# Finding equation of hyperbola from asymptotes and foci

• March 17th 2014, 09:22 PM
calmo11
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...
• March 17th 2014, 09:48 PM
romsek
Re: Finding equation of hyperbola from asymptotes and foci
have a look at this and see if it helps
• March 18th 2014, 06:35 AM
Soroban
Re: Finding equation of hyperbola from asymptotes and foci
Hello, calmo11!

Quote:

Find the equation of a hyperbola with foci $(\text{-}3,\text{-}1)$ and $(\text{-}3,9)$
and asymptotes: . $y\:=\:\text{-}\tfrac{3}{4}x+\tfrac{7}{4}$ and $y\:=\:\tfrac{3}{4}x+\tfrac{25}{4}$

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

Its general equation is: . $\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} \:=\:1$
. . where $(h,k)$ is the center and $\pm\tfrac{b}{a}$ is the slope of the asymptotes.

The foci are $(\text{-}3,\text{-}1)$ and $(\text{-}3,9)$.
The center is the midpoint: . $(h,k) \:=\:(\text{-}3,4).$

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

Got it?
• March 18th 2014, 02:05 PM
calmo11
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!