How to Identify Conic Sections.pdf

10-6 Identifying Conic Sections

Battle of the CST’s

Lesson Presentation

Lesson Quiz

Identify and transform conic functions.

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

Objectives

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.

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.

Maya

Highlight


Example 1: Identifying Conic Sections in Standard

Form

This equation is of the same form as a circle.

C.


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.

Maya

Highlight

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.

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.


Example 2B: Identifying Conic Sections in General

Form

0


9x2 – 12xy + 4y

2 + 6x – 8y = 0.

A = 9, B = –12, C = 4

B2 – 4AC


2 – 4(9)(4)

Simplify.

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

parabola.


Example 2C: Identifying Conic Sections in General

Form

33


8x2 – 15xy + 6y

2 + x – 8y + 12 = 0

A = 8, B = –15, C = 6

B2 – 4AC


2 – 4(8)(6)

Simplify.

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

hyperbola.


–324


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.


A = C


0


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.


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.

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

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.

Example 3B: Finding the Standard Form of the

Equation for a Conic Section


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 + +


Example 3B Continued

Complete both squares.

5(x + 3)2 + 20(y + 1)


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

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


Rearrange to prepare for completing the square in y.

y2 – 9x + 16y + 64 = 0

y2 + 16y + = 9x – 64 +


Add , or 64, to both sides to complete the square.

(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


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 + +



Maya

Highlight

Maya

Highlight

Maya

Highlight


16(x – 4)2 + 9(y + 6)


Divide both sides by

144.


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

2

12

2

2

8

2

2 2 12

2

Maya

Highlight

Maya

Highlight

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).


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


–9x2 + 18x + + 25y

2 +50y + = 209 + +

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.


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

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

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?



–16x2 + 96x + + 9y

2 + 36y + = 252 + +

–16(x – 3)2 + 9(y + 2)2 = 144

Factor –16 from the x terms, and factor 9 from the y terms.


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

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


10-6 Identifying Conic Sections

Day 2

Battle of the CST’s

Review HW Q’s

Work on today’s HW

How to Identify Conic Sections.pdf

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