Advanced design for trusses of steel and concrete-filled tubular sections.pdf
How to Identify Conic Sections.pdf
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Transcript of How to Identify Conic Sections.pdf
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10-6 Identifying Conic Sections
Battle of the CST’s
Lesson Presentation
Lesson Quiz
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Identify and transform conic functions.
Use the method of completing the square to identify and graph conic sections.
Objectives
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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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Identify the conic section that each equation represents.
Example 1: Identifying Conic Sections in Standard
Form
A.
This equation is of the same form as a parabola with a horizontal axis of symmetry.
x + 4 = (y – 2)
2
10
B.
This equation is of the same form as a hyperbola with a horizontal transverse axis.
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Identify the conic section that each equation represents.
Example 1: Identifying Conic Sections in Standard
Form
This equation is of the same form as a circle.
C.
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Identify the conic section that each equation represents.
This equation is of the same form as a hyperbola with a vertical transverse axis.
Check for understanding
This equation is of the same form as a circle.
a. x2 + (y + 14)
2 = 11
2
– = 1 (y – 6)
2
22
(x – 1)2
212
b.
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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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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.
4x2 – 10xy + 5y
2 + 12x + 20y = 0
A = 4, B = –10, C = 5
B2 – 4AC
Substitute into B2 – 4AC. (–10)
2 – 4(4)(5)
Simplify.
Because B2 – 4AC > 0, the equation represents a
hyperbola.
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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.
9x2 – 12xy + 4y
2 + 6x – 8y = 0.
A = 9, B = –12, C = 4
B2 – 4AC
Substitute into B2 – 4AC. (–12)
2 – 4(9)(4)
Simplify.
Because B2 – 4AC = 0, the equation represents a
parabola.
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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.
8x2 – 15xy + 6y
2 + x – 8y + 12 = 0
A = 8, B = –15, C = 6
B2 – 4AC
Substitute into B2 – 4AC. (–15)
2 – 4(8)(6)
Simplify.
Because B2 – 4AC > 0, the equation represents a
hyperbola.
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Identify the conic section that the equation represents.
–324
Identify the values for A, B, and C.
9x2 + 9y
2 – 18x – 12y – 50 = 0
A = 9, B = 0, C = 9
B2 – 4AC Substitute into B
2 – 4AC.
(0)2 – 4(9)(9)
Simplify. The conic is either a
circle or an ellipse.
Because B2 – 4AC < 0 and A = C, the equation
represents a circle.
Check for understanding
A = C
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Identify the conic section that the equation represents.
0
Identify the values for A, B, and C.
12x2 + 24xy + 12y
2 + 25y = 0
A = 12, B = 24, C = 12
B2 – 4AC Substitute into B
2 – 4AC.
–242 – 4(12)(12)
Simplify.
Because B2 – 4AC = 0, the equation represents a
parabola.
Check for understanding
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You must factor out the leading coefficient of x2
and y2 before completing the square.
Remember!
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.
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Find the standard form of the equation by completing the square. Then identify and graph each conic.
Example 3A: Finding the Standard Form of the
Equation for a Conic Section
Rearrange to prepare for completing the square in x and y.
x2 + y
2 + 8x – 10y – 8 = 0
x2 + 8x + + y
2 – 10y + = 8 + +
Complete both squares. 2
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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 = r
2,
it is a circle with center (–4, 5) and radius 7.
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Example 3B: Finding the Standard Form of the
Equation for a Conic Section
Rearrange to prepare for completing the square in x and y.
5x2 + 20y
2 + 30x + 40y – 15 = 0
5x2 + 30x + + 20y
2 + 40y + = 15 + +
Factor 5 from the x terms, and factor 20 from the y terms.
5(x2 + 6x + )+ 20(y
2 + 2y + ) = 15 + +
Find the standard form of the equation by completing the square. Then identify and graph each conic.
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Example 3B Continued
Complete both squares.
5(x + 3)2 + 20(y + 1)
2 = 80 Factor and simplify.
Divide both sides by
80.
6 5 x2 + 6x + + 20 y2 + 2y + = 15 + 5 + 20
2
2
2
2
6
2
2
2
2 2
1
16 4
x + 3 2 y +1 2
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Because the conic is of the form (x – h)2
a2 + = 1,
(y – k)2
b2
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).
Example 3B Continued
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Find the standard form of the equation by completing the square. Then identify and graph each conic.
Rearrange to prepare for completing the square in y.
y2 – 9x + 16y + 64 = 0
y2 + 16y + = 9x – 64 +
Check for understanding
Add , or 64, to both sides to complete the square.
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(y + 8)2 = 9x Factor and simplify.
Check for understanding 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 = –2.25.
1
4p
x = (y + 8)2
1
9
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Rearrange to prepare for completing the square in x and y.
16x2 + 9y
2 – 128x + 108y + 436 = 0
16x2 – 128x + + 9y
2+ 108y + = –436 + +
Factor 16 from the x terms, and factor 9 from the y terms.
16(x2 – 8x + )+ 9(y
2 + 12y + ) = –436 + +
Check for understanding
Find the standard form of the equation by completing the square. Then identify and graph each conic.
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Complete both squares.
16(x – 4)2 + 9(y + 6)
2 = 144 Factor and simplify.
Divide both sides by
144.
Check for understanding
16 x2 + 8x + + 9 y2 + 12y + = –436 + 16 + 9 8
2
12
2
2
8
2
2 2 12
2
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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 for understanding
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The graph of –9x2 +25y
2 + 18x +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
2 + 18x + 50y – 209
= 0 with dimensions in hundreds of feet. How close to the ground does the airplane pass?
Example 4: Aviation Application
Rearrange to prepare for completing the square in x and y.
–9x2 + 18x + + 25y
2 +50y + = 209 + +
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Example 4 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(x2 – 2x + ) + 25(y
2 + 2y + ) = 209 + +
–9 x2 – 2x + + 25 y2 + 2y + = 209 + 9 + 25 –2
2
2
2
2
–2
2
2 2 2
2
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Example 4 Continued
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.
(y + 1)2
9 – = 1
(x – 1)2
25
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The graph of –16x2+ 9y
2 + 96x +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
2 + 96x + 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 for understanding
–16x2 + 96x + + 9y
2 + 36y + = 252 + +
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–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 for understanding Continued
–16(x2 – 6x + ) + 9(y
2 + 4y + ) = 252 + +
–16 x2 – 6x + + 9 y2 + 4y + = 252 + – 16 + 9 –6
2
4
2
2
–6
2
2 2 4
2
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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.
(y + 2)2
16 – = 1
(x – 3)2
9
Check for understanding
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10-6 Identifying Conic Sections
Day 2
Battle of the CST’s
Review HW Q’s
Work on today’s HW