The Conic Sections Index
The Conics
Translations
Completing the Square
Classifying Conics

Parabola
Ellipse
Circle
Click on a Photo
Hyperbola
Back to Index

A parabola is formed when a plane intersects a cone and the base of that cone

 A Parabola is a set of points equidistant from a fixed point and a fixed line.
 The fixed point is called the focus .
 The fixed line is called the directrix .

Parabola
FOCUS
Directrix

Standard form of the equation of a parabola with vertex (0,0)
•Equation •Focus •Directrix •Axis
•x 2 =4py •(0,p) •y = -p
•y 2 =4px •(p,0) •y = p

To Find p
4p is equal to the term in front of x or y. Then solve for p.
Example: x 2 =24y
4p=24 p=6

Find the Focus and Directrix
Example 1 y = 4x 2 x 2 = ( 1 /
4
)y
4p = 1 /
4 p = 1 /
16
FOCUS
(0, 1 /
16
)
Directrix
Y = 1 /
16

Find the Focus and Directrix
Example 2 x = -3y 2 y 2 = ( -1 /
3
)x
4p = -1 /
3 p = -1 /
12
FOCUS
( -1 /
12
, 0)
Directrix x = 1 /
12

Find the Focus and Directrix
Example 3
(try this one on your own) y = -6x 2
FOCUS
????
Directrix
????

Find the Focus and Directrix
Example 3 y = -6x 2
FOCUS
(0, 1 /
24
)
Directrix y = 1 /
24

Find the Focus and Directrix
Example 4
(try this one on your own) x = 8y 2
FOCUS
????
Directrix
????

Find the Focus and Directrix
Example 4 x = 8y 2
FOCUS
(2, 0)
Directrix x = -2

Now write an equation in standard form for each of the following four parabolas

Example 1
Focus at (-4,0)
Identify equation y 2 =4px p = -4 y 2 = 4(-4)x y 2 = -16x

Example 2
With directrix y = 6
Identify equation x 2 =4py p = -6 x 2 = 4(-6)y x 2 = -24y

Example 3 (Now try this one on your own)
With directrix x = -1 y 2 = 4x

Example 4 (On your own)
Focus at (0,3) x 2 = 12y
Back to Conics

A Circle is formed when a plane intersects a cone parallel to the base of the cone.

x 2  y 2  r 2

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Distance formula used to find the radius
( x
1
 x
2
) 2  ( y
1
 y
2
) 2  r

Example 1
Write the equation of the circle with the point (4,5) on the circle and the origin as it’s center.

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Point (4,5) on the circle and the origin as it’s center.
( x
1
 x
2
) 2  ( y
1
 y
2
) 2  r
(4  0) 2  (5  0) 2
16  25  r
 r
41  r x 2  y 2  41

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Find the intersection points on the graph of the following two equations x 2  y 2  25 y  2 x  2 x 2  (2 x  2) 2  25 x 2  4 x 2  8 x  4  25
(5 x  7)( x  3)  0
(5 x  7)  0 ( x  3)  0
5 x 2
5 x 2
 8 x  4  25
 8 x  21  0 x 
7
5 x   3

Now what??!!??!!??

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Find the intersection points on the graph of the following two equations x 2  y 2  25 y  2 x  2 x 
7
5 x   3
Plug these in for x
.

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Find the intersection points on the graph of the following two equations x 2  y 2  25 x 
7
5 y  2( y 
7
5
)  2
24
5 y  2 x  2 x   3 y  2(  3)  2 y   4

7
5
,
24
5 
  3,  4

Back to Conics

Ellipses
Examples of Ellipses

Ellipses
Horizontal Major Axis

FOCI
(-c,0) & (c,0)
CO-VERTICES
(0,b)& (0,-b)
CENTER (0,0) x 2 a 2
 y 2 b 2

Vertices
(-a,0) & (a,0)
1

Ellipses
Vertical Major Axis

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FOCI
(0,-c) & (0,c)
CO-VERTICES
(b, 0)& (-b,0)
CENTER (0,0) x 2 b 2
 y 2 a 2
 1
Vertices
(0,-a) & (0, a)

Ellipse Notes

Length of major axis = a (vertex & larger #)

Length of minor axis = b (co-vertex & smaller#)

To Find the foci (c) use: c 2 = a 2 b 2

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Ellipse Examples
Find the Foci and Vertices x 2
144 y 2

169
 0 a  vertices a   13 vertices  (0,13),(0,  13) c 2 c 2
 a 2  b 2
 169  144 c 2  25 c   5 foci  (0,5),(0,  5)

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Ellipse Examples
Find the Foci and Vertices x 2
81
 y 2
9
 0 a  vertices a   9 vertices  (9,0),(  9,0) c 2 c 2
 a 2 
 81  9 b 2 c 2  72 c   72 foci  ( 72,0),(  72,0)

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Write an equation of an ellipse whose vertices are (-5,0) & (5,0) and whose co-vertices are (0,-3) & (0,3). Then find the foci.
a   5 a 2 b   3 b 2 x 2
 25
 9
 c 2 c c
2
2 foci
 a 2  b 2
 25  9
 16
 c  4
(4,0),(  4,0)
25 y
9
2
 1

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Write the equation in standard form and then find the foci and vertices.
49 x 2  64 y 2  3136 c 2  a 2  b 2
49 x 2
3136 x 2
64

64 y 2
3136

3136
3136 c 2 c 2
 15
 64  49 c   15
 y 2
49
 1 vert .
 (8,0),(  8,0) foci  ( 15,0),(  15,0)

Back to the Conics

The Hyperbola

Hyperbola Examples

Hyperbola Notes
Horizontal Transverse Axis
Center (0,0)
Asymptotes
Vertices (a,0) &
(-a,0)
Foci (c,0) &
(-c, 0)

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Hyperbola Notes
Horizontal Transverse Axis
Equation x 2 a 2
 y 2 b 2
 1
Foci : c 2  a 2  b 2

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Hyperbola Notes
Horizontal Transverse Axis
To find asymptotes y y 
 b a
 b x x a

Hyperbola Notes
Vertical Transverse Axis
Center (0,0)
Vertices (a,0) &
(-a,0)
Asymptotes
Foci (c,0) &
(-c, 0)

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Hyperbola Notes
Vertical Transverse Axis
Equation y 2 a 2
 x 2 b 2
 1
Foci : c 2  a 2  b 2

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Hyperbola Notes
Vertical Transverse Axis
To find asymptotes y y 
 a b
 a x b x

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Write an equation of the hyperbola with foci (-5,0) & (5,0) and vertices
(-3,0) & (3,0) a = 3 c = 5 c 2  a 2  b 2
5 2  3 2  b 2
25  9  b 2 b 2  16 x 2
9
 y
16
2
 1

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Write an equation of the hyperbola with foci (0,-6) & (0,6) and vertices
(0,-4) & (0,4) a = 4 c = 6 c 2  a 2  b 2
6 2  4 2  b 2
36  16  b 2 b 2  20 y 2
16
 x 2
20
 1
The Conics

Translations
What happens when the conic is NOT centered on
(0,0)?
Back
Next

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Translations
Circle
( x  h ) 2  ( y  k ) 2  r 2
Next

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Translations
Parabola
( y  k )
( x  h )
 4 p ( x  h ) or
 4 p ( y  k )
Next

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Translations
Ellipse
( x  h ) 2 a 2

( y  k ) 2 b 2 or
 1
( x  h ) 2 b 2

( y  k ) 2 a 2
 1
Next

Translations
Hyperbola
( x  h ) 2 a 2

( y  k ) 2 b 2 or
 1
( y  k ) 2 a 2

( x  h ) 2 b 2
 1
Next

Translations
Identify the conic and graph
( x  1) 2  ( y  2) 2  3 2 r= 3 center (1,-2)
Next

Translations
Identify the conic and graph
( x  2) 2
3 2

( y  1) 2
2 2
 1
Next

Translations
Identify the conic and graph
( x  3) 2
1 2

( y  2) 2
3 2
 1 center asymptotes vertices
Next

Translations
Identify the conic and graph
( x  2) 2  4 (  1)( y  3)
Conic center
Back to Index

Completing the Square
Here are the steps for completing the square
Steps
1) Group x 2 + x, y 2 +y move constant
2) Take # in front of x, ÷2, square, add to both sides
3) Repeat Step 2 for y if needed
4) Rewrite as perfect square binomial
Next

Completing the Square
Circle: x 2 +y 2 +10x-6y+18=0 x 2 +10x+____ + y 2 -6y=-18
(x 2 +10x+25) + (y 2 -6y+9)=-18+25+9
(x+5) 2 + (y-3) 2 =16
Center (-5,3) Radius = 4
Next

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Completing the Square
(
Ellipse: x 2 +4y 2 +6x-8y+9=0 x 2 +6x+____ + 4y 2 -8y+____=-9
(x 2 +6x+9) + 4(y 2 -2y+1)=-9+9+4
(x+3) 2 + (y-1) 2 =4 x 
4
3) 2

( y  1) 2
1
 1
C: (-3,1) a=2, b=1
Index

Classifying Conics

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Classifying Conics
Given in General Form
Ax 2  Bxy  Cy 2  Dx  Ey  F  0
B 2  4 AC  0
B 2  4 AC  0
B 2  4 AC  0
Next

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Classifying Conics
Given in General Form
Ax 2  Bxy  Cy 2  Dx  Ey  F  0
B 2  4 AC  0
B 2  4 AC  0
B 2  4 AC  0
Examples

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Classifying Conics
Given in general form, classify the conic
5 x 2  2 y 2
A  5
B  0
C  2
 20 x  4 y  24  0
B 2
0 2
 4 AC
 4(5)(2)
 40
Ellipse
Next

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Classifying Conics
Given in general form, classify the conic y 2  8 x  12 y  0
A  0
B  0
C  1
B 2
0 2
 4 AC
 4(0)(1)
0
Parabola
Next

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Classifying Conics
Given in general form, classify the conic
 24 x 2  18 y 2  18  0
A   24
B  18
C  0
B 2  4 AC
0 2  4(  24)(18)
1728
Hyperbola
Next

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Classifying Conics
Given in general form, classify the conic
16  xy  0
A  0
B   1
C  0
B 2  4 AC
(  1) 2  4(0)(0)
1
Hyperbola
Back to Index

Classifying Conics
Given in General Form
Then
Ellipse
Back
Circle

Classifying Conics
Given in General Form
Then
Back

Classifying Conics
Given in General Form
Then
Hyperbola
Back