10-6 Identifying Conic Sections
Reminder: Solve by completing the square.
1. x2 + 6x = 91
x2 + 6x + 9 = 100
(x+3)2 = 100
x = –13 or 7
Objectives
Identify and transform conic functions.
Use the method of completing the square to
identify and graph conic sections.
Holt Algebra 2
10-6 Identifying Conic Sections
Notes: Identify Conic Sections
1. Identify the conic section for each equation
(y – 2)2
A. x + 4 =
10
B.
C.
2. Identify the conic section for each equation. Change
to graphing form. Graph each.
A. x2 + y2 - 16x + 10y + 53 = 0
B. 5x2 + 20y2 + 30x +40y – 15 = 0
C.
16y2 – 4x2 + 32y - 16x – 64 = 0
Holt Algebra 2
10-6 Identifying Conic Sections
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.
Holt Algebra 2
10-6 Identifying Conic Sections
Example 1: Finding the Standard Form of the Equation for a
Conic Section by completing the square
Identify the comic. Find the standard form of
the equation by completing the square. Graph.
x2 + y2 + 8x – 10y – 8 = 0
Rearrange to prepare for completing the square in x and y.
x2 + 8x +
+ y2 – 10y +
Complete both squares.
2
Holt Algebra 2
=8+
+
10-6 Identifying Conic Sections
Example 1 Continued
2
2
(x + 4) + (y – 5) = 49
Factor and simplify.
It is a circle with center (–4, 5) and radius 7.
Holt Algebra 2
10-6 Identifying Conic Sections
Example 2: Identify and write in Standard Form
Identify the comic. Find the standard form of
the equation by completing the square. Graph.
5x2 + 20y2 + 30x + 40y – 15 = 0
Rearrange to prepare for completing the square in x and y.
5x2 + 30x +
+ 20y2 + 40y +
= 15 +
+
Divide everything by 5 and factor 4 from the y terms.
(x2 + 6x +
Holt Algebra 2
)+ 4 (y2 + 2y +
)=3
+
+
10-6 Identifying Conic Sections
Example 2 Continued
Complete both squares.
2
2



6
2
 x2 + 6x +    + 4  y2 + 2y +  
 2  


 2
(x + 3)2 + 4(y + 1)2 = 16
(x + 3)2 + (y +1 )2 =
16
Holt Algebra 2
4
1
2

 
 = 3 + 3 + 4
 

 
 1
 
Factor and simplify.
10-6 Identifying Conic Sections
Example 3: Aviation Application
An airplane makes a dive that can be modeled
2
2
by the equation –9x +25y + 18x + 50y – 209
= 0 with dimensions in hundreds of feet. How
close to the ground does the airplane pass?
2
2
The graph of –9x +25y + 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.
2
–9x + 18x +
Holt Algebra 2
2
+ 25y +50y +
= 209 +
+
10-6 Identifying Conic Sections
Example 3 Continued
Factor –9 from the x terms, and factor 25 from the y terms.
–9(x2 – 2x +
) + 25(y2 + 2y +
) = 209 +
+
Complete both squares.
2
2




–2
2




–9  x2 – 2x +    + 25 y2 + 2y +    = 209 + 9
 2  


 2  
25(y + 1)2 – 9(x – 1)2 = 225
Holt Algebra 2
Simplify.
2
 2
–2 
+
25
 2
2 
 
 
10-6 Identifying Conic Sections
Example 3 Continued
(y + 1)2
–
9
(x – 1)2
=1
25
Divide both sides by 225.
(y – k)2
(x – h)2
–
= 1,
Because the conic is of the form
2
2
a
b
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.
Holt Algebra 2
10-6 Identifying Conic Sections
Notes: Identify Conic Sections
2. Identify the conic section for each equation.
Change to graphing form. Graph each.
A.
x2 + y2 - 16x + 10y + 53 = 0
B. 5x2 + 20y2 + 30x +40y – 15 = 0
C. 16y2 – 4x2 + 32y - 16x – 64 = 0
Holt Algebra 2
10-6 Identifying Conic Sections
Conic Review: Circles
Helpful Hint
If the center of the circle is at the origin, the equation
simplifies to x2 + y2 = r2.
Holt Algebra 2
10-6 Identifying Conic Sections
Conic Review: Ellipses (2 slides)
Holt Algebra 2
10-6 Identifying Conic Sections
The standard form of an ellipse centered at (0, 0)
depends on whether the major axis is horizontal
or vertical.
Holt Algebra 2
10-6 Identifying Conic Sections
Conic Review: Hyperbolas (2 slides)
Holt Algebra 2
10-6 Identifying Conic Sections
The standard form of the equation of a hyperbola
depends on whether the hyperbola’s transverse axis is
horizontal or vertical.
Holt Algebra 2
10-6 Identifying Conic Sections
Conic Review: Parabolas (2 slides)
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.
Holt Algebra 2
10-6 Identifying Conic Sections
Holt Algebra 2
10-6 Identifying Conic Sections
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.
Holt Algebra 2
10-6 Identifying Conic Sections
Check It Out! Example 3a Continued
(y + 8)2 = 9x
x=
1
9
Factor and simplify.
(y + 8)2
Because the conic form is of the
form x – h = 1 (y – k)2, it is a
4p
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.
Holt Algebra 2
10-6 Identifying Conic Sections
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.
16x2 – 128x +
+ 9y2+ 108y +
= –436 +
+
Factor 16 from the x terms, and factor 9 from the y terms.
16(x2 – 8x +
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)+ 9(y2 + 12y +
) = –436 +
+
10-6 Identifying Conic Sections
Check It Out! Example 3b Continued
Complete both squares.
2
2
2




8
8
 12 
 
12 



2
2
16  x + 8x +    + 9 y + 12y +    = –436 + 16   + 9  
2
 2  

 2

 2  
16(x – 4)2 + 9(y + 6)2 = 144
Factor and simplify.
Divide both sides by
144.
Holt Algebra 2
10-6 Identifying Conic Sections
Check It Out! Example 3b Continued
(x – h)2
(y – k)2
+
= 1,
Because the conic is of the form
2
2
b
a
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).
Holt Algebra 2
10-6 Identifying Conic Sections
Check It Out! Example 4
An airplane makes a dive that can be modeled by
2
2
the equation –16x + 9y + 96x + 36y – 252 = 0,
measured in hundreds of feet. How close to the
ground does the airplane pass?
2
2
The graph of –16x + 9y + 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.
–16x2 + 96x +
Holt Algebra 2
+ 9y2 + 36y +
= 252 +
+
10-6 Identifying Conic Sections
Check It Out! Example 4 Continued
Factor –16 from the x terms, and factor 9 from the y terms.
–16(x2 – 6x +
) + 9(y2 + 4y +
) = 252 +
+
Complete both squares.
2
2
2




–6
 4
–6
4
–16  x2 – 6x +    + 9 y2 + 4y +    = 252 + – 16   + 9 
2 
 2  

 2

 2  
–16(x – 3)2 + 9(y + 2)2 = 144
Holt Algebra 2
Simplify.
10-6 Identifying Conic Sections
Check It Out! Example 4 Continued
(y + 2)2
–
16
(x – 3)2
=1
9
Divide both sides by 144.
(y – k)2
(x – h)2
–
= 1,
Because the conic is of the form
2
2
a
b
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.
Holt Algebra 2
10-6 Identifying Conic Sections
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.
Holt Algebra 2
10-6 Identifying Conic Sections
Example 2A: Identifying Conic Sections in General
Form
Identify the conic section that the equation
represents.
4x2 – 10xy + 5y2 + 12x + 20y = 0
A = 4, B = –10, C = 5 Identify the values for A, B, and C.
B2 – 4AC
2
(–10) – 4(4)(5)
Substitute into B2 – 4AC.
20
Simplify.
Because B2 – 4AC > 0, the equation represents a
hyperbola.
Holt Algebra 2
10-6 Identifying Conic Sections
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
2
(–12) – 4(9)(4)
Substitute into B2 – 4AC.
0
Simplify.
Because B2 – 4AC = 0, the equation represents a
parabola.
Holt Algebra 2
10-6 Identifying Conic Sections
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.
33
Simplify.
Because B2 – 4AC > 0, the equation represents a
hyperbola.
Holt Algebra 2
10-6 Identifying Conic Sections
Check It Out! Example 2a
Identify the conic section that the equation
represents.
2
2
9x + 9y – 18x – 12y – 50 = 0
A = 9, B = 0, C = 9
2
B – 4AC
Identify the values for A, B, and C.
Substitute into B2 – 4AC.
(0)2 – 4(9)(9)
–324
Simplify. The conic is either a
circle or an ellipse.
A=C
Because B2 – 4AC < 0 and A = C, the equation
represents a circle.
Holt Algebra 2
10-6 Identifying Conic Sections
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.
2
B – 4AC
2
Substitute into B – 4AC.
–242 – 4(12)(12)
0
Simplify.
Because B2 – 4AC = 0, the equation represents a
parabola.
Holt Algebra 2