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?

Algebra ->  Quadratic-relations-and-conic-sections -> 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?      Log On


   

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) About Me  (Show Source):
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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:
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1, (h,k)=(x,y) coordinates of center.
..
For given problem:
equation: %28x-3%29%5E2%2F4-%28y%2B2%29%5E2%2F9=1
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