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# Chapter 12 - PowerPoint PPT Presentation

Chapter 12 . 12-5 Parabolas. Objectives . Write the standard equation of a parabola and its axis of symmetry. Graph a parabola and identify its focus, directrix , and axis of symmetry. parabolas.

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### Chapter 12

12-5 Parabolas

Write the standard equation of a parabola and its axis of symmetry.

Graph a parabola and identify its focus, directrix, and axis of symmetry.

• 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.

• 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.

• Use the Distance Formula to find the equation of a parabola with focus F(2, 4) and directrixy = –4.

• Sol.

• PF = PD

• Distance Formula.

solution of a Parabola

Substitute (2, 4) for (x1, y1) and (x, –4) for (x2, y2).

Simplify.

(x – 2)2 + (y – 4)2 = (y + 4)2

Square both sides.

(x – 2)2 + y2 – 8y + 16 = y2 + 8y + 16

solution of a Parabola

Subtract y2 and 16 from both sides.

• (x – 2)2 – 8y = 8y

(x – 2)2 = 16y

Solve for y

Check It Out! of a Parabola Example 1

• Use the Distance Formula to find the equation of a parabola with focus F(0, 4) and directrixy = –4.

Parabolas of a Parabola

• 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.

Example 2A: Writing Equations of Parabolas of a Parabola

• Write the equation in standard form for the parabola.

solution of a Parabola

• Step 1 Because the axis of symmetry is vertical and the parabola opens downward, the equation is in the form

• Step 2 The distance from the focus (0, –5) to the vertex (0, 0), is 5, so p = –5 and 4p = –20.

y = 1/4p x2 with p < 0.

solution of a Parabola

• Step 3 The equation of the parabola is y = – 1/20 x2

Example of a Parabola

• Write the equation in standard form for the parabola.

• vertex (0, 0), directrixx = –6

• Solution:

• 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.

solution of a Parabola

• Step 2 Because the directrix is x = –6, p = 6 and 4p = 24.

• Step 3 The equation of the parabola is

x = 1/24 y2

Check It Out! of a Parabola Example 2a

• Write the equation in standard form for the parabola.

• vertex (0, 0), directrixx = 1.25

Parabolas of a Parabola

• 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.

Standard form of a Parabola

Example 3: Graphing Parabolas of a Parabola

• Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola y + 3= 1/8 (x –2)2.Then graph.

• Solution:

• Step 1 The vertex is (2,–3).

• Step 2 1/4p=1/8 , so 4p = 8 and p = 2.

solution of a Parabola

• Step 3 The graph has a vertical axis of symmetry, with equation x = 2, and opens upward.

• 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.

solution of a Parabola

Check it-out Example of a Parabola

• Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola. Then graph.

Parabolas of a Parabola

• 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.

x of a Parabola= y2,

x = y2.

1

1

132

4p

The equation for the cross section is in the form

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.

Example 4: Using the Equation of a Parabola

• The cross section of a larger parabolic microphone can be modeled by the equation What is the length of

• the feedhorn?

• Solution:

Student Guided Practice of a Parabola

• Do problems 2-8 in your book page 849

Homework of a Parabola

• Do problems 14-21 in your book page 849

Closure of a Parabola

• Today we learned about parabolas

• Next class we are going to learn Identifying conic sections