Subtract the right side from both sides to get 0 on the right:



We can tell this is the graph of a hyperbola because the x² and 
y² terms have opposite signs when all non-zero terms are one 
side and all like terms combined. 

Get the y terms together and the x terms together, and add 15
to both sides to get the contant term off the left side:



Factor the coefficient of the y² term, which is 2, out of the 
first two terms on the left, skipping a space at the end of the
parentheses.

Factor the coefficient of the x² term, which is -1, out of the 
last two terms on the left, skipping a space at the end of the
parentheses.


  

Complete the square inside each parentheses:

In your head or on scratch paper,
1. Multiply the coefficient of y, which is -1, by 1/2, getting -1/2.
2. Square -1/2, getting (-1/2)² = +1/4
3. Add +1/4 in the space in the first parentheses, which amounts to
   adding 2 times +1/4 or +1/2 to the left side. So add +1/2 to the right 
   side.

1. Multiply the coefficient of x, which is -6, by 1/2, getting -3.
2. Square -3, getting (-3)² = +9
3. Add +9 in the space in the first parentheses, which amounts to
   adding -1 times +9 or -9 to the left side. So add -9 to the 
   right side.



Factor each trinomial: 
                       
Combine the terms on the right 



Clear the fraction on the right by multiplying 
through by 2:



Get 1 on the right by dividing through by 13:



Divide numerator and denominator by the coefficients of
the squares of the binomials:





This is in standard form:

 which is the equation of a hyperbola which
opens like this:                                  
  
So , , , , , 

The center is (h,k) = , the big green dot below.

We draw the defining rectangle with (h,k) as its center, its vertical
dimension is 2a = , and its horizontal dimension is 2b = .

We extend the diagonals of the defining rectangle, since they are the
asymptotes of the hyperbola. Then we sketch in the hyperbola:



Edwin