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Question 481495: Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes.
y^/25-x^/64=1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. y^/25-x^/64=1
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This is an equation of a hyperbola with vertical transverse axis of the standard form:
(y-k)^2/a^2-(x-h)^2/b^2=1
For given equation:
Center: (0,0)
a^2=25
a=√25=5
Length of vertical transverse axis=2a=10
Vertices (end points of transverse axis): (0,0±a)=(0,0±5)=(0,5) and (0-5)
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b^2=64
b=√64=8
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c^2=a^2+b^2=25+64=89
c=√89=9.4
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Foci: (0,0±c)=(0,±√89)=(0,√89) and (0,-√89)
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Asymptotes:
slope: ±a/b=±5/8
Equation of asymptotes which are straight lines of the standard form: y=mx+b, m=slope,
b=y-intercept. Asymptotes also go thru center at (0,0) in this case, so y-intercept=0.
Equation of asymptotes: y=±5x/8
See graph below as a visual check on answers:
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y=±(25+25x^2/64)^.5
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