holt algebra 2 10-6 identifying conic sections identify and transform conic functions. use the...

Holt Algebra 2

10-6 Identifying Conic Sections

Identify and transform conic functions.

Use the method of completing the square to identify and graph conic sections.

Objectives

Reminder: Solve by completing the square.

1. x2 + 6x = 91

x2 + 6x + 9 = 100

x = –13 or 7 (x+3)2 = 100

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1. Identify the conic section for each equation

Notes: Identify Conic Sections

A. x + 4 = (y – 2)2

10

C.

B.

x2 + y2 - 16x + 10y + 53 = 0

2. Identify the conic section for each equation. Change to graphing form. Graph each.

5x2 + 20y2 + 30x +40y – 15 = 0

16y2 – 4x2 + 32y - 16x – 64 = 0C.

B.

A.

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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 comic. Find the standard form of the equation by completing the square. Graph.

Example 1: Finding the Standard Form of the Equation for a Conic Section by completing the square

Rearrange to prepare for completing the square in x and y.

x2 + y2 + 8x – 10y – 8 = 0

x2 + 8x + + y2 – 10y + = 8 + +

Complete both squares.2

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Example 1 Continued

(x + 4)2 + (y – 5)2 = 49 Factor and simplify.

It is a circle with center (–4, 5) and radius 7.

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Example 2: Identify and write in Standard Form


5x2 + 20y2 + 30x + 40y – 15 = 0

5x2 + 30x + + 20y2 + 40y + = 15 + +

Divide everything by 5 and factor 4 from the y terms.

(x2 + 6x + )+ 4 (y2 + 2y + ) = 3 + +

Identify the comic. Find the standard form of the equation by completing the square. Graph.

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10-6 Identifying Conic SectionsExample 2 Continued

Complete both squares.

(x + 3)2 + 4(y + 1)2 = 16 Factor and simplify.

6 x2 + 6x + + 4 y2 + 2y + = 3 + 3 + 4 1

2

2

2

2

2 2

1

16 4

x + 3 2 y +1 2

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The graph of –9x2 +25y2 + 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 –9x2 +25y2 + 18x + 50y – 209 = 0 with dimensions in hundreds of feet. How close to the ground does the airplane pass?

Example 3: Aviation Application


–9x2 + 18x + + 25y2 +50y + = 209 + +

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Example 3 Continued

25(y + 1)2 – 9(x – 1)2 = 225

Factor –9 from the x terms, and factor 25 from the y terms.


Simplify.

–9(x2 – 2x + ) + 25(y2 + 2y + ) = 209 + +

–9 x2 – 2x + + 25 y2 + 2y + = 209 + 9 + 25–2

222

2

–2

2

2 2 22

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Example 3 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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10-6 Identifying Conic SectionsNotes: Identify Conic Sections

x2 + y2 - 16x + 10y + 53 = 0

2. Identify the conic section for each equation. Change to graphing form. Graph each.

5x2 + 20y2 + 30x +40y – 15 = 0

16y2 – 4x2 + 32y - 16x – 64 = 0C.

B.

A.

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If the center of the circle is at the origin, the equation simplifies to x2 + y2 = r2.

Helpful Hint

Conic Review: Circles

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Conic Review: Ellipses (2 slides)

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

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Conic Review: Hyperbolas (2 slides)

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

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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.

Conic Review: Parabolas (2 slides)

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Holt Algebra 2


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.

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(y + 8)2 = 9x Factor and simplify.

Check It Out! Example 3a 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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16x2 + 9y2 – 128x + 108y + 436 = 0

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

Find the standard form of the equation by completing the square. Then identify and graph each conic.

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16(x – 4)2 + 9(y + 6)2 = 144 Factor and simplify.


Check It Out! Example 3b Continued

16 x2 + 8x + + 9 y2 + 12y + = –436 + 16 + 98

2122

2

8

2

2 2 122

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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 It Out! Example 3b Continued

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The graph of –16x2+ 9y2 + 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 –16x2 + 9y2 + 96x + 36y – 252 = 0, measured in hundreds of feet. How close to the ground does the airplane pass?


Check It Out! Example 4

–16x2 + 96x + + 9y2 + 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.


Simplify.

Check It Out! Example 4 Continued

–16(x2 – 6x + ) + 9(y2 + 4y + ) = 252 + +

–16 x2 – 6x + + 9 y2 + 4y + = 252 + – 16 + 9–6

242

2

–6

2

2 2 42

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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 It Out! Example 4 Continued

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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.

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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 + 5y2 + 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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Example 2B: Identifying Conic Sections in General Form

0


9x2 – 12xy + 4y2 + 6x – 8y = 0.

A = 9, B = –12, C = 4

B2 – 4AC


Simplify.

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

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Example 2C: Identifying Conic Sections in General Form

33


8x2 – 15xy + 6y2 + x – 8y + 12 = 0

A = 8, B = –15, C = 6

B2 – 4AC


Simplify.

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

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–324


9x2 + 9y2 – 18x – 12y – 50 = 0

A = 9, B = 0, C = 9

B2 – 4AC Substitute into B2 – 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 It Out! Example 2a

A = C

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0


12x2 + 24xy + 12y2 + 25y = 0

A = 12, B = 24, C = 12

B2 – 4AC Substitute into B2 – 4AC.

–242 – 4(12)(12)

Simplify.

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

Check It Out! Example 2b

holt algebra 2 10-6 identifying conic sections identify and transform conic functions. use the...

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