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Holt Algebra Identifying Conic Sections Identify and transform conic functions. Use the method of completing the square to identify and graph conic sections. Objectives Reminder: Solve by completing the square. 1. x 2 + 6x = 91 x 2 + 6x + 9 = 100 x = –13 or 7 ( x+3) 2 = 100

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Holt Algebra Identifying Conic Sections 1. Identify the conic section for each equation Notes: Identify Conic Sections A. x + 4 = (y – 2) 2 10 C. B. x 2 + y x + 10y + 53 = 0 2. Identify the conic section for each equation. Change to graphing form. Graph each. 5x y x +40y – 15 = 0 16y 2 – 4x y - 16x – 64 = 0C. B. A.

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Holt Algebra Identifying Conic Sections In Lesson 10-2 through 10-5, you learned about the four conic sections. Recall the equations of conic sections in standard form. In these forms, the characteristics of the conic sections can be identified.

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Holt Algebra Identifying Conic Sections Identify the comic. Find the standard form of the equation by completing the square. Graph. Example 1: Finding the Standard Form of the Equation for a Conic Section by completing the square Rearrange to prepare for completing the square in x and y. x 2 + y 2 + 8x – 10y – 8 = 0 x 2 + 8x + + y 2 – 10y + = Complete both squares. 2

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Holt Algebra Identifying Conic Sections Example 1 Continued (x + 4) 2 + (y – 5) 2 = 49 Factor and simplify. It is a circle with center (–4, 5) and radius 7.

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Holt Algebra Identifying Conic Sections Example 2: Identify and write in Standard Form Rearrange to prepare for completing the square in x and y. 5x y x + 40y – 15 = 0 5x x y y + = Divide everything by 5 and factor 4 from the y terms. (x 2 + 6x + )+ 4 (y 2 + 2y + ) = Identify the comic. Find the standard form of the equation by completing the square. Graph.

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Holt Algebra Identifying Conic Sections Example 2 Continued Complete both squares. (x + 3) 2 + 4(y + 1) 2 = 16 Factor and simplify. 6 x 2 + 6x y 2 + 2y + = x y +1 2

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Holt Algebra Identifying Conic Sections The graph of –9x 2 +25y x +50y – 209 = 0 is a conic section. Write the equation in standard form. An airplane makes a dive that can be modeled by the equation –9x 2 +25y x + 50y – 209 = 0 with dimensions in hundreds of feet. How close to the ground does the airplane pass? Example 3: Aviation Application Rearrange to prepare for completing the square in x and y. –9x x y 2 +50y + =

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Holt Algebra Identifying Conic Sections Example 3 Continued 25(y + 1) 2 – 9(x – 1) 2 = 225 Factor –9 from the x terms, and factor 25 from the y terms. Complete both squares. Simplify. –9(x 2 – 2x + ) + 25(y 2 + 2y + ) = –9 x 2 – 2x y 2 + 2y + = – –

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Holt Algebra Identifying Conic Sections Example 3 Continued Divide both sides by 225. Because the conic is of the form (y – k) 2 a 2 – = 1, (x – h) 2 b 2 it is an a hyperbola with vertical transverse axis length 6 and center (1, –1). The vertices are then (1, 2) and (1, –4). Because distance above ground is always positive, the airplane will be on the upper branch of the hyperbola. The relevant vertex is (1, 2), with y-coordinate 2. The minimum height of the plane is 200 feet. (y + 1) 2 9 – = 1 (x – 1) 2 25

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Holt Algebra Identifying Conic Sections Notes: Identify Conic Sections x 2 + y x + 10y + 53 = 0 2. Identify the conic section for each equation. Change to graphing form. Graph each. 5x y x +40y – 15 = 0 16y 2 – 4x y - 16x – 64 = 0C. B. A.

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Holt Algebra Identifying Conic Sections If the center of the circle is at the origin, the equation simplifies to x 2 + y 2 = r 2. Helpful Hint Conic Review: Circles

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Holt Algebra Identifying Conic Sections Conic Review: Ellipses (2 slides)

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Holt Algebra Identifying Conic Sections The standard form of an ellipse centered at (0, 0) depends on whether the major axis is horizontal or vertical.

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Holt Algebra Identifying Conic Sections Conic Review: Hyperbolas (2 slides)

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Holt Algebra Identifying Conic Sections The standard form of the equation of a hyperbola depends on whether the hyperbolas transverse axis is horizontal or vertical.

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Holt Algebra Identifying Conic Sections 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. Conic Review: Parabolas (2 slides)

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Holt Algebra Identifying Conic Sections

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Holt Algebra Identifying Conic Sections Conic Review: Extra Info The following power-point slides contain extra examples and information. Review of Lesson Objectives: Identify and transform conic functions. Use the method of completing the square to identify and graph conic sections.

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Holt Algebra Identifying Conic Sections (y + 8) 2 = 9x Factor and simplify. Check It Out! Example 3a Continued Because the conic form is of the form x – h = (y – k) 2, it is a parabola with vertex (0, –8), and p = 2.25, and it opens right. The focus is (2.25, –8) and directrix is x = – p x = (y + 8) 2 1 9

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Holt Algebra Identifying Conic Sections Rearrange to prepare for completing the square in x and y. 16x 2 + 9y 2 – 128x + 108y = 0 16x 2 – 128x + + 9y y + = – Factor 16 from the x terms, and factor 9 from the y terms. 16(x 2 – 8x + )+ 9(y y + ) = – Check It Out! Example 3b Find the standard form of the equation by completing the square. Then identify and graph each conic.

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Holt Algebra Identifying Conic Sections Complete both squares. 16(x – 4) 2 + 9(y + 6) 2 = 144 Factor and simplify. Divide both sides by 144. Check It Out! Example 3b Continued 16 x 2 + 8x y y + = –

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Holt Algebra Identifying Conic Sections Because the conic is of the form (x – h) 2 b 2 + = 1, (y – k) 2 a 2 it is an ellipse with center (4, –6), vertical major axis length 8, and minor axis length 6. The vertices are (7, –6) and (1, –6), and the co-vertices are (4, –2) and (4, –10). Check It Out! Example 3b Continued

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Holt Algebra Identifying Conic Sections The graph of –16x 2 + 9y x +36y – 252 = 0 is a conic section. Write the equation in standard form. An airplane makes a dive that can be modeled by the equation –16x 2 + 9y x + 36y – 252 = 0, measured in hundreds of feet. How close to the ground does the airplane pass? Rearrange to prepare for completing the square in x and y. Check It Out! Example 4 –16x x + + 9y y + =

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Holt Algebra Identifying Conic Sections –16(x – 3) 2 + 9(y + 2) 2 = 144 Factor –16 from the x terms, and factor 9 from the y terms. Complete both squares. Simplify. Check It Out! Example 4 Continued –16(x 2 – 6x + ) + 9(y 2 + 4y + ) = –16 x 2 – 6x y 2 + 4y + = – – –

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Holt Algebra Identifying Conic Sections Divide both sides by 144. Because the conic is of the form (y – k) 2 a 2 – = 1, (x – h) 2 b 2 it is an a hyperbola with vertical transverse axis length 8 and center (3, –2). The vertices are (3, 2) and (3, –6). Because distance above ground is always positive, the airplane will be on the upper branch of the hyperbola. The relevant vertex is (3, 2), with y-coordinate 2. The minimum height of the plane is 200 feet. (y + 2) 2 16 – = 1 (x – 3) 2 9 Check It Out! Example 4 Continued

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Holt Algebra Identifying Conic Sections All conic sections can be written in the general form Ax 2 + Bxy + Cy 2 + Dx + Ey+ F = 0. The conic section represented by an equation in general form can be determined by the coefficients.

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Holt Algebra Identifying Conic Sections Identify the conic section that the equation represents. Example 2A: Identifying Conic Sections in General Form 20 Identify the values for A, B, and C. 4x 2 – 10xy + 5y x + 20y = 0 A = 4, B = –10, C = 5 B 2 – 4AC Substitute into B 2 – 4AC. (–10) 2 – 4(4)(5) Simplify. Because B 2 – 4AC > 0, the equation represents a hyperbola.

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Holt Algebra Identifying Conic Sections Identify the conic section that the equation represents. Example 2B: Identifying Conic Sections in General Form 0 Identify the values for A, B, and C. 9x 2 – 12xy + 4y 2 + 6x – 8y = 0. A = 9, B = –12, C = 4 B 2 – 4AC Substitute into B 2 – 4AC. (–12) 2 – 4(9)(4) Simplify. Because B 2 – 4AC = 0, the equation represents a parabola.

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Holt Algebra Identifying Conic Sections Identify the conic section that the equation represents. Example 2C: Identifying Conic Sections in General Form 33 Identify the values for A, B, and C. 8x 2 – 15xy + 6y 2 + x – 8y + 12 = 0 A = 8, B = –15, C = 6 B 2 – 4AC Substitute into B 2 – 4AC. (–15) 2 – 4(8)(6) Simplify. Because B 2 – 4AC > 0, the equation represents a hyperbola.

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Holt Algebra Identifying Conic Sections Identify the conic section that the equation represents. –324 Identify the values for A, B, and C. 9x 2 + 9y 2 – 18x – 12y – 50 = 0 A = 9, B = 0, C = 9 B 2 – 4AC Substitute into B 2 – 4AC. (0) 2 – 4(9)(9) Simplify. The conic is either a circle or an ellipse. Because B 2 – 4AC < 0 and A = C, the equation represents a circle. Check It Out! Example 2a A = C

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Holt Algebra Identifying Conic Sections Identify the conic section that the equation represents. 0 Identify the values for A, B, and C. 12x xy + 12y y = 0 A = 12, B = 24, C = 12 B 2 – 4AC Substitute into B 2 – 4AC. –24 2 – 4(12)(12) Simplify. Because B 2 – 4AC = 0, the equation represents a parabola. Check It Out! Example 2b

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