(x-5)^2-((y+1)^2/4)=1
This is supposed to be a graph of a hyperbola and I'm supposed to find the vertices and foci. I'm really confused, please help!

Question

mertsj · Answer

What is the center?

anonymous · Answer

I believe it would be (5,-1)

mertsj · Answer

That is correct.
What is the value of a^2?

anonymous · Answer

Where did the a come from?

mertsj · Answer

Does this hyperbola open to the right and left or up and down?

anonymous · Answer

Up?

mertsj · Answer

How do you know?  Was that a guess?

anonymous · Answer

Well, the x came first so its up or down. And its positive.

mertsj · Answer

Since x is positive, it opens right and left.

anonymous · Answer

Is that because its on the x axis?

mertsj · Answer

(x-h)^2/a^2-(y-k)^2/b^2=1

mertsj · Answer

It's because x cannot be 0

mertsj · Answer

The equation I posted is the equation of a hyperbola with center (h,k) that opens right and left.  What is the value of a^2 in your problem?

anonymous · Answer

1?

mertsj · Answer

and if a^2 is 1 then a is also 1 and the vertices are 1 unit to the right and 1 unit to the left of the center.  So find the vertices.

anonymous · Answer

(4,-1), (6,-1)

mertsj · Answer

yes.  Very good

mertsj · Answer

Now for a hyperbola that opens right and left, the foci are c units to the right and c units to the left of the center.

mertsj · Answer

The relationship between a, b, and c is that a^2+b^2=c^2
We know that a^2=1.  What is b^2?

anonymous · Answer

How do I find c?

mertsj · Answer

The relationship between a, b, and c is that a^2+b^2=c^2
 We know that a^2=1. 
What is b^2?

anonymous · Answer

c^2-1

mertsj · Answer

What is b^2?

anonymous · Answer

How do I find b without knowing what c is?

mertsj · Answer

The equation of a hyperbola that opens to the right and the left and has center (h,k) is:
(x-h)^2/a^2-(y-k)^2/b^2=1

mertsj · Answer

Compare that to your equation and find out what b^2 is

anonymous · Answer

so b^2=4 and b=2

mertsj · Answer

Yes.  a^2=1 and b^2=4
Now find c knowing that a^2+b^2=c^2

anonymous · Answer

c^2=5 so c=

mertsj · Answer

c=sqrt5

anonymous · Answer

the square root of 5

mertsj · Answer

and the foci are sqrt5 units to the right and left of the center so they are???

anonymous · Answer

(5+sqrt5, -1) and (5-sqrt5, -1)

mertsj · Answer

Very good.