-(y-k)^2/b^2=1 (transverse axis horizontal)
(y-k)^2/a^2-(x-h)^2/b^2=1 (transverse axis vertical)
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given hyperbola:
x^2-6x-9y^2=0
complete the square
(x^2-6x+9)-9y^2=9
(x-3)^2-9y^2=9
divide by 9
((x-3)^2)/9-(y^2)/1=1
This is a hyperbola with horizontal transverse axis, that is, it opens sideways.
a^2=9
a=3
b^2=1
b=1
c=sqrt(a^2+b^2)=sqrt(9+1)=sqrt(10)=3.16
center:(3,0)
vertices:(3+-a,0)=(3+-3,0) or (0,0),(6,0)
foci:(3+-c,0)=(3+-Sqrt(10),0) or (-.16,0),(6.16.0)
equations of asymptotes:
y=(b/a)x=(x-3)/3
y=-(b/a)x=-(3-x)3
Your graph should look much like the graph below
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y=((x^2-6x)/9)^.5
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