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10-4 Hyperbolas Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2

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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

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**Objectives Write the standard equation for a hyperbola.**

Graph a hyperbola, and identify its center, vertices, co-vertices, foci, and asymptotes.

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**Vocabulary hyperbola focus of a hyperbola branch of a hyperbola**

transverse axis vertices of a hyperbola conjugate axis co-vertices of a hyperbola

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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.

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**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.

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**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.

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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.

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

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**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.

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**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

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**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

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**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

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**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

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**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

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**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 .

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**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

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

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**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

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**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.

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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.

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**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

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**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

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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.

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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

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**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

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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.

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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

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**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

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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.

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

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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).

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**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)

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