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Question 416117: I do not understand how to find the center, transverse asxis, vertices, foci and asymptotes for the following equations:
y^2-x^2-4y+4x-1=0 and y^2-4x^2-4y-8x-4=0
(I think I am supposed to complete the square but I do not know how to do that with two variables)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! I do not understand how to find the center, transverse asxis, vertices, foci and asymptotes for the following equations:
y^2-x^2-4y+4x-1=0 and y^2-4x^2-4y-8x-4=0
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y^2-x^2-4y+4x-1=0
completing the square,
(y^2-4y+4)-(x^2-4x+4)=1+4+4
(y-2)^2-(x-2)^2=9
(y-2)^2/9-(x-2)^2/9=1
Standard form for this hyperbola: (y-k)^2/a^2-(x-h)^2/b^2+1, with (h,k) being the (x,y) coordinates of the center. Since the y-term is first, the hyperbola opens upwards (transverse axis vertical) If the x-term is first, the hyperbola opens sideways and the transverse axis is horizontal.
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center, (2,2)
transverse axis, x=2
a^2=9
a=3 (length of vertices)
b^2=9
b=3
c^=a^2+b^2
c=sqrt(a^2+b^2)=sqrt(18)=4.34(length of foci)
foci:(2,6.34),(2,-2.34)
vertices:(2,5),(2-1)
Asymptotes: slope of asymptotes = a/b and -a/b= 3/3 and -3/3=1 and -1
finding equation of asymptotes:
y=mx+b and y=-mx+b (b is y-intercept)
using points of center,
2=2+b
b=0
2=-2+b
b=4
Equations of asymptotes:
y=x
y=-x+4
The graph below should help you understand the algebra above.
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y=2+((9+(x-2)^2)^.5
I'm not going to do the second equation as it pretty much would be a repeat of what I just did. If you get stuck on it send me an email and I will try to help.
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