In order to learn how to graph hyperbolas, it helps to recongize the equation, such as the one below:

  x²     y²
----- - ----- = 1
  16     9

What do you think you should do now?

Take the square root of the number under x², and this will be graphed on the x-axis as +4 and -4.

Then take the square root of the number under the y², and this will be graphed on the y-axis as +3 & -3

What do you think you should do now?

In this particular problem, a² is 16 because it is before the subtraction sign in the original equation and b² is 9 because it is after the subtraction sign. Therefore, you find "c" by substituting those numbers into the above equation and solving for c like this:

9 = c² - 16

25 = c²

5 = c Therefore the foci are (5,0) and (-5,0).

To find the equations of the asymptotes (the dotted lines) we use the equations: y = (b/a)x and y = (-b/a)x. So the two equations for the asymptotes in this problem are: y = (3/4)x and y = (-3/4)x.

Now, try to graph this on your own ... check your answer below.

Now that we have four points on the graph, we can draw a rectangle around these four points. After this we can draw dotted lines through the corners of the rectangle. These dotted lines allow us to create a hyperbola, each half crosses the x-axis and approaces the dotted line (asymptote). Also, label the foci and asymptotes from the information above.

This is the equation when X is the major axis:

This is the equation when Y is the major axis:

What if the equation isn't in one of these two forms?

Then we need to get it it into one of these forms. For example, take the equation : 4x² - 9y² = 36. We must get a "one" on the right side first, so we must divide everything by 36 before we do anything. By dividing, we get the new equation:

Now we are ready to take the square roots again to help us graph the hyperbola. The numbers on the x axis are 3 and -3 and the numbers on the y axis are 2 and -2. Then we must use the equation with all 3 variables again: b2 = c2 - a2 to find c for the foci. They are: (√13,0) and (-√13,0). The asymptotes are: y = (2/3)x and y = (-2/3)x. Now we are ready to put them on a graph again.