help with conics please!! exam tomorrow :|

[ggdotcom]

hey there everyone

so ive got an exam tomorrow, and i know pretty much everything perfectly except for the conics, they give me a tough time!
so right now im trying to solve this problem, to begin solving it really, but i'm not sure about some things, the problem is:

find the equation of the hyperbola with vertices in (-1,3) and (3,3) and with excentricity of 3/2.

see im really lost here cause i don't know what i'm supposed to use in the formula, is it the centre? one of the vertices? i know the formula you use is

(x-h)[sup:32lrph65]2[/sup:32lrph65] / a[sup:32lrph65]2[/sup:32lrph65] - (y-k)[sup:32lrph65]2[/sup:32lrph65] / b[sup:32lrph65]2[/sup:32lrph65] = 1

but i don't know what to substitute h and k for.

also how do you find the equation of an hyperbola when given the asymptotes? and vice versa, how do you get the asymptotes when given the equation?

also, how do you get the centre of an hyperbola?

please and thank you in advance! and would it be okay if i post some more questions that pop up here?

 

[BigGlenntheHeavy]

$\displaystyle \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2} \ = \ 1$

$\displaystyle Center \ of \ hyperbola \ (h,k), \ use \ midpoint \ formula.$

$\displaystyle Also, \ e \ = \ \frac{c}{a} \ and \ b^2 \ = \ c^2-a^2$

$\displaystyle You \ should \ be \ able \ to \ take \ it \ from \ here, \ see \ graph.$

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