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  1. Identifying Conic Sections 10-6 Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2

  2. Warm Up Solve by completing the square. x = –13 or 7 1. x2 + 6x = 91 x = –9 or 5 2. 2x2 + 8x –90 = 0

  3. Objectives Identify and transform conic functions. Use the method of completing the square to identify and graph conic sections.

  4. 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.

  5. (y – 2)2 10 x + 4= Example 1: Identifying Conic Sections in Standard Form Identify the conic section that each equation represents. A. This equation is of the same form as a parabola with a horizontal axis of symmetry. B. This equation is of the same form as a hyperbola with a horizontal transverse axis.

  6. Example 1: Identifying Conic Sections in Standard Form Identify the conic section that each equation represents. C. This equation is of the same form as a circle.

  7. (y – 6)2 (x – 1)2 22 212 –= 1 Check It Out! Example 1 Identify the conic section that each equation represents. a. x2 + (y + 14)2 = 112 This equation is of the same form as a circle. b. This equation is of the same form as a hyperbola with a vertical transverse axis.

  8. 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.

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

  10. 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 Substitute into B2– 4AC. (–12)2 – 4(9)(4) Simplify. 0 Because B2 – 4AC = 0, the equation represents a parabola.

  11. 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. Simplify. 33 Because B2 – 4AC > 0, the equation represents a hyperbola.

  12. 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.

  13. 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. 0 Because B2 – 4AC = 0, the equation represents a parabola.

  14. Remember! You must factor out the leading coefficient of x2 and y2 before completing the square. If you are given the equation of a conic in standard form, you can write the equation in general form by expanding the binomials. If you are given the general form of a conic section, you can use the method of completing the square from Lesson 5-4 to write the equation in standard form.

  15. x2 + 8x + + y2 – 10y + = 8 + + 2 Example 3A: Finding the Standard Form of the Equation for a Conic Section Find the standard form of the equation by completing the square. Then identify and graph each conic. x2 + y2 + 8x – 10y – 8 = 0 Rearrange to prepare for completing the square in x and y. Complete both squares.

  16. Example 3A Continued (x + 4)2 + (y –5)2 = 49 Factor and simplify. Because the conic is of the form (x – h)2 + (y – k)2 = r2, it is a circle with center (–4, 5) and radius 7.

  17. 5x2 + 30x + + 20y2 + 40y + = 15 + + 5(x2 + 6x + )+ 20(y2 + 2y + ) = 15 + + Example 3B: Finding the Standard Form of the Equation for a Conic Section Find the standard form of the equation by completing the square. Then identify and graph each conic. 5x2 + 20y2 + 30x + 40y – 15 = 0 Rearrange to prepare for completing the square in x and y. Factor 5 from the x terms, and factor 20 from the y terms.

  18. é é ù ù 2 2 6 2 6 2 2 ö æ æ ö æ ö ö æ 5 x2 + 6x + + 20 y2 + 2y + = 15 + 5 + 20 ê ê ú ú ÷ ç ç ÷ ç ÷ ÷ ç 2 2 2 è 2 ø ø è ø ê ê ú è ú ø è ë ë û û ( ) ( ) x + 3 2y +1 2 + = 1 16 4 Example 3B Continued Complete both squares. 5(x + 3)2 + 20(y + 1)2 = 80 Factor and simplify. Divide both sides by 80.

  19. Because the conic is of the form it is an ellipse with center (–3, –1), horizontal major axis length 8, and minor axis length 4. The co-vertices are (–3, –3) and (–3, 1), and the vertices are (–7, –1) and (1, –1). (x – h)2 (y – k)2 a2 b2 + = 1, Example 3B Continued

  20. y2 + 16y + = 9x – 64 + Add , or 64, to both sides to complete the square. Check It Out! Example 3a Find the standard form of the equation by completing the square. Then identify and graph each conic. y2 – 9x + 16y + 64 = 0 Rearrange to prepare for completing the square in y.

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

  22. 16x2 – 128x + + 9y2+ 108y + = –436 + + 16(x2 – 8x + )+ 9(y2 + 12y + ) = –436 + + 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. Factor 16 from the x terms, and factor 9 from the y terms.

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

  24. Because the conic is of the form 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). (x – h)2 (y – k)2 b2 a2 + = 1, Check It Out! Example 3b Continued

  25. –9x2 + 18x + + 25y2 +50y + = 209 + + Example 4: 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.

  26. –9(x2 – 2x + ) + 25(y2 + 2y + )= 209 + + é é ù ù 2 2 –2 2 –2 2 ö æ æ 2 ö æ ö ö æ –9 x2 – 2x + + 25 y2 + 2y + = 209 + 9 + 25 ê ê ú ú ÷ ç ç ÷ ç ÷ ÷ ç 2 2 2 2 è ø ø è ê ø è ê ú ú ø è ë ë û û Example 4 Continued Factor –9 from the x terms, and factor 25 from the y terms. Complete both squares. 25(y + 1)2 – 9(x – 1)2= 225 Simplify.

  27. Because the conic is of the form 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 – k)2 (x – h)2 (y + 1)2 (x –1)2 a2 b2 9 25 – = 1, – = 1 Example 4 Continued Divide both sides by 225.

  28. –16x2 + 96x + + 9y2 + 36y + = 252 + + 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.

  29. –16(x2 – 6x + ) + 9(y2 + 4y + )= 252 + + é é ù ù 2 2 –6 2 –6 4 ö æ æ 4 ö æ ö ö æ –16 x2 – 6x + + 9 y2 + 4y + = 252 + – 16 + 9 ê ê ú ú ÷ ç ç ÷ ç ÷ ÷ ç 2 2 2 2 è ø ø è ê ø è ê ú ú ø è ë ë û û Check It Out! Example 4 Continued Factor –16 from the x terms, and factor 9 from the y terms. Complete both squares. –16(x – 3)2 + 9(y + 2)2= 144 Simplify.

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

  31. Lesson Quiz: Part I Identify the conic section that each equation represents. parabola 1. 2x2 – 8xy + 8y2 + 11x – 5y = 0 ellipse 2. 4(x – 1)2 = 100 – 25(y – 2)2 3. 2x2 – 6xy + 3y2 + 12x + 18y –2 = 0 hyperbola

  32. ; ellipse Lesson Quiz: Part II 3. Find the standard form of 4x2 + y2 + 8x – 8y + 16 = 0 by completing the square. Then identify and graph the conic.