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Question 817914: Name the coordinates of the two vertices,the two foci, and find the equations of the asymptotes for the hyperbola 16x^2-y^2+96x+8y+112=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Name the coordinates of the two vertices,the two foci, and find the equations of the asymptotes for the hyperbola
16x^2-y^2+96x+8y+112=0
rearrange terms:
16x^2+96x-y^2+8y=-112
complete the square:
16(x^2+6x+9)-(y^2-8y+16)=-112+144-16
16(x+3)^2-(y-4)^2=16
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form of equation:  , (h,k)=(x,y) coordinates of center.
center:(-3,4)
a^2=1
a=1
vertices: (-3±a,4)=(-3±1,4)=(-4,4) and (-2,4)
b^2=16
b=√16=4
c^2=a^2+b^2=1+4=5
c=√5≈2.2
Foci:(-3±c,4)=(-3±2.2,4)=(-5.2,4) and (-.8,4)
slopes of asymptotes with horizontal transverse axis=±b/a=±4/1=±4
asymptotes are equations of straight lines that go through the center(-3,4)
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asymptote with positive slope:
y=mx+b
y=4x+b
solve for b using coordinates of center which are on the line.
4=4*(-3)+b
b=16
equation:y=4x+16
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asymptote with negative slope:
y=mx+b
y=-4x+b
solve for b using coordinates of center which are on the line.
4=-4*(-3)+b
b=-8
equation:y=-4x-8
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