Conic Sections

1
Conic Sections

• Parabola

2
Conic Sections - Parabola
The intersection of a plane with one nappe of the
cone is a parabola.
3
Conic Sections - Parabola
The parabola has the characteristic shape shown
above. A parabola is defined to be the set of
points the same distance from a point and a line.
4
Conic Sections - Parabola
Focus
Directrix
The line is called the directrix and the point is
called the focus.
5
Conic Sections - Parabola
Axis of Symmetry
Focus
Directrix
Vertex
The line perpendicular to the directrix passing
through the focus is the axis of symmetry. The
vertex is the point of intersection of the axis
of symmetry with the parabola.
6
Conic Sections - Parabola
Focus
d1
Directrix
d2
The definition of the parabola is the set of
points the same distance from the focus and
directrix. Therefore, d1 d2 for any point (x,
y) on the parabola.
7
Finding the Focus and Directrix

• Parabola

8
Conic Sections - Parabola
Focus
y ax2
p
Directrix
p
We know that a parabola has a basic equation y
ax2. The vertex is at (0, 0). The distance from
the vertex to the focus and directrix is the
same. Lets call it p.
9
Conic Sections - Parabola
Focus( ?, ?)
y ax2
p
Directrix ???
( 0, 0)
p
Find the point for the focus and the equation of
the directrix if the vertex is at (0, 0).
10
Conic Sections - Parabola
Focus( 0, p)
y ax2
p
Directrix ???
( 0, 0)
p
The focus is p units up from (0, 0), so the focus
is at the point (0, p).
11
Conic Sections - Parabola
Focus( 0, p)
y ax2
p
Directrix ???
( 0, 0)
p
The directrix is a horizontal line p units below
the origin. Find the equation of the directrix.
12
Conic Sections - Parabola
Focus( 0, p)
y ax2
p
Directrixy -p
( 0, 0)
p
The directrix is a horizontal line p units below
the origin or a horizontal line through the point
(0, -p). The equation is y -p.
13
Conic Sections - Parabola
( x, y)
Focus( 0, p)
d1
y ax2
d2
Directrix y -p
( 0, 0)
The definition of the parabola indicates the
distance d1 from any point (x, y) on the curve to
the focus and the distance d2 from the point to
the directrix must be equal.
14
Conic Sections - Parabola
( x, ax2)
Focus( 0, p)
d1
y ax2
d2
Directrix y -p
( 0, 0)
However, the parabola is y ax2. We can
substitute for y in the point (x, y). The point
on the curve is (x, ax2).
15
Conic Sections - Parabola
( x, ax2)
Focus( 0, p)
d1
y ax2
d2
Directrix y -p
( 0, 0)
( ?, ?)
What is the coordinates of the point on the
directrix immediately below the point (x, ax2)?
16
Conic Sections - Parabola
( x, ax2)
Focus( 0, p)
d1
y ax2
d2
Directrix y -p
( 0, 0)
( x, -p)
The x value is the same as the point (x, ax2) and
the y value is on the line y -p, so the point
must be (x, -p).
17
Conic Sections - Parabola
( x, ax2)
Focus( 0, p)
d1
y ax2
d2
Directrix y -p
( 0, 0)
( x, -p)
d1 is the distance from (0, p) to (x, ax2). d2
is the distance from (x, ax2) to (x, -p) and d1
d2. Use the distance formula to solve for p.
18
Conic Sections - Parabola
d1 is the distance from (0, p) to (x, ax2). d2
is the distance from (x, ax2) to (x, -p) and d1
d2. Use the distance formula to solve for p.
d1 d2 You finish the rest.
19
Conic Sections - Parabola
d1 is the distance from (0, p) to (x, ax2). d2
is the distance from (x, ax2) to (x, -p) and d1
d2. Use the distance formula to solve for p.
d1 d2
20
Conic Sections - Parabola
Therefore, the distance p from the vertex to the
focus and the vertex to the directrix is given by
the formula
21
Conic Sections - Parabola
Using transformations, we can shift the parabola
yax2 horizontally and vertically. If the
parabola is shifted h units right and k units up,
the equation would be
The vertex is shifted from (0, 0) to (h, k).
Recall that when a is positive, the graph opens
up. When a is negative, the graph reflects
about the x-axis and opens down.
22
Example 1

• Graph a parabola.
• Find the vertex, focus and directrix.

23
Parabola Example 1
Make a table of values. Graph the function.
Find the vertex, focus, and directrix.
24
Parabola Example 1
The vertex is (-2, -3). Since the parabola opens
up and the axis of symmetry passes through the
vertex, the axis of symmetry is x -2.
25
Parabola Example 1
x y -2 -1 0 1 2 3 4
-3
Make a table of values.
-1
Plot the points on the graph!Use the line of
symmetry to plot the other side of the graph.
26
Parabola Example 1
Find the focus and directrix.
27
Parabola Example 1
The focus and directrix are p units from the
vertex where
The focus and directrix are 2 units from the
vertex.
28
Parabola Example 1
2 Units
Focus (-2, -1) Directrix y -5
29
Latus Rectum

• Parabola

30
Conic Sections - Parabola
The latus rectum is the line segment passing
through the focus, perpendicular to the axis of
symmetry with endpoints on the parabola.
LatusRectum
Focus
y ax2
Vertex(0, 0)
31
Conic Sections - Parabola
In the previous set, we learned that the distance
from the vertex to the focus is 1/(4a).
Therefore, the focus is at
LatusRectum
Focus
y ax2
Vertex(0, 0)
32
Conic Sections - Parabola
Using the axis of symmetry and the y-value of the
focus, the endpoints of the latus rectum must be
LatusRectum
Vertex(0, 0)
y ax2
33
Conic Sections - Parabola
Since the equation of the parabola is y ax2,
substitute for y and solve for x.
34
Conic Sections - Parabola
Replacing x, the endpoints of the latus rectum are
and
LatusRectum
Vertex(0, 0)
y ax2
35
Conic Sections - Parabola
The length of the latus rectum is
LatusRectum
Vertex(0, 0)
y ax2
36
Conic Sections - Parabola
Given the value of a in the quadratic
equation y a (x h)2 k, the length of the
latus rectum is

• An alternate method to graphing a parabola with
  the latus rectum is to
• Plot the vertex and axis of symmetry
• Plot the focus and directrix.
• Use the length of the latus rectum to plot two
  points on the parabola.
• Draw the parabola.

37
Example 2

• Graph a parabola using the vertex, focus, axis of
  symmetry and latus rectum.

38
Parabola Example 2
Find the vertex, axis of symmetry, focus,
directrix, endpoints of the latus rectum and
sketch the graph.
39
Parabola Example 2
The vertex is at (1, 2) with the parabola opening
down.
The focus is 4 units down and the directrix is 4
units up. The length of the latus rectum is
40
Parabola Example 2
Find the vertex, axis of symmetry, focus,
directrix, endpoints of the latus rectum and
sketch the graph.
Directrixy6
Axisx1
V(1, 2)
Focus(1, -2)
Latus Rectum
41
Parabola Example 2
The graph of the parabola
Axisx1
Directrixy6
V(1, 2)
Focus(1, -2)
Latus Rectum
42
x ay2 Parabola

• Graphing and finding the vertex, focus,
  directrix, axis of symmetry and latus rectum.

43
Parabola Graphing x ay2

• Graph x 2y2 by constructing
• a table of values.

x y
-3 -2 -1 0 1 2 3
18
8
2
0
Graph x 2y2 by plotting the points in the table.
2
8
18
44
Parabola Graphing x ay2

• Graph the table of values.

45
Parabola Graphing x ay2

• One could follow a similar proof to show the
  distancefrom the vertex to the focus and
  directrix to be .

Similarly, the length of the latus rectum can be
shown to be .
46
Parabola Graphing x ay2

• Graphing the axis of symmetry, vertex, focus,
  directrix and latus rectum.

Directrix
V(0,0)
Focus
Axis y0
47
x a(y k)2 h

• Graphing and finding the vertex, focus,
  directrix, axis of symmetry and latus rectum.

48
Parabola x a(y k)2 h
We have just seen that a parabola x ay2 opens
to the right when a is positive. When a is
negative, the graph will reflect about the y-axis
and open to the left.
When horizontal and vertical transformations are
applied, a vertical shift of k units and a
horizontal shift of h units will result in the
equation x a(y
k)2 h
Note In both cases of the parabola, the x
always goes with h and the y always goes with k.
49
Example 3

• Graphing and finding the vertex, focus,
  directrix, axis of symmetry and latus rectum.

50
Parabola Example 3
Graph the parabola by finding the vertex, focus,
directrix and length of the latus rectum.
What is the vertex?
Remember that inside the function we always do
the opposite. So the graph moves -1 in the y
direction and -2 in the x direction. The vertex
is (-2, -1)
What is the direction of opening?
The parabola opens to the left since it is x and
a is negative.
51
Parabola Example 3
Graph the parabola by finding the vertex, focus,
directrix and length of the latus rectum.
What is the distance to the focus and directrix?

The distance is
The parabola opens to the left with a vertex of
(-2, -1) and a distance to the focus and
directrix of ½. Begin the sketch of the
parabola.
52
Parabola Example 3
The parabola opens to the left with a vertex of
(-2, -1) and a distance to the focus and
directrix of ½. Begin the sketch of the
parabola.
Vertex?
(-2, -1)
Focus?
(-2.5, -1)
Directrix?
x -1.5
53
Parabola Example 3

• What is the length of the latus rectum?

54
Parabola Example 3
Construct the latus rectum with a length of 2.
Construct the parabola.
Vertex?
(-2, -1)
Focus?
(-2.5, -1)
Directrix?
x -1.5
Latus Rectum?
2
55
Parabola Example 3
The parabola is
Vertex?
(-2, -1)
Focus?
(-2.5, -1)
Directrix?
x -1.5
Latus Rectum?
2
56
Building a Table of Rules

• Parabola

57
Table of Rules - y a(x - h)2 k
a gt 0
a lt 0
x h
Up
Opens
Down
Vertex
(h, k)
(h, k)
(h, k)
Focus
x h
x h
Axis
Directrix
(h, k)
Latus Rectum
x h
58
Table of Rules - x a(y - k)2 h
a gt 0
a lt 0
Right
Opens
Left
y k
(h, k)
Vertex
(h, k)
(h, k)
Focus
y k
y k
Axis
Directrix
(h, k)
y k
Latus Rectum
59
Paraboloid Revolution

• Parabola

60
Paraboloid Revolution

• A paraboloid revolution results from rotating a
  parabola around its axis of symmetry as shown at
  the right.

http//commons.wikimedia.org/wiki/ImageParaboloid
OfRevolution.pngGNU Free Documentation License
61
Paraboloid Revolution

• They are commonly used today in satellite
  technology as well as lighting in motor vehicle
  headlights and flashlights.

62
Paraboloid Revolution

• The focus becomes an important point. As waves
  approach a properly positioned parabolic
  reflector, they reflect back toward the focus.
  Since the distance traveled by all of the waves
  is the same, the wave is concentrated at the
  focus where the receiver is positioned.

63
Example 4 Satellite Receiver

• A satellite dish has a diameter of 8 feet. The
  depth of the dish is 1 foot at the center of the
  dish. Where should the receiver be placed?

8 ft
(?, ?)
1 ft
V(0, 0)
Let the vertex be at (0, 0). What are the
coordinates of a point at the diameter of the
dish?
64
Example 4 Satellite Receiver
8 ft
(4, 1)
1 ft
V(0, 0)
With a vertex of (0, 0), the point on the
diameter would be (4, 1). Fit a parabolic
equation passing through these two points.
y a(x h)2 kSince the vertex is (0, 0), h
and k are 0.y ax2
65
Example 4 Satellite Receiver
8 ft
(4, 1)
1 ft
V(0, 0)
y ax2 The parabola must pass through the point
(4, 1).
1 a(4)2 Solve for a.
1 16a
66
Example 4 Satellite Receiver
8 ft
(4, 1)
1 ft
V(0, 0)
The model for the parabola is
The receiver should be placed at the focus.
Locate the focus of the parabola.
Distance to the focus is
67
Example 4 Satellite Receiver
8 ft
(4, 1)
1 ft
V(0, 0)
The receiver should be placed 4 ft. above the
vertex.
68
Sample Problems

• Parabola

69
Sample Problems

• 1. (y 3)2 12(x -1)

• Find the vertex, focus, directrix, and length of
  the latus rectum.

b. Sketch the graph.
c. Graph using a grapher.
70
Sample Problems

• 1. (y 3)2 12(x -1)

• Find the vertex, focus, directrix, axis of
  symmetry and length of the latus rectum.

Since the y term is squared, solve for x.
71
Sample Problems

• Find the direction of opening and vertex.

The parabola opens to the right with a vertex at
(1, -3).
Find the distance from the vertex to the focus.
72
Sample Problems

• Find the length of the latus rectum.

73
Sample Problems

• b. Sketch the graph given

• The parabola opens to the right.
• The vertex is (1, -3)
• The distance to the focus and directrix is 3.
• The length of the latus rectum is 12.

74
Sample Problems
Vertex (1, -3)Opens RightAxis y -3Focus
(4, -3)Directrix x -2
75
Sample Problems

• 1. (y 3)2 12(x -1)

c. Graph using a grapher.
Solve the equation for y.
Graph as 2 separate equations in the grapher.
76
Sample Problems

• 1. (y 3)2 12(x -1)

c. Graph using a grapher.
77
Sample Problems

• 2. 2x2 8x 3 y 0

• Find the vertex, focus, directrix, axis of
  symmetry and length of the latus rectum.

b. Sketch the graph.
c. Graph using a grapher.
78
Sample Problems

• 2. 2x2 8x 3 y 0

• Find the vertex, focus, directrix, axis of
  symmetry and length of the latus rectum.

Solve for y since x is squared.
y -2x2 - 8x 3
Complete the square.
y -2(x2 4x ) 3
y -2(x2 4x 4 ) 3 8 (-24) is -8.
To balance the side, we must add 8.
y -2(x 2) 2 11
79
Sample Problems

• 2. 2x2 8x 3 y 0

• Find the vertex, focus, directrix, and length of
  the latus rectum.

y -2(x 2) 2 11
Find the direction of opening and the vertex.
The parabola opens down with a vertex at (-2, 11).
Find the distance to the focus and directrix.
80
Sample Problems

• 2. 2x2 8x 3 y 0 or y -2(x 2) 2 11

• Find the vertex, focus, directrix, and length of
  the latus rectum.

x y
Since the latus rectum is quite small, make a
table of values to graph.
-2 11
-1 9
0 3
Graph the table of values and use the axis of
symmetry to plot the other side of the parabola.
1 -7
81
Sample Problems

• 2. 2x2 8x 3 y 0 or y -2(x 2) 2
  11

b. Sketch the graph using the axis of symmetry.
x y
-2 11
-1 9
0 3
1 -7
82
Sample Problems

• 2. 2x2 8x 3 y 0 or y -2(x 2) 2
  11

c. Graph with a grapher.
Solve for y. y -2x2 - 8x 3
83
Sample Problems

• 3. Write the equation of a parabola with vertex
  at (3, 2) and focus at (-1, 2).

Plot the known points.
What can be determined from these points?
84
Sample Problems

• 3. Write the equation of a parabola with vertex
  at (3, 2) and focus at (-1, 2).

The parabola opens the the left and has a model
of x a(y k)2 h.
Can you determine any of the values a, h, or k in
the model?
The vertex is (3, 2) so h is 3 and k is 2.
x a(y 3)2 2
85
Sample Problems

• 3. Write the equation of a parabola with vertex
  at (3, 2) and focus at (-1, 2).

How can we find the value of a?
x a(y 3)2 2
The distance from the vertex to the focus is 4.
86
Sample Problems

• 3. Write the equation of a parabola with vertex
  at (3, 2) and focus at (-1, 2).

How can we find the value of a?
x a(y 3)2 2
The distance from the vertex to the focus is 4.
How can this be used to solve for a?
87
Sample Problems

• 3. Write the equation of a parabola with vertex
  at (3, 2) and focus at (-1, 2).

x a(y 3)2 2
88
Sample Problems

• 3. Write the equation of a parabola with vertex
  at (3, 2) and focus at (-1, 2).

x a(y 3)2 2
Which is the correct value of a?
Since the parabola opens to the left, a must be
negative.
89
Sample Problems

• 4. Write the equation of a parabola with focus at
  (4, 0) and directrix y 2.

Graph the known values.
What can be determined from the graph?
The parabola opens down and has a model ofy
a(x h)2 k
What is the vertex?
90
Sample Problems

• 4. Write the equation of a parabola with focus at
  (4, 0) and directrix y 2.

The vertex must be on the axis of symmetry, the
same distance from the focus and directrix. The
vertex must be the midpoint of the focus and the
intersection of the axis and directrix.
The vertex is (4, 1)
91
Sample Problems

• 4. Write the equation of a parabola with focus at
  (4, 0) and directrix y 2.

The vertex is (4, 1).
How can the value of a be found?
The distance from the focus to the vertex is 1.
Therefore
92
Sample Problems

• 4. Write the equation of a parabola with focus at
  (4, 0) and directrix y 2.

Which value of a?
Since the parabola opens down, a must be negative
and the vertex is (4, 1). Write the model.