Upcoming SlideShare
×

# 3.3 conic sections circles

1,200 views
1,041 views

Published on

Published in: Technology, Education
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
1,200
On SlideShare
0
From Embeds
0
Number of Embeds
84
Actions
Shares
0
0
0
Likes
0
Embeds 0
No embeds

No notes for slide

### 3.3 conic sections circles

1. 1. Conic Sections
2. 2. Conic Sections One way to study a solid is to slice it open.
3. 3. Conic Sections One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area.
4. 4. Conic Sections A right circular cone One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area. Conic sections are the borders of the cross sectional areas of a right circular cone as shown.
5. 5. Conic Sections A right circular cone and conic sections (wikipedia “Conic Sections”) One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area. Conic sections are the borders of the cross sectional areas of a right circular cone as shown.
6. 6. Conic Sections A Horizontal Section A right circular cone and conic sections (wikipedia “Conic Sections”) One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area. Conic sections are the borders of the cross sectional areas of a right circular cone as shown.
7. 7. Conic Sections A Horizontal Section A right circular cone and conic sections (wikipedia “Conic Sections”) One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area. Conic sections are the borders of the cross sectional areas of a right circular cone as shown.
8. 8. Conic Sections A Moderately Tilted Section A right circular cone and conic sections (wikipedia “Conic Sections”) One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area. Conic sections are the borders of the cross sectional areas of a right circular cone as shown.
9. 9. Conic Sections A Moderately Tilted Section A right circular cone and conic sections (wikipedia “Conic Sections”) One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area. Conic sections are the borders of the cross sectional areas of a right circular cone as shown.
10. 10. Conic Sections A Horizontal Section A Moderately Tilted Section A right circular cone and conic sections (wikipedia “Conic Sections”) One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area. Conic sections are the borders of the cross sectional areas of a right circular cone as shown. Circles and ellipsis are enclosed.
11. 11. Conic Sections A right circular cone and conic sections (wikipedia “Conic Sections”) A Parallel–Section One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area. Conic sections are the borders of the cross sectional areas of a right circular cone as shown.
12. 12. Conic Sections A right circular cone and conic sections (wikipedia “Conic Sections”) A Parallel–Section One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area. Conic sections are the borders of the cross sectional areas of a right circular cone as shown.
13. 13. Conic Sections A right circular cone and conic sections (wikipedia “Conic Sections”) One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area. Conic sections are the borders of the cross sectional areas of a right circular cone as shown. An Cut-away Section
14. 14. Conic Sections A right circular cone and conic sections (wikipedia “Conic Sections”) One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area. Conic sections are the borders of the cross sectional areas of a right circular cone as shown. An Cut-away Section
15. 15. Conic Sections A right circular cone and conic sections (wikipedia “Conic Sections”) An Cut-away Section One way to study a solid is to slice it open. The exposed area of the sliced solid is called a cross sectional area. Conic sections are the borders of the cross sectional areas of a right circular cone as shown. Parabolas and hyperbolas are open. A Horizontal Section A Moderately Tilted Section Circles and ellipsis are enclosed. A Parallel–Section
16. 16. We summarize the four types of conics sections here. Circles Ellipses Parabolas Hyperbolas Conic Sections
17. 17. Conic Sections
18. 18. Conic Sections Conic sections are the cross sections of right circular cones. There are four different types of curves: • circles • ellipses
19. 19. Conic Sections Conic sections are the cross sections of right circular cones. There are four different types of curves: • circles • ellipses • parabolas
20. 20. Conic Sections Conic sections are the cross sections of right circular cones. There are four different types of curves: • circles • ellipses • parabolas • hyperbolas Where as straight lines are the graphs of 1st degree equations Ax + By = C, conic sections are the graphs of 2nd degree equations in x and y.
21. 21. Conic Sections Conic sections are the cross sections of right circular cones. There are four different types of curves: • circles • ellipses • parabolas • hyperbolas Where as straight lines are the graphs of 1st degree equations Ax + By = C, conic sections are the graphs of 2nd degree equations in x and y. In particular, the conic sections that are parallel to the axes (not tilted) have equations of the form Ax2 + By2 + Cx + Dy = E, where A, B, C, D, and E are numbers (not both A and B equal to 0).
22. 22. Conic Sections Conic sections are the cross sections of right circular cones. There are four different types of curves: • circles • ellipses • parabolas • hyperbolas Where as straight lines are the graphs of 1st degree equations Ax + By = C, conic sections are the graphs of 2nd degree equations in x and y. In particular, the conic sections that are parallel to the axes (not tilted) have equations of the form Ax2 + By2 + Cx + Dy = E, where A, B, C, D, and E are numbers (not both A and B equal to 0). We are to match these 2nd degree equations with the different conic sections.
23. 23. Conic Sections Conic sections are the cross sections of right circular cones. There are four different types of curves: • circles • ellipses • parabolas • hyperbolas Where as straight lines are the graphs of 1st degree equations Ax + By = C, conic sections are the graphs of 2nd degree equations in x and y. In particular, the conic sections that are parallel to the axes (not tilted) have equations of the form Ax2 + By2 + Cx + Dy = E, where A, B, C, D, and E are numbers (not both A and B equal to 0). We are to match these 2nd degree equations with the different conic sections. The algebraic technique that enable us to sort out which equation corresponds to which conic section is called "completing the square".
24. 24. Conic Sections Conic sections are the cross sections of right circular cones. There are four different types of curves: • circles • ellipses • parabolas • hyperbolas Where as straight lines are the graphs of 1st degree equations Ax + By = C, conic sections are the graphs of 2nd degree equations in x and y. In particular, the conic sections that are parallel to the axes (not tilted) have equations of the form Ax2 + By2 + Cx + Dy = E, where A, B, C, D, and E are numbers (not both A and B equal to 0). We are to match these 2nd degree equations with the different conic sections. The algebraic technique that enable us to sort out which equation corresponds to which conic section is called "completing the square". We start with the Distance Formula.
25. 25. The Distance Formula: Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane, the distance r between P and Q is: Conic Sections
26. 26. The Distance Formula: Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane, the distance r between P and Q is: r = √(y2 – y1)2 + (x2 – x1)2 Conic Sections
27. 27. The Distance Formula: Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane, the distance r between P and Q is: r = √(y2 – y1)2 + (x2 – x1)2 = √Δy2 + Δx2 Conic Sections
28. 28. The Distance Formula: Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane, the distance r between P and Q is: r = √(y2 – y1)2 + (x2 – x1)2 = √Δy2 + Δx2 where Conic Sections Δy = the difference between the y's = y2 – y1
29. 29. The Distance Formula: Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane, the distance r between P and Q is: r = √(y2 – y1)2 + (x2 – x1)2 = √Δy2 + Δx2 where Conic Sections Δy = the difference between the y's = y2 – y1 Δx = the difference between the x's = x2 – x1
30. 30. The Distance Formula: Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane, the distance r between P and Q is: r = √(y2 – y1)2 + (x2 – x1)2 = √Δy2 + Δx2 where Example A. Find the distance between (2, –1) and (–2, 2). Conic Sections Δy = the difference between the y's = y2 – y1 Δx = the difference between the x's = x2 – x1
31. 31. The Distance Formula: Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane, the distance r between P and Q is: r = √(y2 – y1)2 + (x2 – x1)2 = √Δy2 + Δx2 where Example A. Find the distance between (2, –1) and (–2, 2). Δy = (–1) – (2) = –3 Δy=-3 Conic Sections Δy = the difference between the y's = y2 – y1 Δx = the difference between the x's = x2 – x1
32. 32. The Distance Formula: Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane, the distance r between P and Q is: r = √(y2 – y1)2 + (x2 – x1)2 = √Δy2 + Δx2 where Example A. Find the distance between (2, –1) and (–2, 2). Δy = (–1) – (2) = –3 Δx = (2) – (–2) = 4 Δy=-3 Δx=4 Conic Sections Δy = the difference between the y's = y2 – y1 Δx = the difference between the x's = x2 – x1
33. 33. The Distance Formula: Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane, the distance r between P and Q is: r = √(y2 – y1)2 + (x2 – x1)2 = √Δy2 + Δx2 where Example A. Find the distance between (2, –1) and (–2, 2). Δy = (–1) – (2) = –3 Δx = (2) – (–2) = 4 r = √(–3)2 + 42 = √25 = 5 Δy=-3 Δx=4 r=5 Conic Sections Δy = the difference between the y's = y2 – y1 Δx = the difference between the x's = x2 – x1
34. 34. The Distance Formula: Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane, the distance r between P and Q is: r = √(y2 – y1)2 + (x2 – x1)2 = √Δy2 + Δx2 where Example A. Find the distance between (2, –1) and (–2, 2). Δy = (–1) – (2) = –3 Δx = (2) – (–2) = 4 r = √(–3)2 + 42 = √25 = 5 Δy=-3 Δx=4 r=5 Conic Sections The geometric definition of all four types of conic sections are distance relations between points. Δy = the difference between the y's = y2 – y1 Δx = the difference between the x's = x2 – x1
35. 35. The Distance Formula: Given two points P = (x1, y1) and Q = (x2, y2) in the xy-plane, the distance r between P and Q is: r = √(y2 – y1)2 + (x2 – x1)2 = √Δy2 + Δx2 where Example A. Find the distance between (2, –1) and (–2, 2). Δy = (–1) – (2) = –3 Δx = (2) – (–2) = 4 r = √(–3)2 + 42 = √25 = 5 Δy=-3 Δx=4 r=5 Conic Sections The geometric definition of all four types of conic sections are distance relations between points. We start with the circles. Δy = the difference between the y's = y2 – y1 Δx = the difference between the x's = x2 – x1
36. 36. Circles A circle is the set of all the points that have equal distance r, called the radius, to a fixed point C which is called the center.
37. 37. r r Circles A circle is the set of all the points that have equal distance r, called the radius, to a fixed point C which is called the center. C
38. 38. r r The radius and the center completely determine the circle. Circles center A circle is the set of all the points that have equal distance r, called the radius, to a fixed point C which is called the center.
39. 39. r The radius and the center completely determine the circle. Circles Let (h, k) be the center of a circle and r be the radius. (h, k) A circle is the set of all the points that have equal distance r, called the radius, to a fixed point C which is called the center.
40. 40. r The radius and the center completely determine the circle. Circles (x, y) Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center is r. (h, k) A circle is the set of all the points that have equal distance r, called the radius, to a fixed point C which is called the center.
41. 41. r The radius and the center completely determine the circle. Circles (x, y) Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center is r. Hence, (h, k) r = √ (x – h)2 + (y – k)2 A circle is the set of all the points that have equal distance r, called the radius, to a fixed point C which is called the center.
42. 42. r The radius and the center completely determine the circle. Circles (x, y) Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center is r. Hence, (h, k) r = √ (x – h)2 + (y – k)2 or r2 = (x – h)2 + (y – k)2 A circle is the set of all the points that have equal distance r, called the radius, to a fixed point C which is called the center.
43. 43. r The radius and the center completely determine the circle. Circles (x, y) Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center is r. Hence, (h, k) r = √ (x – h)2 + (y – k)2 or r2 = (x – h)2 + (y – k)2 This is called the standard form of circles. A circle is the set of all the points that have equal distance r, called the radius, to a fixed point C which is called the center.
44. 44. r The radius and the center completely determine the circle. Circles (x, y) Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center is r. Hence, (h, k) r = √ (x – h)2 + (y – k)2 or r2 = (x – h)2 + (y – k)2 This is called the standard form of circles. Given an equation of this form, we can easily identify the center and the radius. A circle is the set of all the points that have equal distance r, called the radius, to a fixed point C which is called the center.
45. 45. r2 = (x – h)2 + (y – k)2 Circles
46. 46. r2 = (x – h)2 + (y – k)2 must be “ – ” Circles
47. 47. r2 = (x – h)2 + (y – k)2 r is the radius must be “ – ” Circles
48. 48. r2 = (x – h)2 + (y – k)2 r is the radius must be “ – ” (h, k) is the center Circles
49. 49. r2 = (x – h)2 + (y – k)2 r is the radius must be “ – ” (h, k) is the center Circles Example B. Write the equation of the circle as shown.
50. 50. r2 = (x – h)2 + (y – k)2 r is the radius must be “ – ” (h, k) is the center Circles Example B. Write the equation of the circle as shown. The center is (–1, 3) and the radius is 5. (–1, 3)
51. 51. r2 = (x – h)2 + (y – k)2 r is the radius must be “ – ” (h, k) is the center Circles Example B. Write the equation of the circle as shown. The center is (–1, 3) and the radius is 5. Hence the equation is: 52 = (x – (–1))2 + (y – 3)2 (–1, 3)
52. 52. r2 = (x – h)2 + (y – k)2 r is the radius must be “ – ” (h, k) is the center Circles Example B. Write the equation of the circle as shown. The center is (–1, 3) and the radius is 5. Hence the equation is: 52 = (x – (–1))2 + (y – 3)2 or 25 = (x + 1)2 + (y – 3 )2 (–1, 3)
53. 53. Example C. Identify the center and the radius of 16 = (x – 3)2 + (y + 2)2 . Label the top, bottom, left and right most points. Graph it. Circles
54. 54. Example C. Identify the center and the radius of 16 = (x – 3)2 + (y + 2)2 . Label the top, bottom, left and right most points. Graph it. Put 16 = (x – 3)2 + (y + 2)2 into the standard form: 42 = (x – 3)2 + (y – (–2))2 Circles
55. 55. Example C. Identify the center and the radius of 16 = (x – 3)2 + (y + 2)2 . Label the top, bottom, left and right most points. Graph it. Put 16 = (x – 3)2 + (y + 2)2 into the standard form: 42 = (x – 3)2 + (y – (–2))2 Hence r = 4, center = (3, –2) Circles
56. 56. Example C. Identify the center and the radius of 16 = (x – 3)2 + (y + 2)2 . Label the top, bottom, left and right most points. Graph it. Put 16 = (x – 3)2 + (y + 2)2 into the standard form: 42 = (x – 3)2 + (y – (–2))2 Hence r = 4, center = (3, –2) (3,-2) Circles
57. 57. Example C. Identify the center and the radius of 16 = (x – 3)2 + (y + 2)2 . Label the top, bottom, left and right most points. Graph it. Put 16 = (x – 3)2 + (y + 2)2 into the standard form: 42 = (x – 3)2 + (y – (–2))2 Hence r = 4, center = (3, –2) (3,-2) Circles (3, 2) (3,-6)
58. 58. Example C. Identify the center and the radius of 16 = (x – 3)2 + (y + 2)2 . Label the top, bottom, left and right most points. Graph it. Put 16 = (x – 3)2 + (y + 2)2 into the standard form: 42 = (x – 3)2 + (y – (–2))2 Hence r = 4, center = (3, –2) (3,-2) Circles (3, 2) (–1,-2) (7,-2) (3,-6)
59. 59. Example C. Identify the center and the radius of 16 = (x – 3)2 + (y + 2)2 . Label the top, bottom, left and right most points. Graph it. Put 16 = (x – 3)2 + (y + 2)2 into the standard form: 42 = (x – 3)2 + (y – (–2))2 Hence r = 4, center = (3, –2) (3,-2) Circles (3, 2) (–1,-2) (7,-2) (3,-6)
60. 60. Example C. Identify the center and the radius of 16 = (x – 3)2 + (y + 2)2 . Label the top, bottom, left and right most points. Graph it. Put 16 = (x – 3)2 + (y + 2)2 into the standard form: 42 = (x – 3)2 + (y – (–2))2 Hence r = 4, center = (3, –2) Circles When equations are not in the standard form, we have to rearrange them into the standard form. We do this by "completing the square". (3,-2) (3, 2) (–1,-2) (7,-2) (3,-6)
61. 61. Example C. Identify the center and the radius of 16 = (x – 3)2 + (y + 2)2 . Label the top, bottom, left and right most points. Graph it. Put 16 = (x – 3)2 + (y + 2)2 into the standard form: 42 = (x – 3)2 + (y – (–2))2 Hence r = 4, center = (3, –2) Circles When equations are not in the standard form, we have to rearrange them into the standard form. We do this by "completing the square". To complete the square means to add a number to an expression so the sum is a perfect square. (3,-2) (3, 2) (–1,-2) (7,-2) (3,-6)
62. 62. Example C. Identify the center and the radius of 16 = (x – 3)2 + (y + 2)2 . Label the top, bottom, left and right most points. Graph it. Put 16 = (x – 3)2 + (y + 2)2 into the standard form: 42 = (x – 3)2 + (y – (–2))2 Hence r = 4, center = (3, –2) Circles When equations are not in the standard form, we have to rearrange them into the standard form. We do this by "completing the square". To complete the square means to add a number to an expression so the sum is a perfect square. This procedure is the main technique in dealing with 2nd degree equations. (3,-2) (3, 2) (–1,-2) (7,-2) (3,-6)
63. 63. (Completing the Square) Circles
64. 64. (Completing the Square) If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, Circles
65. 65. (Completing the Square) If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is the perfect square (x + b/2)2 . Circles
66. 66. (Completing the Square) If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is the perfect square (x + b/2)2 . Circles Example D. Fill in the blank to make a perfect square. a. x2 – 6x + (–6/2)2
67. 67. (Completing the Square) If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is the perfect square (x + b/2)2 . Circles Example D. Fill in the blank to make a perfect square. a. x2 – 6x + (–6/2)2
68. 68. (Completing the Square) If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is the perfect square (x + b/2)2 . Circles Example D. Fill in the blank to make a perfect square. a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2
69. 69. (Completing the Square) If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is the perfect square (x + b/2)2 . Circles Example D. Fill in the blank to make a perfect square. a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2 b. y2 + 12y + (12/2)2
70. 70. (Completing the Square) If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is the perfect square (x + b/2)2 . Circles Example D. Fill in the blank to make a perfect square. a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2 b. y2 + 12y + (12/2)2
71. 71. (Completing the Square) If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is the perfect square (x + b/2)2 . Circles Example D. Fill in the blank to make a perfect square. a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2 b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2
72. 72. (Completing the Square) If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is the perfect square (x + b/2)2 . Circles Example D. Fill in the blank to make a perfect square. a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2 b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2 The following are the steps in putting a 2nd degree equation into the standard form.
73. 73. (Completing the Square) If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is the perfect square (x + b/2)2 . Circles Example D. Fill in the blank to make a perfect square. a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2 b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2 The following are the steps in putting a 2nd degree equation into the standard form. 1. Group the x2 and the x-terms together, group the y2 and y terms together, and move the number term the the other side of the equation.
74. 74. (Completing the Square) If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is the perfect square (x + b/2)2 . Circles Example D. Fill in the blank to make a perfect square. a. x2 – 6x + (–6/2)2 = x2 – 6x + 9 = (x – 3)2 b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = ( y + 6)2 The following are the steps in putting a 2nd degree equation into the standard form. 1. Group the x2 and the x-terms together, group the y2 and y terms together, and move the number term the the other side of the equation. 2. Complete the square for the x-terms and for the y-terms. Make sure add the necessary numbers to both sides.
75. 75. Example E. Use completing the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom, left and right most points. Graph it. Circles
76. 76. Example E. Use completing the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom, left and right most points. Graph it. We use completing the square to put the equation into the standard form: Circles
77. 77. Example E. Use completing the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom, left and right most points. Graph it. We use completing the square to put the equation into the standard form: x2 – 6x + + y2 + 12y + = –36 Circles
78. 78. Example E. Use completing the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom, left and right most points. Graph it. We use completing the square to put the equation into the standard form: x2 – 6x + + y2 + 12y + = –36 ;complete squares x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 Circles
79. 79. Example E. Use completing the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom, left and right most points. Graph it. We use completing the square to put the equation into the standard form: x2 – 6x + + y2 + 12y + = –36 ;complete squares x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 Circles
80. 80. Example E. Use completing the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom, left and right most points. Graph it. We use completing the square to put the equation into the standard form: x2 – 6x + + y2 + 12y + = –36 ;complete squares x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 ( x – 3 )2 + (y + 6)2 = 9 Circles
81. 81. Example E. Use completing the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom, left and right most points. Graph it. We use completing the square to put the equation into the standard form: x2 – 6x + + y2 + 12y + = –36 ;complete squares x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 ( x – 3 )2 + (y + 6)2 = 9 ( x – 3 )2 + (y + 6)2 = 32 Circles
82. 82. Example E. Use completing the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom, left and right most points. Graph it. We use completing the square to put the equation into the standard form: x2 – 6x + + y2 + 12y + = –36 ;complete squares x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 ( x – 3 )2 + (y + 6)2 = 9 ( x – 3 )2 + (y + 6)2 = 32 Hence the center is (3 , –6), and radius is 3. Circles
83. 83. Example E. Use completing the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom, left and right most points. Graph it. We use completing the square to put the equation into the standard form: x2 – 6x + + y2 + 12y + = –36 ;complete squares x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 ( x – 3 )2 + (y + 6)2 = 9 ( x – 3 )2 + (y + 6)2 = 32 Hence the center is (3 , –6), and radius is 3. Circles (3, –6)
84. 84. Example E. Use completing the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom, left and right most points. Graph it. We use completing the square to put the equation into the standard form: x2 – 6x + + y2 + 12y + = –36 ;complete squares x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 ( x – 3 )2 + (y + 6)2 = 9 ( x – 3 )2 + (y + 6)2 = 32 Hence the center is (3 , –6), and radius is 3. Circles (3, –6) (3, –3) (3, –9) (6, –6)(0, –6)
85. 85. Example E. Use completing the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Find the top, bottom, left and right most points. Graph it. We use completing the square to put the equation into the standard form: x2 – 6x + + y2 + 12y + = –36 ;complete squares x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 ( x – 3 )2 + (y + 6)2 = 9 ( x – 3 )2 + (y + 6)2 = 32 Hence the center is (3 , –6), and radius is 3. Circles (3, –6) (3, –3) (3, –9) (6, –6)(0, –6)
86. 86. Conic Sections
87. 87. Conic Sections
88. 88. Conic Sections
89. 89. Conic Sections 21. x2 + y2 = 4 Use the completing the square method to write the equation in the standard form of the circles. Where is the center? What is the radius? Draw the circle and label the four cardinal points on the circle. 22. x2 + y2 = 16 23. x2 – 4x + y2 = 0 24. x2 + y2 + 2y = 24 25. x2 – 2x + y2 + 4y = –4 26. x2 – 6x + y2 + 4y = 3 27. x2 + 12y + y2 – 8x = – 3 28. x2 + y2 – 10x + 2y = –1 29. x2 – 18y + y2 – 8x = 3 30. x2 + y2 – 14x + 2y = 14
90. 90. Conic Sections
91. 91. Conic Sections