SOLUTION: Equation of a hyberbola (x-3)^2 over 4 - (y+2)^2 over 9 = 1 How do I find the vertices, foci, and asymptote slope?

Question 712796: Equation of a hyberbola
(x-3)^2 over 4 - (y+2)^2 over 9 = 1
How do I find the vertices, foci, and asymptote slope?
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Equation of a hyberbola
(x-3)^2 over 4 - (y+2)^2 over 9 = 1
How do I find the vertices, foci, and asymptote slope?
**
Standard form of equation for a hyperbola with horizontal transverse axis:
, (h,k)=(x,y) coordinates of center.
..
For given problem:
equation:
center: (3,-2)
a^2=4
a=2
vertices: (3�a,-2)=(3�2,-2)=(1,-2) and (5,-2)
..
b^2=9
b=3
..
c^2=a^2+b^2=4+9=13
c=√13≈3.6
foci: (3�c,-2)=(3�3.6,-2)=(-0.6,-2) and (6.6,-2)
..
slopes of asymptotes of hyperbolas with horizontal transverse axis: �b/a=�3/2

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