Precalc
posted by Grace .
A hyperbola has this Cartesian equation:
((x2)/5)^2 + ((y+1)/3)^2 = 1
A. Find direction in which it opens.
Horizontal?
B. Find the coordinates of the center.
I think it's (2,1) or (1,2)?
C. Find the slopes of the asymptotes: m=+ ?

change to more standard way of writing the hyperbola
(x2)^2 /25  (y+1)^2 /9 = 1
So the major axis is parallel to the yaxis and we have a vertical hyperbola
centre is ( 2,1) , (there is only one centre)
a = 5 , b = 3
I then sketch a rectangle that has (2,1) as its centre, having a height of 10 and a width of 6 , (2a and 2b)
So the coordinates of the main diagonal, which would be one of the asymptotes, are
(5,4) and (1,6)
Slope of that asymptote = (64)/(1,5) = 10/6 = 5/3
So the slope of the other one is 5/3
Quick way:
slope of asymptotes = ± a/b = ± 5/3