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Parabola. Warm Up 132. from (0, 2) to (12, 7) Find each distance. 3. from the line y = –6 to (12, 7) 13 1. Given, solve for p when c =

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Presentation on theme: "Parabola. Warm Up 132. from (0, 2) to (12, 7) Find each distance. 3. from the line y = –6 to (12, 7) 13 1. Given, solve for p when c ="— Presentation transcript:

1 parabola

2 Warm Up 132. from (0, 2) to (12, 7) Find each distance. 3. from the line y = –6 to (12, 7) 13 1. Given, solve for p when c =

3 Write the standard equation of a parabola and its axis of symmetry. Graph a parabola and identify its focus, directrix, and axis of symmetry. Objectives

4 focus of a parabola directrix Vocabulary

5 In Chapter 5, you learned that the graph of a quadratic function is a parabola. Because a parabola is a conic section, it can also be defined in terms of distance.

6 A parabola is the set of all points P(x, y) in a plane that are an equal distance from both a fixed point, the focus, and a fixed line, the directrix. A parabola has a axis of symmetry perpendicular to its directrix and that passes through its vertex. The vertex of a parabola is the midpoint of the perpendicular segment connecting the focus and the directrix.

7 The distance from a point to a line is defined as the length of the line segment from the point perpendicular to the line. Remember!

8 Use the Distance Formula to find the equation of a parabola with focus F(2, 4) and directrix y = –4. Example 1: Using the Distance Formula to Write the Equation of a Parabola Definition of a parabola. PF = PD Substitute (2, 4) for (x 1, y 1 ) and (x, –4) for (x 2, y 2 ). Distance Formula.

9 Example 1 Continued (x – 2) 2 + (y – 4) 2 = (y + 4) 2 Square both sides. Expand. Subtract y 2 and 16 from both sides. (x – 2) 2 + y 2 – 8y + 16 = y 2 + 8y + 16 (x – 2) 2 – 8y = 8y (x – 2) 2 = 16y Add 8y to both sides. Solve for y. Simplify.

10 Use the Distance Formula to find the equation of a parabola with focus F(0, 4) and directrix y = –4. Definition of a parabola. PF = PD Substitute (0, 4) for (x 1, y 1 ) and (x, –4) for (x 2, y 2 ). Distance Formula Check It Out! Example 1

11 x 2 + (y – 4) 2 = (y + 4) 2 Square both sides. Expand. Subtract y 2 and 16 from both sides. x 2 + y 2 – 8y + 16 = y 2 + 8y +16 x 2 – 8y = 8y x 2 = 16y Add 8y to both sides. Solve for y. Check It Out! Example 1 Continued Simplify.

12 Previously, you have graphed parabolas with vertical axes of symmetry that open upward or downward. Parabolas may also have horizontal axes of symmetry and may open to the left or right. The equations of parabolas use the parameter p. The |p| gives the distance from the vertex to both the focus and the directrix.

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14 Write the equation in standard form for the parabola. Example 2A: Writing Equations of Parabolas Step 1 Because the axis of symmetry is vertical and the parabola opens downward, the equation is in the form y = x 2 with p < 0. 1 4p4p

15 Example 2A Continued Step 2 The distance from the focus (0, –5) to the vertex (0, 0), is 5, so p = –5 and 4p = –20. Step 3 The equation of the parabola is. y = – x 2 1 20 Check Use your graphing calculator. The graph of the equation appears to match.

16 Example 2B: Writing Equations of Parabolas vertex (0, 0), directrix x = –6 Write the equation in standard form for the parabola. Step 1 Because the directrix is a vertical line, the equation is in the form. The vertex is to the right of the directrix, so the graph will open to the right.

17 Example 2B Continued Step 2 Because the directrix is x = –6, p = 6 and 4p = 24. Step 3 The equation of the parabola is. x = y 2 1 24 Check Use your graphing calculator.

18 vertex (0, 0), directrix x = 1.25 Check It Out! Example 2a Write the equation in standard form for the parabola. Step 1 Because the directrix is a vertical line, the equation is in the form of. The vertex is to the left of the directrix, so the graph will open to the left.

19 Step 2 Because the directrix is x = 1.25, p = –1.25 and 4p = –5. Check Check It Out! Example 2a Continued Step 3 The equation of the parabola is Use your graphing calculator.

20 Write the equation in standard form for each parabola. vertex (0, 0), focus (0, –7) Check It Out! Example 2b Step 1 Because the axis of symmetry is vertical and the parabola opens downward, the equation is in the form

21 Step 2 The distance from the focus (0, –7) to the vertex (0, 0) is 7, so p = –7 and 4p = –28. Check Check It Out! Example 2b Continued Use your graphing calculator. Step 3 The equation of the parabola is

22 The vertex of a parabola may not always be the origin. Adding or subtracting a value from x or y translates the graph of a parabola. Also notice that the values of p stretch or compress the graph.

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24 Example 3: Graphing Parabolas Step 1 The vertex is (2, –3). Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola Then graph. y + 3 = (x – 2) 2. 1 8 Step 2, so 4p = 8 and p = 2. 1 4p 1 8 =

25 Example 3 Continued Step 4 The focus is (2, –3 + 2), or (2, –1). Step 5 The directrix is a horizontal line y = –3 – 2, or y = –5. Step 3 The graph has a vertical axis of symmetry, with equation x = 2, and opens upward.

26 Step 1 The vertex is (1, 3). Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola. Then graph. Step 2, so 4p = 12 and p = 3. 1 4p 1 12 = Check It Out! Example 3a

27 Step 4 The focus is (1 + 3, 3), or (4, 3). Step 5 The directrix is a vertical line x = 1 – 3, or x = –2. Step 3 The graph has a horizontal axis of symmetry with equation y = 3, and opens right. Check It Out! Example 3a Continued

28 Step 1 The vertex is (8, 4). Find the vertex, value of p axis of symmetry, focus, and directrix of the parabola. Then graph. Check It Out! Example 3b Step 2, so 4p = –2 and p = –. 1 4p 1 2 = – 1 2

29 Step 3 The graph has a vertical axis of symmetry, with equation x = 8, and opens downward. Check It Out! Example 3b Continued Step 4 The focus is or (8, 3.5). Step 5 The directrix is a horizontal line or y = 4.5.

30 Light or sound waves collected by a parabola will be reflected by the curve through the focus of the parabola, as shown in the figure. Waves emitted from the focus will be reflected out parallel to the axis of symmetry of a parabola. This property is used in communications technology.

31 The cross section of a larger parabolic microphone can be modeled by the equation What is the length of the feedhorn? Example 4: Using the Equation of a Parabola x = y 2. 1 132 The equation for the cross section is in the form x = y 2, 1 4p so 4p = 132 and p = 33. The focus should be 33 inches from the vertex of the cross section. Therefore, the feedhorn should be 33 inches long.

32 Check It Out! Example 4 Find the length of the feedhorn for a microphone with a cross section equation x = y 2. 1 44 The equation for the cross section is in the form x = y 2, 1 4p so 4p = 44 and p = 11. The focus should be 11 inches from the vertex of the cross section. Therefore, the feedhorn should be 11 inches long.

33 Lesson Quiz 1. Write an equation for the parabola with focus F(0, 0) and directrix y = 1. 2. Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola y – 2 = (x – 4) 2, 1 12 then graph. vertex: (4, 2); focus: (4,5); directrix: y = –1; p = 3; axis of symmetry: x = 4


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