Transcript Conic Sections

Conic Sections
Parabola
Conic Sections - Parabola
The intersection of a
plane with one nappe
of the cone is a
parabola.
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”.
Conic Sections - Parabola
Focus
Directrix
The line is called the directrix and the point is called
the focus.
Conic Sections - Parabola
Focus
Vertex
Axis of
Symmetry
Directrix
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.
Conic Sections - Parabola
Focus
d1
d2
Directrix
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.
Finding the Focus
and Directrix
Parabola
Conic Sections - Parabola
Focus
y = ax2
p
p
Directrix
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. Let’s call it p.
Conic Sections - Parabola
Focus
( ?, ?)
y = ax2
p
Directrix ???
p
( 0, 0)
Find the point for the focus and the equation of the
directrix if the vertex is at (0, 0).
Conic Sections - Parabola
Focus
( 0, p)
y = ax2
p
Directrix ???
p
( 0, 0)
The focus is p units up from (0, 0), so the focus is at
the point (0, p).
Conic Sections - Parabola
Focus
( 0, p)
y = ax2
p
Directrix ???
p
( 0, 0)
The directrix is a horizontal line p units below the
origin. Find the equation of the directrix.
Conic Sections - Parabola
Focus
( 0, p)
y = ax2
p
Directrix
y = -p
p
( 0, 0)
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.
Conic Sections - Parabola
Focus
( 0, p)
( x, y)
d1
y = ax2
Directrix
y = -p
( 0, 0)
d2
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.
Conic Sections - Parabola
Focus
( 0, p)
( x, ax2)
d1
y = ax2
Directrix
y = -p
( 0, 0)
d2
However, the parabola is y = ax2. We can substitute
for y in the point (x, y). The point on the curve is
(x, ax2).
Conic Sections - Parabola
Focus
( 0, p)
( x, ax2)
d1
y = ax2
Directrix
y = -p
( 0, 0)
d2
( ?, ?)
What is the coordinates of the point on the directrix
immediately below the point (x, ax2)?
Conic Sections - Parabola
Focus
( 0, p)
( x, ax2)
d1
y = ax2
Directrix
y = -p
( 0, 0)
d2
( 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).
Conic Sections - Parabola
Focus
( 0, p)
( x, ax2)
d1
y = ax2
Directrix
y = -p
( 0, 0)
d2
( 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.
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.
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
( x  0) 2  ( ax 2  p ) 2 
( x  x ) 2  ( ax 2  p ) 2
( x ) 2  ( ax 2  p ) 2  ( ax 2  p ) 2
x 2  a 2 x 4  2 ax 2 p  p 2  a 2 x 4  2 ax 2 p  p 2
x 2  4 ax 2 p
1  4 ap
p
1
4a
Conic Sections - Parabola
Therefore, the distance p from the vertex to the focus
and the vertex to the directrix is given by the formula
p
1
4a
or p  1 /(4 a )
Conic Sections - Parabola
Using transformations, we can shift the parabola y=ax2
horizontally and vertically. If the parabola is shifted h
units right and k units up, the equation would be
y  a( x  h)  k
2
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.
Example 1
Graph a parabola.
Find the vertex, focus and
directrix.
Parabola – Example 1
Make a table of values. Graph the function. Find the
vertex, focus, and directrix.
y   x2
1
8
2
3
Parabola – Example 1
y   x2
1
2
3
8
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.
Parabola – Example 1
y   x2
1
2
3
8
Make a table of values.
Plot the points on the graph!
Use the line of symmetry to
plot the other side of the
graph.
x
y
-2
-3
-1
2 7
0
2 1
1
1 7
2
-1
3
1
4
8
2
8
8
11
2
Parabola – Example 1
Find the focus and directrix.
Parabola – Example 1
y   x2
1
2
3
8
The focus and directrix are “p” units from the vertex
1
where p 
4a
p
1
 8
4 1
1

2
1
2
The focus and directrix are 2 units from the vertex.
Parabola – Example 1
2 Units
Focus: (-2, -1) Directrix: y = -5
Latus Rectum
Parabola
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.
Focus
Latus
Rectum
y = ax2
Vertex
(0, 0)
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
1 

at 0,
 4a 


Focus
Latus
Rectum
y = ax2
Vertex
(0, 0)
Conic Sections - Parabola
Using the axis of symmetry and the y-value of the focus,
1 


x
,
the endpoints of the latus rectum must be 

4
a


Latus
Rectum
1 

  x, 4 a 


Vertex
(0, 0)
1 

 x, 4a 


y = ax2
Conic Sections - Parabola
Since the equation of the parabola is y = ax2, substitute
1
4a
for y and solve for x.
y  ax 2
1
 ax 2
4a
1
4a
2
x
 x2
1
2a
Conic Sections - Parabola
Replacing x, the endpoints of the latus rectum are
 1 1 
  2a , 4a 


and
 1 1 
 2a , 4a 


Latus
Rectum
 1 1 
  2a , 4a 


Vertex
(0, 0)
 1 1 
 2a , 4a 


y = ax2
Conic Sections - Parabola
The length of the latus rectum is
1  1 
1
1
2
1
 



 
2a  2a 
2a 2a
2a
a
Latus
Rectum
 1 1 
  2a , 4a 


Vertex
(0, 0)
 1 1 
 2a , 4a 


y = ax2
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
1
a
An alternate method to graphing a parabola with the
latus rectum is to:
1. Plot the vertex and axis of symmetry
2. Plot the focus and directrix.
3. Use the length of the latus rectum to plot two points
on the parabola.
4. Draw the parabola.
Example 2
Graph a parabola using the
vertex, focus, axis of
symmetry and latus rectum.
Parabola – Example 2
Find the vertex, axis of symmetry, focus, directrix,
endpoints of the latus rectum and sketch the graph.
y
1
16
 x 1 
2
2
Parabola – Example 2
y
1
16
 x 1 
2
2
The vertex is at (1, 2) with the parabola opening down.
p

1
4 1
16


1
4
1
4
The focus is 4 units down and the directrix is 4 units up.
The length of the latus rectum is
1
a

1
1
16
 16
Parabola – Example 2
Find the vertex, axis of symmetry, focus, directrix,
endpoints of the latus rectum and sketch the graph.
Axis
x=1
Focus
(1, -2)
V(1, 2)
Directrix
y=6
Latus
Rectum
Parabola – Example 2
The graph of the parabola
y
1
16
 x 1   2
2
Axis
x=1
V(1, 2)
Focus
(1, -2)
Directrix
y=6
Latus
Rectum
x=
2
ay
Parabola
Graphing and finding the
vertex, focus, directrix, axis of
symmetry and latus rectum.
Parabola – Graphing x = ay2
Graph x = 2y2 by constructing
a table of values.
Graph x = 2y2 by plotting the
points in the table.
x
y
18
-3
8
-2
2
-1
0
0
2
1
8
2
18
3
Parabola – Graphing x = ay2
Graph the table of values.
Parabola – Graphing x = ay2
One could follow a similar proof to show the distance
from the vertex to the focus and directrix to be
1

4a
1

4(2)
1 .
4a
1
8
Similarly, the length of the latus rectum can be shown to
be
1
a
.
1
a

1
2

1
2
Parabola – Graphing x = ay2
Graphing the axis of symmetry, vertex, focus, directrix
and latus rectum.
Directrix
x  1
8
V(0,0)
Focus

1 ,0
8

Axis
y=0
x = a(y –
2
k)
+h
Graphing and finding the
vertex, focus, directrix, axis of
symmetry and latus rectum.
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.
Example 3
Graphing and finding the
vertex, focus, directrix, axis of
symmetry and latus rectum.
Parabola – Example 3
Graph the parabola by finding the vertex, focus, directrix
and length of the latus rectum.
x    y 1 
1
2
2
2
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.
Parabola – Example 3
Graph the parabola by finding the vertex, focus, directrix
and length of the latus rectum.
x    y 1 
1
2
2
2
What is the distance to the focus and directrix?
The distance is
1
4a

1
1
1


2 2
4(  1 )
2
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.
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.
2
1
x    y 1   2
2
Vertex?
(-2, -1)
Focus?
(-2.5, -1)
Directrix? x = -1.5
Parabola – Example 3
x    y 1 
1
2
2
2
What is the length of the latus rectum?
1
1

1
a
 2  2
2
Parabola – Example 3
Construct the latus rectum with a length of 2.
x    y 1 
1
Construct the parabola.
2
2
2
Vertex?
(-2, -1)
Focus?
(-2.5, -1)
Directrix? x = -1.5
Latus
2
Rectum?
Parabola – Example 3
The parabola is:
x    y 1 
1
2
2
2
Vertex?
(-2, -1)
Focus?
(-2.5, -1)
Directrix? x = -1.5
Latus
2
Rectum?
Building a Table of
Rules
Parabola
Table of Rules - y = a(x - h)2 + k
a>0
a<0
Opens
Up
Down
Vertex
(h, k)
(h, k)
Focus
Axis

 h,


k 1
4a
x=h
Directrix y  k  1
Latus
Rectum





 h,


k 1
a

 h,


4a




x=h
yk 1
4a
1
1
x=h
a
k 1
4a




(h, k)
yk 1
yk 1
4a
4a
(h, k)
4a

 h,


x=h
k 1
4a




Table of Rules - x = a(y - k)2 + h
a>0
Right
Opens
Vertex
Focus
a<0
Left
(h, k)

h


 1
4a
,k





h


 1
4a
y=k
Directrix x  h  1
xh 1
Latus
Rectum
4a
1
1
a
(h, k)
(h, k)
y=k
Axis
y=k
a
,k




xh 1

h


 1

4a
,k


4a
xh 1
4a
(h, k)
4a

h


 1

4a
,k


y=k
Paraboloid
Revolution
Parabola
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/Image:ParaboloidOfRevolution.png
GNU Free Documentation License
Paraboloid Revolution
They are commonly
used today in satellite
technology as well as
lighting in motor vehicle
headlights and
flashlights.
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.
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?
Example 4 – Satellite Receiver
8 ft
1 ft
(4, 1)
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 + k
Since the vertex is (0, 0), h and k are 0.
y = ax2
Example 4 – Satellite Receiver
8 ft
1 ft
(4, 1)
V(0, 0)
y = ax2
The parabola must pass through the point (4, 1).
1 = a(4)2
1 = 16a
a 1
16
Solve for a.
Example 4 – Satellite Receiver
8 ft
(4, 1)
1 ft
V(0, 0)
The model for the parabola is:
y  1 x2
16
The receiver should be placed at the focus. Locate
the focus of the parabola.
Distance to the focus is:
1
1


4
1
4a
4 1
4
16

1

Example 4 – Satellite Receiver
8 ft
1 ft
(4, 1)
V(0, 0)
The receiver should be placed 4 ft. above the vertex.
Sample Problems
Parabola
Sample Problems
1. (y + 3)2 = 12(x -1)
a.
Find the vertex, focus, directrix, and length
of the latus rectum.
b. Sketch the graph.
c. Graph using a grapher.
Sample Problems
1. (y + 3)2 = 12(x -1)
a.
Find the vertex, focus, directrix, axis of
symmetry and length of the latus rectum.
Since the y term is squared, solve for x.
1
( y  3) 2  x  1
12
x 1
12
( y  3) 2  1
Sample Problems
x 1
12
( y  3) 2  1
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.
1
1
1


3
1
4a
4 1
3
12


Sample Problems
x 1
12
( y  3) 2  1
Find the length of the latus rectum.
1
1

 12
1
a
12
Sample Problems
x 1
12
( y  3) 2  1
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.
Sample Problems
x 1
12
( y  3) 2  1
Vertex (1, -3)
Opens Right
Axis y = -3
Focus (4, -3)
Directrix x = -2
Sample Problems
1. (y + 3)2 = 12(x -1)
c. Graph using a grapher.
Solve the equation for y.
12( x  1)  ( y  3) 2
12( x 1)  ( y  3) 2
y  3   12( x 1)
y   3  12( x 1)
Graph as 2 separate equations in the
grapher.
Sample Problems
1. (y + 3)2 = 12(x -1)
c. Graph using a grapher.
y   3  12( x 1)
y   3  12( x 1)
Sample Problems
2. 2x2 + 8x – 3 + y = 0
a.
Find the vertex, focus, directrix, axis of
symmetry and length of the latus rectum.
b. Sketch the graph.
c. Graph using a grapher.
Sample Problems
2. 2x2 + 8x – 3 + y = 0
a.
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 (-2*4) is -8.
To balance the side, we must add 8.
y = -2(x + 2) 2 + 11
Sample Problems
2. 2x2 + 8x – 3 + y = 0
a.
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.
1
1
1


4a
4 2  8
Sample Problems
2. 2x2 + 8x – 3 + y = 0 or y = -2(x + 2) 2 + 11
a.
Find the vertex, focus, directrix, and length
of the latus rectum.
Since the latus rectum is
quite small, make a table
of values to graph.
Graph the table of values and
use the axis of symmetry to plot
the other side of the parabola.
x
y
-2
-1
0
11
9
3
1
-7
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
-1
0
11
9
3
1
-7
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
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?
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
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”?
The distance from
the vertex to the
focus is 4.
1
4
4a
1  16a
a  1
16
x = a(y – 3)2 + 2
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”?
The distance from
the vertex to the
focus is 4.
How can this be
used to solve for “a”?
x = a(y – 3)2 + 2
Sample Problems
3. Write the equation of a parabola with vertex at
(3, 2) and focus at (-1, 2).
1
x = a(y – 3)2 + 2
4
4a
1  16 a
a  1
a 1
16
or
16
a  1
16
Sample Problems
3. Write the equation of a parabola with vertex at
(3, 2) and focus at (-1, 2).
a 1
16
or
a  1
x = a(y – 3)2 + 2
16
Which is the correct
value of “a”?
Since the parabola
opens to the left, a
must be negative.
x  1
y  3

16
2
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 of
y = a(x – h)2 + k
What is the vertex?
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)
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
1
4a
1
4a  1
a 1
4
or
a  1
4
Sample Problems
4. Write the equation of a parabola with focus at
(4, 0) and directrix y = 2.
Which value of a?
a 1
4
or
a  1
4
Since the parabola opens
down, a must be negative and
the vertex is (4, 1). Write the
model.
y  1
x  4

4
2
1