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September 2, 2014

September 2, 2014

Posted by **Grace** on Friday, January 20, 2012 at 12:39am.

-((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 -
**Reiny**, Friday, January 20, 2012 at 9:32amchange 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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