Expert: Richard J. Raridon - 4/4/2009

Question
Richard,
I am taking college algebra and have a few question on an assignment. These are sample problems, but I'm not sure how to even work them out..if you can show my step by step I would appreciate it, then I should be able to do the assignment. This is an online class and the book is horrible and doesn't give hardly any examples. Thanks!

1. Find the standard equation of the hyperbola with foci at (0,-4) and (0,4) and asymptotes y= +/- x
It should be the plus sign over the minus sign, but I didn't know how to do that.

2. Find the vertices of the hyperbola defined by the equation below:
9y^2 - 4x^2 + 36=0

I appreciate your help.
Thanks,
Tara

Answer
The standard equation for an hyperbola is
(x-h)^2/a^2 -(y-k)^2/b^2 = 1 where the center is at (h,k)
1. Here the center is at (0,0) so that reduces it to
x^2/a^2 -y^2/b^2 = 1
Since the slopes of the asymptotes are +/- 1, then a = b.  
The distance from the center to either focus is (a^2+b^2)^1/2,
so that equals 4 and a^2+b^2 = 16 making a=b=8
and your equation is x^2/8 -y^2/8 = 1
2. divide each term by 36 and rearrange to get
x^2/9 -y^2/4 = 1
According to my book, hyperbola have foci but not vertices