EXAMPLE 1
Consider a hyperbola described by the general equation 9x2 - 16y2 - 36x - 32y -124 = 0. Rewrite this equation into standard form to find out what the graph of the hyperbola looks like.
STEP 1:	Collect the x terms together. Collect the y terms together. (9x2 - 36x) + (-16y2 - 32y) - 124 = 0
STEP 2:	Move any constant terms to the right hand side of the equation: (9x2 - 36x) + (-16y2 - 32y) = 124
STEP 3:	Factor out the coefficient of the x2-term from every term containing an x. Factor out the coefficient of the y2-term from every term containing a y. 9(x2 - 4x) -16(y2 + 2) = 124
STEP 4:	Find the numbers which complete each square. Add these numbers to both sides of the equation. 9(x2 - 4x + 4) -16(y2 + 2 + 1) = 124 + 9(4) - 16(1) = 124 + 36 - 16
STEP 5:	Simplify the equation. 9(x - 2)2 -16(y + 1)2 = 144
STEP 6:	Make the right hand side of the equation equal to 1 by dividing both sides by an appropriate number. 9(x - 2)2/ 144 -16(y + 1)2/ 144 = 144/144
STEP 7:	Simplify any fractions. (x - 2)2/ 16 -(y + 1)2/ 9 = 1
This equation is now in standard form. It is easy to see that the hyperbola opens left and right, is centred at (2, -1), and has asymptotes with slopes of +3/4 and -3/4.