# Precalc

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A hyperbola has this Cartesian equation:
-((x-2)/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=+- ?

• Precalc -

change to more standard way of writing the hyperbola

(x-2)^2 /25 - (y+1)^2 /9 = -1

So the major axis is parallel to the y-axis 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 = (-6-4)/(-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

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