To understand what this curve might look like, we have to work towards a 
standard form.  This is best accomplished by completing the square in the x 
terms and in the y terms.

9x2 - 4y2 - 18x + 32 y - 91 = 0
9(x2 - 2x) - 4(y2 - 8y) - 91 = 0
9(x2 - 2x + 1) -9(1) - 4(y2 - 8y + 16) + 4(16) - 91 = 0
9(x - 1)2 - 4(y - 4)2 = 36

(x - 1)2     (y - 4)2
--------  -  -------- = 1
   4             9

(x - 1)2     (y - 4)2
--------  -  -------- = 1
   22             32

From this, we see that the curve is a hyperbola centered at (1, 4).  When 
y = 4 we have:  ( x - 1)2
                --------- = 1
                    4
So,
(x - 1)2 = 22
(x - 1) = 2  or (x - 1) = -2
Thus, x = 3 or x = -1.

Therefore, (3, 4) and ( -1, 4) are both on the curve.  The asymptotes are the 
lines y = (3/2)x and y = -(3/2)x and they pass through the center (1, 4). 
See diagram below: