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Question 455972: I have to find the center, vertices, co-vertices, foci, and asymptotes of the hyperbola  . Can someone help me please?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! I have to find the center, vertices, co-vertices, foci, and asymptotes of the hyperbola
x^2-2x-y^2-4y-2=0
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Standard form of a hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
Standard form of a hyperbola with vertical transverse axis: (y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
The difference between the two forms is the interchange of (x-h) and (y-k)
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x^2-2x-y^2-4y-2=0
completing the square
(x^2-2x+1)-(y^2+4y+4)=2+1+4=7
(x-1)^2-(y+2)^2=7
(x-1)^2/7-(y+2)^2/7=1
This is a hyperbola with horizontal transverse axis (first form listed)
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center (1,-2)
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a^2=7
a=√7
vertices: (1±√7,-2)
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b^2=7
b=√7
co-vertices:(1,-2±√7)
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c^2=a^2+b^2=7+7=14
c=√14
foci: (1±√14,-2)
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asymptotes:
slope=±b/a=±7/7=±1
y=mx+b
using coordinates of center to find b since center is on asymptotes
-2=±1+b
b=-3 and -1
equations of asymptotes
y=x-3 (positive slope)
y=-x-1 (negative slope)
see graph below as a visual check on the answers above
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y=((x-1)^2-7)^.5-2
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