Hyperbolas 12-4 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

Warm Up Multiply both sides of each equation by the least common multiple to eliminate the denominators. 1. x2 9 – = 1 y2 4 4x2 – 9y2 = 36 2. y2 25 – = 1 x2 16 16y2 – 25x2 = 400

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

Vocabulary hyperbola focus of a hyperbola branch of a hyperbola transverse axis vertices of a hyperbola conjugate axis co-vertices of a 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.

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). d = |PF1 – PF2 | Definition of the constant difference of a hyperbola. Distance Formula Substitute. Simplify. d = 4 The constant difference is 4.

Definition of the constant difference of a hyperbola. Check It Out! Example 1 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). d = |PF1 – PF2 | Definition of the constant difference of a hyperbola. Distance Formula Substitute. Simplify. d = 12 The constant difference is 12.

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.

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

The values a, b, and c, are related by the equation c2 = a2 + b2 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.

Example 2A: Writing Equations of Hyperbolas Write an equation in standard form for each hyperbola. 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

Step 2 Identify the center and the vertices. Example 2A Continued 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

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). 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

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

Write an equation in standard form for each hyperbola. Check It Out! Example 2a Write an equation in standard form for each hyperbola. Vertex (0, 9), co-vertex (7, 0) Step 1 Because the vertex is on the vertical axis, the transverse axis is vertical and the equation is in the form . y2 a2 – = 1 x2 b2 Step 2 a = 9 and b = 7. Step 3 Write the equation. Because a = 9 and b = 7, the equation of the graph is , or . y2 92 – = 1 x2 72 81 49

Write an equation in standard form for each hyperbola. Check It Out! Example 2b Write an equation in standard form for each hyperbola. Vertex (8, 0), focus (10, 0) x2 a2 – = 1 y2 b2 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 .

Check It Out! Example 2b Continued Step 2 a = 8 and c = 10; Use c2 = a2 + b2 to solve for b2. 102 = 82 + b2 Substitute 10 for c, and 8 for a. 36 = b2 Step 3 The equation of the hyperbola is . x2 64 – = 1 y2 36

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

Example 3A: Graphing a Hyperbola Find the vertices, co-vertices, and asymptotes of each hyperbola, and then graph. 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

Step 3 The equations of the asymptotes are y = x and y = – x. Example 3A Continued 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.

Example 3A Continued 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.

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 Step 1 The equation is in the form , so the transverse axis is horizontal with center (3, –5). (x – h)2 a2 – = 1 (y – k)2 b2

Step 3 The equations of the asymptotes are Example 3B Continued Step 2 Because a = 3 and b =7, the vertices are (0, –5) and (6, –5) and the co-vertices are (3, –12) and (3, 2) . Step 3 The equations of the asymptotes are y + 5 = (x – 3) and y = – (x – 3). 7 3

Example 3B Continued 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.

Check It Out! Example 3a Find the vertices, co-vertices, and asymptotes of each hyperbola, and then graph. x2 16 – = 1 y2 36 Step 1 The equation is in the form so the transverse axis is horizontal with center (0, 0). x2 a2 – = 1 y2 b2

Check It Out! Example 3a Continued Step 2 Because a = 4 and b = 6, the vertices are (4, 0) and (–4, 0) and the co-vertices are (0, 6) and . (0, –6). Step 3 The equations of the asymptotes are y = x and y = – x . 3 2

Check It Out! Example 3 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.

Check It Out! Example 3b Find the vertices, co-vertices, and asymptotes of each hyperbola, and then graph. (y + 5)2 1 – = 1 (x – 1)2 9 Step 1 The equation is in the form so the transverse axis is vertical with center (1, –5). (y – k)2 a2 – = 1 (x – h)2 b2

Check It Out! Example 3b Continued Step 2 Because a = 1 and b =3, the vertices are (1, –4) and (1, –6) and the co-vertices are (4, –5) and (–2, –5). Step 3 The equations of the asymptotes are y + 5 = (x – 1) and y + 5 = – (x – 3). 1 3

Check It Out! Continued 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.

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

Lesson Quiz: Part I 1. Find the constant difference for a hyperbola with foci (–3.5, 0) and (3.5, 0) and a point on the hyperbola (3.5, 24). 1 2. Write an equation in standard form for a hyperbola with center hyperbola (4, 0), vertex (10, 0), and focus (12, 0).

3. Find the vertices, co-vertices, and asymptotes of , then graph. Lesson Quiz Part II 3. Find the vertices, co-vertices, and asymptotes of , then graph. asymptotes: vertices: (–6, ±5); co-vertices (6, 0), (–18, 0); 5 12 y = ± (x + 6)