P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 892

892 Chapter 9 Conic Sections and Analytic Geometry

EXAMPLE 4 Graphing a Hyperbola

Graph and locate the foci: 9y2 - 4x2 = 36. What are the equations of the
asymptotes?
Solution We begin by writing the equation in standard form. The right side
should be 1, so we divide both sides by 36.
9y2 4x 2 36
- =
36 36 36
y2 x2
- = 1 Simplify. The right side is now 1.
4 9
Now we are ready to use our four-step procedure for graphing hyperbolas.

Step 1 Locate the vertices. The equation that we obtained is in the form
y2 x2
2
- 2 = 1, with a2 = 4 and b2 = 9.
a b
y2 x2
- =1
4 9
a2 = 4 b2 = 9

Based on the standard form of the equation with the transverse axis on the y-axis,
we know that the vertices are 10, - a2 and 10, a2. Because a2 = 4, a = 2. Thus, the
vertices are 10, - 22 and (0, 2), shown in Figure 9.23.
Step 2 Draw a rectangle. Because a2 = 4 and b2 = 9, a = 2 and b = 3. We
construct a rectangle to find the asymptotes, using -2 and 2 on the y-axis (the
vertices are located here) and -3 and 3 on the x-axis. The rectangle passes through
these four points, shown using dashed lines in Figure 9.23.
Step 3 Draw extended diagonals of the rectangle to obtain the asymptotes. We
draw dashed lines through the opposite corners of the rectangle, shown in Figure 9.23,
to obtain the graph of the asymptotes. Based on the standard form of the hyperbola’s
equation, the equations of these asymptotes are
a 2
y = ; x or y = ; x.
b 3
Step 4 Draw the two branches of the hyperbola by starting at each vertex and
approaching the asymptotes. The hyperbola is shown in Figure 9.24.

y y

5 5
Asymptote
4 Vertex (0, 2) 4
3 3

1 1
x x
−5 −4 −2 −1−1 1 2 4 5 −5 −4 −2 −1−1 1 2 4 5

−3 −3
−4 Vertex (0, −2) −4
Asymptote
−5 −5

Figure 9.23 Preparing to graph Figure 9.24 The graph of

y2 x2 y2 x2
- = 1 - = 1
4 9 4 9

We now consider the foci, located at 10, -c2 and 10, c2. We find c using
c2 = a2 + b2.
c2 = 4 + 9 = 13