Hi, there--

THE PROBLEM:
Graph hyperbola 9y^2-x^2=1. Find the foci and asymptotes.

A SOLUTION:
The general equation for a hyperbola is  when the transverse axis is horizontal.

If the equation is of the form  then the transverse axis is vertical.

Your equation is . We see that it is the second type, so the transverse axis is vertical. The graph looks kind of like two parabolas, one opening up, the other down.


Find the Center:
Next we see that h=0 and k=0. This tells us the the hyperbola is centered at the point
 (h,k) = (0,0).

Find Vertices and Foci:
To find the vertices and foci, we need to calculate a, b, and c. We see right away that b^2 = 1 
and b=1,  but we need to do a little work to find a. We need to move the coefficient of y^2 
without changing its value.

 so  and .

Now find c.






The vertices and foci are located to the above and below the center (because the transverse 
axis is vertical.)

The foci are the points (0,c)=(0,1) and (0.-c)=(0,-1).
The vertices are the points (0,a)=( 0, 1/3) and (0,-a)=(0,-1/3).
 
Find Asymptotes:
When the transverse axis is vertical, the formulas for the two asymptotes are
 and  

You asymptotes are
 and 

Graph the Hyperbola:
This is the graph of the hyperbola and the asymptotes.
y=(1/3)sqrt(x^2+1)



Hope this helps! Feel free to email if you have any questions about the solution.

Good luck with your math,

Mrs. F
math.in.the.vortex@gmail.com