You can put this solution on YOUR website! how do i find the asymptotes for
(y^2/16)-(x^2/81)=1
This is an equation of a hyperbola with vertical transverse axis.
Its standard form of equation: , (h,k)=(x,y) coordinates of center.
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Asymptotes of hyperbolas are straight-line equations that go through the center.
Their standard form: y=mx+b, m=slope, b=y-intercept
slopes of hyperbolas with vertical transverse axis=±a/b
slopes of hyperbolas with horizontal transverse axis=±b/a
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For given hyperbola:
center: (0,0)
a^2=16
a=√16=4
b^2=81
b=√81=9
slopes of asymptotes=±a/b=±4/9
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Equation of asymptote with positive slope:
y=(4/9)x+b
solve for b using coordinates of center:
b=y-intercept=0
equation: y=4x/9
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Equation of asymptote with negative slope:
y=(-4/9)x+b
b=0
equation:y=-4x/9
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