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Question 975398: For the following hyperbola:
(x + 1)2 (y 3)2 = 1
16 9
Find: a)Center :___________
b)Vertices:___________
___________
c) Foci: ___________
___________
d) Asymptotes: (list separately)
_________________
_________________
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! For the following hyperbola:
(x+1)^2/16 (y3)^2/9 = 1
Find: a)Center : (-1, 3)
b)Vertices: (-3, 3) and (5, 3)
c) Foci: (-4, 3) and (6, 3)
d) Asymptotes: (list separately)
y=-3x/4+9/4
y=3x/4+15/4
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Given hyperbola has a horizontal transverse axis.
Its standard form of equation:  , (h,k)=coordinates of center.
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center: (-1, 3)
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a^2=16
a=√16=4
vertices: (-1ħa, 3)=(1ħ4, 3)= (-3, 3) and (5, 3)
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b^2=9
b=√9=3
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c^2=a^2+b^2=16+9=25
c=√25=5
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foci: (-1ħc, 3)= (1ħ5, 3= (-4, 3) and (6, 3)
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Asymptotes are straight lines that go thru center.
slopes of asymptotes=ħb/a=ħ3/4
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Equation of asymptote with negative slope:
y=-3x/4+b
solve for b using coordinates of center (-1, 3) which are on the line
b=y+3x/4=3+(3*(-1)/4)=9/4
b=9/4
equation: y=-3x/4+9/4
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Equation of asymptote with positive slope:
y=3x/4+b
solve for b using coordinates of center (-1, 3) which are on the line
b=y-3x/4=3-(3*(-1)/4)=9/4
b=15/4
equation: y=3x/4+15/4
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