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Question 620844: Help stuck again. I have an hyperbola with the formula of 36x^2+144x+27-9y^2+36y=0 I need to find the formula plus the asymtopes foci and vertices.
I make the formula out to be [(2/5)(x+2)^2]-(1/10)(y+2)^2 =1.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Help stuck again. I have an hyperbola with the formula of 36x^2+144x+27-9y^2+36y=0 I need to find the formula plus the asymtopes foci and vertices.
I make the formula out to be [(2/5)(x+2)^2]-(1/10)(y+2)^2 =1.
**
36x^2+144x+27-9y^2+36y=0
36x^2+144x-9y^2+36y=-27
complete the square
36(x^2+4+4)-9(y^2-4y+4)=-27-36+144
36(x+2)^2-9(y-2)^2=81
(x+2)^2/(81/36)-(y-2)^2/9=1
(x+2)^2/(9/4)-(y-2)^2/9=1
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center
For given equation:
center: (-2,2)
a^2=9/4
a=√(9/4)=3/2=1.5
vertices: (-2±a,2)=(-2±1.5,2)=(-3.5,2) and (-.5,2)
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b^2=9
b=√9=3
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c^2=a^2+b^2=9/4+9=9/4+36/4=45/4
c=√45/2≈3.35
Foci: (-2±c,2)=(-2±3.35,2)=(-5.35,2) and (1.35,2)
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Asymptotes are two straight lines that intersect at center (-2,2)
Slopes of asymptotes for hyperbolas with horizontal transverse axis=±b/a=±3/1.5=±2
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Equation of asymptote with slope<0:
y=-2x+b
solve for b using coordinates of center
2=-2*-2+b
b=-2
equation: y=-2x-2
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Equation of asymptote with slope>0:
y= 2x+b
solve for b using coordinates of center
2=-2*2+b
b=6
equation: y= 2x+6
see graph below as a visual check:
y=±(9(x+2)^2/2.25-9)^.5+2
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