Question 604752: Graph the equation. Identify the vertices, foci, and asymptotes of the hyperbola. y^2/16-x^2/36=1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Graph the equation. Identify the vertices, foci, and asymptotes of the hyperbola.
y^2/16-x^2/36=1
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This is an equation of a hyperbola with vertical transverse axis.
Its standard form: (y-k)^2/a^2-(x-h)^2/b^2=1, (h,k)=(x,y) coordinates of center
For given hyperbola: y^2/16-x^2/36=1
center: (0,0)
a^2=16
a=√16=4
vertices:(0,0±a)=(0,0±4)=(0,-4) and (0,4)
..
b^2=36
b=√36=6
..
c^2=a^2+b^2=16+36=52
c=√52≈7.2
foci: (0,0±c)=(0,0±7.2)=(0,-7.2) and (0,7.2)
..
Asymptotes:
slope=±a/b=±4/6=±2/3
equation of asymptotes:
y=±2x/3
..
y=±(16+4x^2/9)^.5
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