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Question 491838: graph the hyperbola and label the vertices, foci and asymptotes:
(((y+2)^2)/25))-(((x+11)^2)/4)=1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! graph the hyperbola and label the vertices, foci and asymptotes:
(((y+2)^2)/25))-(((x+11)^2)/4)=1
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graph the hyperbola and label the vertices, foci and asymptotes:
((y+2)^2)/25)-((x+11)^2)/4)=1
This is a hyperbola with vertical transverse axis of the standard form:
(y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
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From given data:
Center: (-11,-2)
a^2=25
a=√25=5
length of transverse axis=2a=10
vertices: (end-points of transverse axis)=(-11,-2±a)=(-11,-2±5)=(-11,3) and (-11,-7)
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b^2=4
b=√4=2
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c^2=a^2+b^2=25+4=29
c=√29=5.39
Foci: (-11,-2±c)=(-11,-2±√29)=(-11,-2±5.39)=(-11,3.39) and (-11,-7.39)
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Asymptotes are straight lines thru the center with equation of the standard form: y=mx+b, with m=slope and b=y-intercept.
slope of asymptotes=±a/b=5/2 and -5/2
Equations of asymptotes:
y=mx+b
using coordinates of center to find b
-2=5/2*-11+b
-2=-55/2+b
-4=-55+2b
2b=51
b=51/2
Equation: y=5x/2+51/2
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-2=-5/2*-11+b
-2=55/2+b
-4=55+2b
2b=-59
b=-59/2
Equation: y=-5x/2-59/2
Ans:
Vertices:(-11,3) and (-11,-7)
Foci: (-11,3.39) and (-11,-7.39)
Equations of asymptotes:y=5x/2+51/2 and y=-5x/2-59/2
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see the graph below as a visual check on the answers above
y=±(25+25(x+11)^2/4)^.5-2
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