1. Chapter 12
12-4 Hyperbolas

2. Objectives Write the standard equation for a hyperbola.
Graph a hyperbola, and identify its center, vertices, co-vertices, foci, and asymptotes.

3. Hyperbola
What would happen if you pulled the two foci of an ellipse so far apart that they moved outside the ellipse?
The result would be a hyperbola, another conic section.
A hyperbola is a set of points P(x, y) in a plane such that the difference of the distances from P to fixed points F1 and F2, the foci, is constant. For a hyperbola, d = |PF1 – PF2 |, where d is the constant difference. You can use the distance formula to find the equation of a hyperbola.

4. Example 1: Using the Distance Formula to Find the Constant Difference of a Hyperbola
Find the constant difference for a hyperbola with foci F1 (–8, 0) and F2 (8, 0) and the point on the hyperbola (8, 30).
Sol:
d = |PF1 – PF2 | Definition of the constant difference of a hyperbola.
Distance Formula

5. solution
Substitute
d = 4
The constant difference is 4.

6. Check it out! Example
Find the constant difference for a hyperbola with foci F1 (0, –10) and F2 (0, 10) and the point on the hyperbola (6, 7.5).
Sol:
The constant difference is 12.

7. Hyperbolas
As the graphs in the following table show, a hyperbola contains two symmetrical parts called branches.
A hyperbola also has two axes of symmetry. The transverse axis of symmetry contains the vertices and, if it were extended, the foci of the hyperbola. The vertices of a hyperbola are the endpoints of the transverse axis.
The conjugate axis of symmetry separates the two branches of the hyperbola. The co-vertices of a hyperbola are the endpoints of the conjugate axis. The transverse axis is not always longer than the conjugate axis.

8. Hyperbolas
The standard form of the equation of a hyperbola depends on whether the hyperbola’s transverse axis is horizontal or vertical.

9. Hyperbolas
The values a, b, and c, are related by the equation c2 = a2 + b2. Also note that the length of the transverse axis is 2a and the length of the conjugate is 2b.

10. Example 2A: Writing Equations of Hyperbolas
Write an equation in standard form for each hyperbola.
Sol:
Step 1 Identify the form of the equation.
The graph opens horizontally, so the equation will be in the form of x2 a2
– = 1 y2 b2

11. solution Step 2 Identify the center and the vertices.
The center of the graph is (0, 0), and the vertices are (–6, 0) and (6, 0), and the co-vertices are (0, –6), and (0, 6). So a = 6, and b = 6.
Step 3 Write the equation. x2 36
– = 1. y2
Because a = 6 and b = 6, the equation of the graph is 62
– = 1, or

12. Example 2B: Writing Equations of Hyperbolas
Write an equation in standard form for each hyperbola.
The hyperbola with center at the origin, vertex (4, 0), and focus (10, 0).
Sol:
Step 1 Because the vertex and the focus are on the horizontal axis, the transverse axis is horizontal and the equation is in the form x2 a2
– = 1 y2 b2

13. solution
Step 2 Use a = 4 and c = 10; Use c2 = a2 + b2 to solve for b2.
102 = 42 + b2
84 = b2
Step 3 The equation of the hyperbola is x2 16
– = 1 y2 84

14. Check it out! Example
Write an equation in standard form for each hyperbola.
Vertex (0, 9), co-vertex (7, 0)
Sol:
Because a = 9 and b = 7, the equation of the graph is , or y2 92
– = 1 x2 72 81 49

15. Hyperbolas
As with circles and ellipses, hyperbolas do not have to be centered at the origin.

16. Example 3A: Graphing a Hyperbola
Find the vertices, co-vertices, and asymptotes of each hyperbola, and then graph.
Sol: x2 49
– = 1 y2 9
Step 1 The equation is in the form so the transverse axis is horizontal with center (0, 0). x2 a2
– = 1 y2 b2

17. solution
Step 2 Because a = 7 and b = 3, the vertices are (–7, 0) and (7, 0) and the co-vertices are (0, –3) and (0, 3). 3 7
Step 3 The equations of the asymptotes are
y = x and y = – x.

18. solution
Step 4 Draw a box by using the vertices and co-vertices. Draw the asymptotes through the corners of the box
Step 5 Draw the hyperbola by using the vertices and the asymptotes.

19. Example 3B: Graphing a Hyperbola
Find the vertices, co-vertices, and asymptotes of each hyperbola, and then graph.
(x – 3)2 9
– = 1
(y + 5)2 49

20. Parameter
Notice that as the parameters change, the graph of the hyperbola is transformed.