Question

Find an equation in standard form for the hyperbola with vertices at (0, ±8) and asymptotes at y=+- 1/2x

Answer 1

What do you know about hyperbolas currently? Maybe you have some kind of equation? Tell me what you know and we can work together to try to understand it better.

Answer 2

I have the equation \[\frac{ (y-k)^2 }{ a^2}-\frac{ (x-h)^2 }{ b^2 }=1\]

Answer 3

So what does h, k, a, and b mean if anything to you?

Answer 4

From the information given, can you sketch out a rough idea of what the hyperbola will look like?

Answer 5

Would k and h be the vertices?

Answer 6

Well, the point (h,k) will be the "center" of the hyperbola, but it'll sort of be equally between both vertices.

Answer 7

So I would plug that in to look like this?
\[\frac{ (y-0)^2 }{ a^2 }-\frac{ (x-8)^2 }{ b^2 }\]

Answer 8

Yeah, now you just need to find the a and b, which is from the asymptotes they give you.

Answer 9

How would I find a and b or is it just -1/2 and 1/2?

Answer 10

Well actually the slope of the asymptotes turns out to be (b/a) so b=1 and a=2 in this case.

Answer 11

Oh, okay. Let me try to solve this and then you can double check for me.

Answer 12

So, I don't know what I'm doing wrong because what I have is no where near to my answer choices. :/
\[\frac{ y^2 }{ 4}-\frac{ 8x^2 }{ 1 }=1\]

Answer 13

These are my choices

Answer 14

Give me a second, I'm checking to make sure what's going on.

Answer 15

Okay, thanks.

Answer 16

I hate to dot his, but I'm just sort of brain dead from staying up all night.

http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx

Here's a good start to checking their graphs rather than doing it by hand to see if it matches. I plugged in the first one to give you an example of how it works. It's a useful tool, don't abuse it.

http://www.wolframalpha.com/input/?i=y%5E2%2F16%5E2-x%5E2%2F8%5E2%3D1&t=crmtb01

I think I'm going, good luck!

Answer 17

It's okay; thank you so much for your help! Have a great day! :)