10 6 Identifying Conic Sections Reminder Solve by

10 -6 Identifying Conic Sections Reminder: Solve by completing the square. 1. x 2 + 6 x = 91 x 2 + 6 x + 9 = 100 (x+3)2 = 100 x = – 13 or 7 Objectives Identify and transform conic functions. Use the method of completing the square to identify and graph conic sections. Holt Algebra 2

10 -6 Identifying Conic Sections Notes: Identify Conic Sections 1. Identify the conic section for each equation A. x + 4 = (y – 2)2 10 B. C. 2. Identify the conic section for each equation. Change to graphing form. Graph each. A. x 2 + y 2 - 16 x + 10 y + 53 = 0 B. 5 x 2 + 20 y 2 + 30 x +40 y – 15 = 0 C. 16 y 2 – 4 x 2 + 32 y - 16 x – 64 = 0 Holt Algebra 2

10 -6 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. Holt Algebra 2

10 -6 Identifying Conic Sections Example 1: Finding the Standard Form of the Equation for a Conic Section by completing the square Identify the comic. Find the standard form of the equation by completing the square. Graph. x 2 + y 2 + 8 x – 10 y – 8 = 0 Rearrange to prepare for completing the square in x and y. x 2 + 8 x + + y 2 – 10 y + Complete both squares. 2 Holt Algebra 2 =8+ +

10 -6 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. Holt Algebra 2

10 -6 Identifying Conic Sections Example 2: Identify and write in Standard Form Identify the comic. Find the standard form of the equation by completing the square. Graph. 5 x 2 + 20 y 2 + 30 x + 40 y – 15 = 0 Rearrange to prepare for completing the square in x and y. 5 x 2 + 30 x + + 20 y 2 + 40 y + = 15 + + Divide everything by 5 and factor 4 from the y terms. (x 2 + 6 x + Holt Algebra 2 )+ 4 (y 2 + 2 y + )=3 + +

10 -6 Identifying Conic Sections Example 2 Continued Complete both squares. 2 2 é é ù 6 2ö ê x 2 + 6 x + æç ö÷ ú + 4 ê y 2 + 2 y + æç ÷ è 2 ø úû êë êë è 2ø (x + 3)2 + 4(y + 1)2 = 16 (x + 3)2 + (y +1 )2 = 16 Holt Algebra 2 4 1 2 ù æ ú = 3 + ç 3 ö÷ + 4 è ø úû æ ö ç 1÷ è ø Factor and simplify.

10 -6 Identifying Conic Sections Example 3: Aviation Application An airplane makes a dive that can be modeled by the equation – 9 x 2 +25 y 2 + 18 x + 50 y – 209 = 0 with dimensions in hundreds of feet. How close to the ground does the airplane pass? The graph of – 9 x 2 +25 y 2 + 18 x +50 y – 209 = 0 is a conic section. Write the equation in standard form. Rearrange to prepare for completing the square in x and y. – 9 x 2 + 18 x + Holt Algebra 2 + 25 y 2 +50 y + = 209 + +

10 -6 Identifying Conic Sections Example 3 Continued Factor – 9 from the x terms, and factor 25 from the y terms. – 9(x 2 – 2 x + ) + 25(y 2 + 2 y + ) = 209 + + Complete both squares. 2 2 2 é é ù ù – 2 æ 2ö æ– 2 2 – 9 ê x 2 – 2 x + æç ö÷ ú + 25 êy 2 + 2 y + æç ö÷ ú= 209 + 9 ç ö÷ + 25 ç ÷ è 2 ø úû êë è 2ø êë è 2 ø úû 25(y + 1)2 – 9(x – 1)2 = 225 Holt Algebra 2 Simplify.

10 -6 Identifying Conic Sections Example 3 Continued (y + 1)2 – 9 (x – 1)2 =1 25 Divide both sides by 225. Because the conic is of the form 2 (y – k) – 2 a 2 (x – h) = 1, 2 b 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. Holt Algebra 2

10 -6 Identifying Conic Sections Notes: Identify Conic Sections 2. Identify the conic section for each equation. Change to graphing form. Graph each. A. x 2 + y 2 - 16 x + 10 y + 53 = 0 B. 5 x 2 + 20 y 2 + 30 x +40 y – 15 = 0 C. 16 y 2 – 4 x 2 + 32 y - 16 x – 64 = 0 Holt Algebra 2

10 -6 Identifying Conic Sections Conic Review: Circles Helpful Hint If the center of the circle is at the origin, the equation simplifies to x 2 + y 2 = r 2. Holt Algebra 2

10 -6 Identifying Conic Sections Conic Review: Ellipses (2 slides) Holt Algebra 2

10 -6 Identifying Conic Sections The standard form of an ellipse centered at (0, 0) depends on whether the major axis is horizontal or vertical. Holt Algebra 2

10 -6 Identifying Conic Sections Conic Review: Hyperbolas (2 slides) Holt Algebra 2

10 -6 Identifying Conic Sections The standard form of the equation of a hyperbola depends on whether the hyperbola’s transverse axis is horizontal or vertical. Holt Algebra 2

10 -6 Identifying Conic Sections Conic Review: Parabolas (2 slides) 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. Holt Algebra 2

10 -6 Identifying Conic Sections Holt Algebra 2

10 -6 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. Holt Algebra 2

10 -6 Identifying Conic Sections Check It Out! Example 3 a Continued (y + 8)2 = 9 x x= 1 9 Factor and simplify. (y + 8)2 Because the conic form is of the 1 form x – h = (y – k)2, it is a 4 p parabola with vertex (0, – 8), and p = 2. 25, and it opens right. The focus is (2. 25, – 8) and directrix is x = – 2. 25. Holt Algebra 2

10 -6 Identifying Conic Sections Check It Out! Example 3 b Find the standard form of the equation by completing the square. Then identify and graph each conic. 16 x 2 + 9 y 2 – 128 x + 108 y + 436 = 0 Rearrange to prepare for completing the square in x and y. 16 x 2 – 128 x + + 9 y 2+ 108 y + = – 436 + + Factor 16 from the x terms, and factor 9 from the y terms. 16(x 2 – 8 x + Holt Algebra 2 )+ 9(y 2 + 12 y + ) = – 436 + +

10 -6 Identifying Conic Sections Check It Out! Example 3 b Continued Complete both squares. 2 2 2 é é ù ù 8 8 æ 12 ö æ ö æ 2 2 16 ê x + 8 x + ç ÷ ú + 9 êy + 12 y + ç ÷ ú = – 436 + 16 ç ÷ + 9 ç ÷ è 2ø è 2 ø úû êë è 2ø êë è 2 ø úû 16(x – 4)2 + 9(y + 6)2 = 144 Factor and simplify. Divide both sides by 144. Holt Algebra 2

10 -6 Identifying Conic Sections Check It Out! Example 3 b Continued Because the conic is of the form (x – h)2 (y – k)2 + = 1, 2 2 b a 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). Holt Algebra 2

10 -6 Identifying Conic Sections Check It Out! Example 4 An airplane makes a dive that can be modeled by the equation – 16 x 2 + 9 y 2 + 96 x + 36 y – 252 = 0, measured in hundreds of feet. How close to the ground does the airplane pass? The graph of – 16 x 2+ 9 y 2 + 96 x +36 y – 252 = 0 is a conic section. Write the equation in standard form. Rearrange to prepare for completing the square in x and y. – 16 x 2 + 96 x + Holt Algebra 2 + 9 y 2 + 36 y + = 252 + +

10 -6 Identifying Conic Sections Check It Out! Example 4 Continued Factor – 16 from the x terms, and factor 9 from the y terms. – 16(x 2 – 6 x + ) + 9(y 2 + 4 y + ) = 252 + + Complete both squares. 2 2 2 é é ù ù – 6 æ 4ö æ– 6 4 – 16 ê x 2 – 6 x + æç ö÷ ú + 9 êy 2 + 4 y + æç ö÷ ú = 252 + – 16 ç ö÷ + 9ç ÷ è 2 ø úû êë è 2ø êë è 2 ø úû – 16(x – 3)2 + 9(y + 2)2 = 144 Holt Algebra 2 Simplify.

10 -6 Identifying Conic Sections Check It Out! Example 4 Continued (y + 2)2 – 16 (x – 3)2 =1 9 Because the conic is of the form Divide both sides by 144. 2 (y – k) – 2 a 2 (x – h) = 1, 2 b 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. Holt Algebra 2

10 -6 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. Holt Algebra 2

10 -6 Identifying Conic Sections Example 2 A: Identifying Conic Sections in General Form Identify the conic section that the equation represents. 4 x 2 – 10 xy + 5 y 2 + 12 x + 20 y = 0 A = 4, B = – 10, C = 5 Identify the values for A, B, and C. B 2 – 4 AC 2 (– 10) – 4(4)(5) Substitute into B 2 – 4 AC. 20 Simplify. Because B 2 – 4 AC > 0, the equation represents a hyperbola. Holt Algebra 2

10 -6 Identifying Conic Sections Example 2 B: Identifying Conic Sections in General Form Identify the conic section that the equation represents. 9 x 2 – 12 xy + 4 y 2 + 6 x – 8 y = 0. A = 9, B = – 12, C = 4 Identify the values for A, B, and C. B 2 – 4 AC 2 (– 12) – 4(9)(4) Substitute into B 2 – 4 AC. 0 Simplify. Because B 2 – 4 AC = 0, the equation represents a parabola. Holt Algebra 2

10 -6 Identifying Conic Sections Example 2 C: Identifying Conic Sections in General Form Identify the conic section that the equation represents. 8 x 2 – 15 xy + 6 y 2 + x – 8 y + 12 = 0 A = 8, B = – 15, C = 6 Identify the values for A, B, and C. B 2 – 4 AC (– 15)2 – 4(8)(6) Substitute into B 2 – 4 AC. 33 Simplify. Because B 2 – 4 AC > 0, the equation represents a hyperbola. Holt Algebra 2

10 -6 Identifying Conic Sections Check It Out! Example 2 a Identify the conic section that the equation represents. 9 x 2 + 9 y 2 – 18 x – 12 y – 50 = 0 A = 9, B = 0, C = 9 2 B – 4 AC Identify the values for A, B, and C. Substitute into B 2 – 4 AC. (0)2 – 4(9)(9) – 324 Simplify. The conic is either a circle or an ellipse. A=C Because B 2 – 4 AC < 0 and A = C, the equation represents a circle. Holt Algebra 2

10 -6 Identifying Conic Sections Check It Out! Example 2 b Identify the conic section that the equation represents. 12 x 2 + 24 xy + 12 y 2 + 25 y = 0 A = 12, B = 24, C = 12 Identify the values for A, B, and C. B 2 – 4 AC Substitute into B 2 – 4 AC. – 242 – 4(12) 0 Simplify. Because B 2 – 4 AC = 0, the equation represents a parabola. Holt Algebra 2