Objectives Identify and transform conic functions. Reminder: Solve by completing the square. 1. x2 + 6x = 91 x2 + 6x + 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.

Notes: Identify Conic Sections 1. Identify the conic section for each equation x + 4 = (y – 2)2 10 B. A. C. 2. Identify the conic section for each equation. Change to graphing form. Graph each. A. x2 + y2 - 16x + 10y + 53 = 0 B. 5x2 + 20y2 + 30x +40y – 15 = 0 C. 16y2 – 4x2 + 32y - 16x – 64 = 0

In Lesson 10-2 through 10-5, you learned about the four 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.

Rearrange to prepare for completing the square in x and y. 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. x2 + y2 + 8x – 10y – 8 = 0 Rearrange to prepare for completing the square in x and y. x2 + 8x + + y2 – 10y + = 8 + + Complete both squares. 2

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

Example 2: Identify and write in Standard Form Identify the comic. Find the standard form of the equation by completing the square. Graph. 5x2 + 20y2 + 30x + 40y – 15 = 0 Rearrange to prepare for completing the square in x and y. 5x2 + 30x + + 20y2 + 40y + = 15 + + Divide everything by 5 and factor 4 from the y terms. (x2 + 6x + )+ 4 (y2 + 2y + ) = 3 + +

( ) x + 3 2 y +1 2 + = 1 16 4 Example 2 Continued Complete both squares. 6 x2 + 6x + + 4 y2 + 2y + = 3 + 3 + 4 1 2 ö é ÷ ø ê ë ù ú û æ ç è (x + 3)2 + 4(y + 1)2 = 16 Factor and simplify. ( ) 1 16 4 x + 3 2 y +1 2 + =

Example 3: Aviation Application An airplane makes a dive that can be modeled by the equation –9x2 +25y2 + 18x + 50y – 209 = 0 with dimensions in hundreds of feet. How close to the ground does the airplane pass? The graph of –9x2 +25y2 + 18x +50y – 209 = 0 is a conic section. Write the equation in standard form. Rearrange to prepare for completing the square in x and y. –9x2 + 18x + + 25y2 +50y + = 209 + +

Factor –9 from the x terms, and factor 25 from the y terms. Example 3 Continued Factor –9 from the x terms, and factor 25 from the y terms. –9(x2 – 2x + ) + 25(y2 + 2y + ) = 209 + + Complete both squares. –9 x2 – 2x + + 25 y2 + 2y + = 209 + 9 + 25 –2 2 ö é ÷ ø ê ë ù ú û æ ç è 25(y + 1)2 – 9(x – 1)2 = 225 Simplify.

Because the conic is of the form Example 3 Continued (y + 1)2 9 – = 1 (x – 1)2 25 Divide both sides by 225. Because the conic is of the form (y – k)2 a2 – = 1, (x – h)2 b2 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.

Notes: Identify Conic Sections 2. Identify the conic section for each equation. Change to graphing form. Graph each. A. x2 + y2 - 16x + 10y + 53 = 0 B. 5x2 + 20y2 + 30x +40y – 15 = 0 C. 16y2 – 4x2 + 32y - 16x – 64 = 0

Conic Review: Circles Helpful Hint If the center of the circle is at the origin, the equation simplifies to x2 + y2 = r2. Helpful Hint 12

Conic Review: Ellipses (2 slides) 13

The standard form of an ellipse centered at (0, 0) depends on whether the major axis is horizontal or vertical. 14

Conic Review: Hyperbolas (2 slides) 15

The standard form of the equation of a hyperbola depends on whether the hyperbola’s transverse axis is horizontal or vertical. 16

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. 17

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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. 19

Check It Out! Example 3a Continued (y + 8)2 = 9x Factor and simplify. x = (y + 8)2 1 9 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 = –2.25. 1 4p

Check It Out! Example 3b Find the standard form of the equation by completing the square. Then identify and graph each conic. 16x2 + 9y2 – 128x + 108y + 436 = 0 Rearrange to prepare for completing the square in x and y. 16x2 – 128x + + 9y2+ 108y + = –436 + + Factor 16 from the x terms, and factor 9 from the y terms. 16(x2 – 8x + )+ 9(y2 + 12y + ) = –436 + +

Check It Out! Example 3b Continued Complete both squares. 16 x2 + 8x + + 9 y2 + 12y + = –436 + 16 + 9 8 2 12 ö é ÷ ø ê ë ù ú û æ ç è 16(x – 4)2 + 9(y + 6)2 = 144 Factor and simplify. Divide both sides by 144.

Check It Out! Example 3b Continued Because the conic is of the form (x – h)2 b2 + = 1, (y – k)2 a2 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 4 An airplane makes a dive that can be modeled by the equation –16x2 + 9y2 + 96x + 36y – 252 = 0, measured in hundreds of feet. How close to the ground does the airplane pass? The graph of –16x2+ 9y2 + 96x +36y – 252 = 0 is a conic section. Write the equation in standard form. Rearrange to prepare for completing the square in x and y. –16x2 + 96x + + 9y2 + 36y + = 252 + +

Check It Out! Example 4 Continued Factor –16 from the x terms, and factor 9 from the y terms. –16(x2 – 6x + ) + 9(y2 + 4y + ) = 252 + + Complete both squares. –16 x2 – 6x + + 9 y2 + 4y + = 252 + – 16 + 9 –6 2 4 ö é ÷ ø ê ë ù ú û æ ç è –16(x – 3)2 + 9(y + 2)2 = 144 Simplify.

Check It Out! Example 4 Continued (y + 2)2 16 – = 1 (x – 3)2 9 Divide both sides by 144. Because the conic is of the form (y – k)2 a2 – = 1, (x – h)2 b2 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.

All conic sections can be written in the general form Ax2 + Bxy + Cy2 + Dx + Ey+ F = 0. The conic section represented by an equation in general form can be determined by the coefficients.

Example 2A: Identifying Conic Sections in General Form Identify the conic section that the equation represents. 4x2 – 10xy + 5y2 + 12x + 20y = 0 A = 4, B = –10, C = 5 Identify the values for A, B, and C. B2 – 4AC (–10)2 – 4(4)(5) Substitute into B2 – 4AC. 20 Simplify. Because B2 – 4AC > 0, the equation represents a hyperbola.

Example 2B: Identifying Conic Sections in General Form Identify the conic section that the equation represents. 9x2 – 12xy + 4y2 + 6x – 8y = 0. A = 9, B = –12, C = 4 Identify the values for A, B, and C. B2 – 4AC (–12)2 – 4(9)(4) Substitute into B2 – 4AC. Simplify. Because B2 – 4AC = 0, the equation represents a parabola.

Example 2C: Identifying Conic Sections in General Form Identify the conic section that the equation represents. 8x2 – 15xy + 6y2 + x – 8y + 12 = 0 A = 8, B = –15, C = 6 Identify the values for A, B, and C. B2 – 4AC (–15)2 – 4(8)(6) Substitute into B2 – 4AC. 33 Simplify. Because B2 – 4AC > 0, the equation represents a hyperbola.

Check It Out! Example 2a Identify the conic section that the equation represents. 9x2 + 9y2 – 18x – 12y – 50 = 0 A = 9, B = 0, C = 9 Identify the values for A, B, and C. Substitute into B2 – 4AC. B2 – 4AC (0)2 – 4(9)(9) Simplify. The conic is either a circle or an ellipse. –324 A = C Because B2 – 4AC < 0 and A = C, the equation represents a circle.

Check It Out! Example 2b Identify the conic section that the equation represents. 12x2 + 24xy + 12y2 + 25y = 0 A = 12, B = 24, C = 12 Identify the values for A, B, and C. B2 – 4AC Substitute into B2 – 4AC. –242 – 4(12)(12) Simplify. Because B2 – 4AC = 0, the equation represents a parabola.