SOLUTION: 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), a

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
thank you
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
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
**
(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
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