Question 878034: Find the vertices, foci, and equations of the asymptotes of the hyperbola. 4x^2-3y^2+8x+16=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the vertices, foci, and equations of the asymptotes of the hyperbola.
4x^2-3y^2+8x+16=0
4x^2+8x-3y^2=-16
complete the square
4(x^2+2x+1)-3y^2=-16+4
4(x+1)^2-3y^2=-12
divide by -12
 , (h,k)=coordinates of center.
For given hyperbola:
center: (-1,0)
a^2=4
a=2
vertices: (-1,0±a)=(-1,0±2)=(-1,-2) and (-1,2)
b^2=3
b=√3
c^2=a^2+b^2=4+3=7
c=√7≈2.7
foci: (-1,0±c)=(-1,0±2.7)=(-1,-2.7) and (-1,2.7)
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Asymptotes are two straight line equations that go thru the center and are of the form: y=mx+b, m=slope, b=y-intercept.
slopes of asymptotes of hyperbolas with vertical transverse axis=±a/b=±2/√3
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Equation of asymptote with negative slope, -2/√3:
y=-2x/√3+b
solving for b using coordinates of center(-1,0)
0=-2*-1/√3+b
b=-2/√3
equation: y=-2x/√3-2/√3
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Equation of asymptote with positive slope, 2/√3:
y=2x/√3+b
solving for b using coordinates of center(-1,0)
0=2*-1/√3+b
b=2/√3
equation: y=2x/√3+2/√3
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