SOLUTION: Graph the equation. Identify the vertices, foci, and asymptotes of the hyperbola. y^2/16-x^2/36=1

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
**
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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