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Question 619455: I have the equation :
(y+5)^2/6^2-(x-2)^2/5^2=1
i need to identify the vertices and foci of the hyperbola. I'm not sure how to do that with my numerators. And I also need to graph this.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! (y+5)^2/6^2-(x-2)^2/5^2=1
i need to identify the vertices and foci of the hyperbola. I'm not sure how to do that with my numerators. And I also need to graph this
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Given equation is that of a hyperbola with vertical transverse axis.
Its standard form: (y-k)^2/a^2-(x-h)^2/b^2=1, (h,k)=(x,y) coordinates of the center
For given equation:
center: (2,-5)
a=6
a^2=36
Vertices: (2,-5±a)=(2,-5±6)=(2,-11) and (2,1)
b=5
b^2=25
c^2=a^2+b^2=36+25=61
c=√61≈7.8
Foci: (2,-5±c)=(2,-5±7.8)=(2,-12.8) and (2,2.8)
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Slopes of asymptotes for hyperbolas with vertical transverse axis=a/b=6/5
Asymptotes are straight lines that intersect at the center: Equation: y=mx+b, m=slope, b=y-intercept
Equation for asymptote with slope<0: y=-6x/5+b
solve for b using coordinates of center (2,-5)
-5=-6*2/5+b
-5=-12/5+b
b=-5+12/5=-25/5+12/5=-13/5
Equation: y=-6x/5-13/5
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Equation for asymptote with slope>0: y=6x/5+b
solve for b using coordinates of center (2,-5)
-5=6*2/5+b
-5=12/5+b
b=-5-12/5=-25/5-12/5=-37/5
Equation: y=6x/5-37/5
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see graph below:
y=±(36+36(x-2)^2/25)^.5-5
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