# Easy speasy but i dont get it.

1. Jun 9, 2005

### shimmeringlight

okay, so math

it says...

Determine an equation in standard from for each hyperbola...
centre (-2,1), one focus (-4,1), length of conjugate axis 4

what i figured out... but its wrong... cuz i cant figure out the answer!!

(x+2)^2 - (y-1)^2 = 1
a^2_____b^2

conjugate axis is 4 so co-ordinates for the co-vertices or vertices should be
(-2,3), (-2,-1)

so then to find find the a or b value... i use
c^2=a^2+b^2
2^2=a^2+2^2

but 4-4=0.... and thats impossible??? i got stuck here...

help plz... thank you in advance.

Last edited: Jun 9, 2005
2. Jun 9, 2005

### moose

I always thought that in these problems, you would use c^2=a^2-b^2....Could someone check that?

3. Jun 9, 2005

### shimmeringlight

i checked it but my book says c^2=a^2+b^2 unless the book is wrong... but i've done other problems and its worked out... im just stuck on this one question.

4. Jun 9, 2005

### Ouabache

$$\frac {(x+2)^2}{a^2}- \frac{(y-1)^2}{b^2} =1$$ your first equation looks fine.

I needed to review conjuage axis as well as transverse axis as it pertains to hyperbolae.
You may want to review that page too, as it may offer some clues on how to solve this problem.
They also confirm your relation $c^2=a^2+b^2$

Another good hyperbola reference from mathworld.

5. Jun 9, 2005

### theCandyman

Moose is corret, $$c^2 = a^2 - b^2$$. You will have to post the orginal problem for me to help you anymore, I do not really remember ellipses well.

For reference, standard form: $$\frac{x-h^2}{a^2} - \frac{y-k^2}{b^2} = 1$$, where the ellipse is centered at $$(h, k)$$.

Last edited: Jun 9, 2005
6. Jun 9, 2005

### shimmeringlight

ok, i figured it out. thx

7. Jun 9, 2005

### Ouabache

That equation is great for an ellipse, this is a hyperbola question.

8. Jun 10, 2005

### theCandyman

Hahaha :tongue2:

I was wondering why I could not do much of the problem, I think mentioning two vertices threw me off the hyperbola, I always associate them with ellipses.