Question 479836
1. Find the center 
2. Slopes of Asymptotes 
Hyperbola 
(x-3)^2/9-(y+4)^2/16=1 
Find the equation of the hyperbola described. 
center at (1, -4), one vertex at (4, -4), and b = 5 

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(x-3)^2/9-(y+4)^2/16=1
This is an equation of a hyperbola with horizontal transverse axis of the standard form:
y=(x-h)^2/a^2-(y-k)^2/b^2=1
For given equation:
1.center: (3,-4)
a^2=9
a=3
..
b^2=16
b=4
..
2.asymptotes:
slope=±b/a=±4/3
..
Find the equation of the hyperbola with given information
Center: (1,-4) (given)
a=distance from center to one vertex on the horizontal transverse axis=4-1=3
a^2=9
b=5 (given)
b^2=25
This is an equation of a hyperbola with horizontal transverse axis of the standard form:
y=(x-h)^2/a^2-(y-k)^2/b^2=1
Given equation:
y=(x-1)^2/9-(y+4)^2/25=1