Question

(x^2/25)-[(y-2)^2/4]=1
Find the foci and vertices

Answer 1

Since it is a Hyperbola

c^2 = a^2 + b^2

where c = Focii

Answer 2

@Yahoo! We're practicing (study session) for a supervised "in the flesh" test. So, it is okay to provide more details if you like. Just saying.

Answer 3

I have my answers, but I'm not sure if it's correct

Answer 4

Ok...Then 1st show ur Work @sakigirl

Answer 5

I guess the others left.

V(5,2) and V(-5,2)
F(-√29,2) and (√29, 2)

Did anybody find the asymtotes?

Answer 6

I get the same foci, but different vertices

Answer 7

Did you agree with the x part of the vertices?

Answer 8

Yes

Answer 9

What do you mean by y= 0 or 4?

Answer 10

Let me see your work.

Answer 11

My work on how I found my vertices?

Answer 12

I don't really do any work, I just look at it, and write them down haha

Answer 13

Well, what did you get?

Answer 14

Here is some good info on hyperbolas.

Answer 15

http://www.purplemath.com/modules/hyperbola2.htm

Answer 16

V(5,4) and V(-5,4)

Answer 17

How do you get 2? Even if you sq rt 4, you end up getting 2 or -6

Answer 18

Do you agree that on this hyperbola that the y-coordinates of the vertices and foci should be the same number, whatever that number may be?

Answer 19

Oh wait, I understand

Answer 20

I got mixed up. I needed to find the center to get it right

Answer 21

The center is at (0,2). Agree?

Answer 22

Yes