Finding equation of hyperbola from asymptotes and foci

Hi I am very confused as to how this question actually works out...

I am asked to find the equation of a hyperbola with foci at (-3,-1) and (-3,9)

and asymptotes at y=(-3x/4)+(7/4) and y=(3x/4)+(25/4)

Every time I try to visualize this my foci are on a different side of the x-axis than my foci which completely throws all my thoughts out

I also don't particularly know how to get the equation when there is asymptotic shifts...

PLEASE HELP ME

Re: Finding equation of hyperbola from asymptotes and foci

have a look at this and see if it helps

Re: Finding equation of hyperbola from asymptotes and foci

Hello, calmo11!

Quote:

Find the equation of a hyperbola with foci $\displaystyle (\text{-}3,\text{-}1)$ and $\displaystyle (\text{-}3,9)$

and asymptotes: .$\displaystyle y\:=\:\text{-}\tfrac{3}{4}x+\tfrac{7}{4}$ and $\displaystyle y\:=\:\tfrac{3}{4}x+\tfrac{25}{4}$

Since the foci are oriented vertically, the hyperbola is "vertical": .$\displaystyle \begin{array}{c}\cup \\ \cap \end{array}$

Its general equation is: .$\displaystyle \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} \:=\:1$

. . where $\displaystyle (h,k)$ is the center and $\displaystyle \pm\tfrac{b}{a}$ is the slope of the asymptotes.

The foci are $\displaystyle (\text{-}3,\text{-}1)$ and $\displaystyle (\text{-}3,9)$.

The center is the midpoint: .$\displaystyle (h,k) \:=\:(\text{-}3,4).$

The slope of the asymptotes is: .$\displaystyle \pm\frac{b}{a} \:=\:\pm\frac{3}{4}$

Hence: .$\displaystyle a = 4.\;b = 3$

Got it?

Re: Finding equation of hyperbola from asymptotes and foci

Thank you very much guys, I used romsek's link and the content on that site helped me heaps!