You can put this solution on YOUR website! Find the vertices and asymptotes of the hyperbola.
9y^2-16x^2=144
y^2/16-x^2/9=1
This is a hyperbola with vertical transverse axis.
Its standard form of equation:
center(0,0)
a^2=16
a=√16=4
vertices: (0, 0±a)=((), 0±4)=(0,-4) and (0,4)
b^2=9
b=3
..
asymptotes: straight line equations of the form: y=mx+b, m=slope, b=y-intercept
line goes thru center of hyperbola, y-intercept=0
slopes of asymptotes: ±a/b=±4/3 (for hyperbolas with vertical transverse axis)
equation of asymptote with negative slope: y=-4x/3
equation of asymptote with positive slope: y=4x/3
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