Title: Conic Sections
1Conic Sections
2Conic 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.
3Conic Sections - Parabola
Focus
Directrix
The line is called the directrix and the point is
called the focus.
4Conic 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.
5Conic 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.
6Finding the Focus and Directrix
7Conic 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.
8Conic 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).
9Conic 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).
10Conic 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.
11Conic 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.
12Conic Sections - Parabola
Therefore, the distance p from the vertex to the
focus and the vertex to the directrix is given by
the formula
13Conic 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.
14Example 1
- Graph a parabola.
- Find the vertex, focus and directrix.
15Parabola Example 1
Make a table of values. Graph the function.
Find the vertex, focus, and directrix.
16Parabola 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.
17Parabola Example 1
Find the focus and directrix.
18Parabola Example 1
The focus and directrix are p units from the
vertex where
The focus and directrix are 2 units from the
vertex.
19Parabola Example 1
2 Units
Focus (-2, -1) Directrix y -5
20Example 2
- Graph a parabola using the vertex, focus, axis of
symmetry and directrix.
21Parabola Example 2
Find the vertex, axis of symmetry, focus,
directrix and sketch the graph.
22Parabola 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.
23Parabola 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)
24Parabola Example 2
The graph of the parabola
Axisx1
Directrixy6
V(1, 2)
Focus(1, -2)
25x ay2 Parabola
- Graphing and finding the vertex, focus,
directrix, axis of symmetry.
26Parabola Graphing x ay2
- Graph the table of values.
27Parabola Graphing x ay2
- One could follow a similar proof to show the
distancefrom the vertex to the focus and
directrix to be .
28Parabola Graphing x ay2
- Graphing the axis of symmetry, vertex, focus,
directrix and latus rectum.
Directrix
V(0,0)
Focus
Axis y0
29x a(y k)2 h
- Graphing and finding the vertex, focus,
directrix, axis of symmetry.
30Parabola 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.
31Example 3
- Graphing and finding the vertex, focus,
directrix, axis of symmetry and latus rectum.
32Parabola Example 3
Graph the parabola by finding the vertex, focus,
directrix.
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.
33Parabola Example 3
Graph the parabola by finding the vertex, focus,
directrix.
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.
34Parabola 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
35Parabola Example 3
36Parabola Example 3
The parabola is
Vertex?
(-2, -1)
Focus?
(-2.5, -1)
Directrix?
x -1.5
37Building a Table of Rules
38Table 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)
x h
39Table 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
40Sample Problems
41Sample Problems
- Find the vertex, focus, directrix.
b. Sketch the graph.
c. Graph using a grapher.
42Sample Problems
- Find the vertex, focus, directrix, axis of
symmetry.
Since the y term is squared, solve for x.
43Sample 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.
44Sample Problems
- Find the length of the latus rectum.
45Sample 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.
46Sample Problems
Vertex (1, -3)Opens RightAxis y -3Focus
(4, -3)Directrix x -2
47Sample Problems
c. Graph using a grapher.
Solve the equation for y.
Graph as 2 separate equations in the grapher.
48Sample Problems
c. Graph using a grapher.
49Sample Problems
- Find the vertex, focus, directrix, axis of
symmetry.
b. Sketch the graph.
c. Graph using a grapher.
50Sample Problems
- Find the vertex, focus, directrix, axis of
symmetry.
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
51Sample Problems
- Find the vertex, focus, directrix.
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.
52Sample 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
5311.2
54Ellipse
An ellipse is the set of all points in a plane
whose distance from two fixed points in the plane
have a constant sum. The fixed points are the
foci (plural of focus) of the ellipse.
55Parts of an Ellipse
56(No Transcript)
57(No Transcript)
58What is the equation of an ellipse in standard
form?
What is always true about the coordinate (x,y)?
Write an equation based on that fact.
59Another way to look at ellipses
60(No Transcript)
61Also acceptable
62Practice
1.
Center ( -1, 1) a 3, b 2 c
Vertices (-1, 4) (-1, -2) Foci
63Practice
Practice
- Find an equation of the ellipse with foci (
-4, 0) and ( 0, 0 ) and whose minor axis has a
length of 2.
Center ( -2, 0) c 2, b 1 a
641. Write the equation of the ellipse with a
major axis length of 10 and with foci (-5,
3) (-5, 9).
a 5, c 3, Center ( -5, 6 )
b 4
652. Sketch a graph of the above ellipse, Label
foci, vertices, center and end points of
minor axis.
663. Write the equation of a parabola in standard
form whose focus is located at the center
of the above ellipse and whose directrix is y
b. (use b from the above ellipse).
Focus ( -5, 6) Directrix y 4
p 1 Vertex (-5, 5)
674. Sketch the graph of the above parabola, label
all important parts.
68a) Determine the type of conic the equation
represents. b) Rewrite into standard form. c)
Graph and label all major parts of the conic
69(No Transcript)
70(No Transcript)
71(No Transcript)
72(No Transcript)
73Eccentricity
Circle e 1 Ellipse 0 lt e lt 1 Hyperbola e
gt 1
74Section 11.3 Hyperbolas
75Hyperbolas
A hyperbola is a set of points in a plane the
difference of whose distances from two fixed
points, called foci, is a constant.
For any point P that is on the hyperbola, d2 d1
is always the same.
P
d2
d1
F1
F2
In this example, the origin is the center of the
hyperbola. It is midway between the foci.
76Hyperbolas
A line through the foci intersects the hyperbola
at two points, called the vertices.
The segment connecting the vertices is called the
transverse axis of the hyperbola.
V
V
F
C
F
The center of the hyperbola is located at the
midpoint of the transverse axis.
As x and y get larger the branches of the
hyperbola approach a pair of intersecting lines
called the asymptotes of the hyperbola. These
asymptotes pass through the center of the
hyperbola.
77Hyperbolas
The figure at the left is an example of a
hyperbola whose branches open up and down instead
of right and left.
F
V
Since the transverse axis is vertical, this type
of hyperbola is often referred to as a vertical
hyperbola.
C
V
F
When the transverse axis is horizontal, the
hyperbola is referred to as a horizontal
hyperbola.
78Standard Form Equation of a Hyperbola
(x h)2 (y k)2
(y k)2 (x h)2
1
1
a2
b2
a2
b2
Horizontal Hyperbola
Vertical Hyperbola
The center of a hyperbola is at the point (h, k)
in either form
For either hyperbola, c2 a2 b2
Where c is the distance from the center to a
focus point.
The equations of the asymptotes are
and
y
x
y
x
79Graphing a Hyperbola
Graph x2 y2 4 9
1
Center (0, 0)
The x-term comes first in the subtraction so this
is a horizontal hyperbola
From the center locate the points that are two
spaces to the right and two spaces to the left
From the center locate the points that are up
three spaces and down three spaces
Draw a dotted rectangle through the four points
you have found.
Draw the asymptotes as dotted lines that pass
diagonally through the rectangle.
c2 9 4 13
c Ö13 3.61
Draw the hyperbola.
Foci (3.61, 0) and (-3.61, 0)
Vertices (2, 0) and (-2, 0)
Asymptotes y
x
80Graphing a Hyperbola
Graph (x 2)2 (y 1)2 9
25
1
Horizontal hyperbola
Center (-2, 1)
Vertices (-5, 1) and (1, 1)
c2 9 25 34
c Ö34 5.83
Foci (-7.83, 1) and (3.83, 1)
53
Asymptotes y x
81Converting an Equation
Graph 9y2 4x2 18y 24x 63 0
9(y2 2y ___) 4(x2 6x ___) 63 ___
___
9
1
9
36
9(y 1)2 4(x 3)2 36
(y 1)2 (x 3)2 4
9
1
The hyperbola is vertical
Center (3, 1)
c2 9 4 13
c Ö13 3.61
Foci (3, 4.61) and (3, -2.61)
23
Asymptotes y x
82Finding an Equation
Find the standard form of the equation of a
hyperbola given
Foci (-7, 0) and (7, 0)
Vertices (-5, 0) and (5, 0)
8
Horizontal hyperbola
Center (0, 0)
10
F
F
V
V
a2 25 and c2 49
C
c2 a2 b2
49 25 b2
b2 24
(x h)2 (y k)2
1
a2
b2
x2 y2
1
25
24
83Finding an Equation
Find the standard form equation of the hyperbola
that is graphed at the right
Vertical hyperbola
(y k)2 (x h)2
1
b2
a2
Center (-1, -2)
a 3 and b 5
(y 2)2 (x 1)2
1
25
9
84Using Polar Coordinates
- Graphing and converting polar and rectangular
coordinates
85Graphing Polar Coordinates
The grid at the left is a polar grid. The
typical angles of 30o, 45o, 90o, are shown on
the graph along with circles of radius 1, 2, 3,
4, and 5 units.
A
Points in polar form are given as (r, ß ) where r
is the radius to the point and ß is the angle of
the point.
On one of your polar graphs, plot the point (3,
90o)?
The point on the graph labeled A is correct.
86Graphing Polar Coordinates
Now, try graphing .
C
A
Did you get point B?
Polar points have a new aspect. A radius can be
negative! A negative radius means to go in the
exact opposite direction of the angle.
B
To graph (-4, 240o), find 240o and move 4 units
in the opposite direction. The opposite
direction is always a 180o difference.
Point C is at (-4, 240o). This point could also
be labeled as (4, 60o).
87Graphing Polar Coordinates
How would you write point A with a negative
radius?
C
A
A correct answer would be (-3, 270o) or (-3,
-90o).
In fact, there are an infinite number of ways to
label a single polar point. Is (3, 450o) the
same point?
B
Dont forget, you can also use radian angles as
well as angles in degrees.
On your own, find at least 4 different polar
coordinates for point B.
88Graphing Polar Coordinates
On your own, find at least 4 different polar
coordinates for point B.
C
A
There are many possible answers.Here are just a
few.(2, 225o), (2, -135o)(-2, 45o), (-2,
-315o)One could add or subtract 360o to any
ofthe above angles.Radians would result in(2,
5p/4), (2, -3p/4)(-2, p/4), (-2, -7p/4)One
could add or subtract 2p to any previous radian
angle.
B
89Converting from Rectangular to Polar
Find the polar form for the rectangular point (4,
3).
To find the polar coordinate, we must calculate
the radius and angle to the given point.
(4, 3)
r
We can use our knowledge of right triangle
trigonometry to find the radius and angle.
3
ß
4
r2 32 42r2 25r 5
tan ß ¾ ß tan-1(¾) ß 36.87o or 0.64 rad
The polar form of the rectangular point (4, 3)
is (5, 36.87o)
90Converting from Rectangular to Polar
In general, the rectangular point (x, y) is
converted to polar form (r, ?) by
1. Finding the radius
r2 x2 y2
(x, y)
2. Finding the angle
r
y
ß
tan ß y/x or ß tan-1(y/x)Recall that some
angles require the angle to be converted to the
appropriate quadrant.
x
Note This is just like finding the length and
direction angle of a vector!
91Converting from Rectangular to Polar
On your own, find polar form for the point (-2,
3).
(-2, 3)
r2 (-2)2 32r2 4 9r2 13r
However, the angle must be in the second
quadrant, so we add 180o to the answer and get an
angle of 123.70o. The polar form is ( ,
123.70o)
92Converting from Polar to Rectanglar
Convert the polar point (4, 30o) to rectangular
coordinates.
We are given the radius of 4 and angle of 30o.
Find the values of x and y.
Using trig to find the values of x and y, we know
that cos ß x/r or x r cos ß. Also, sin ß
y/r ory r sin ß.
4
y
30o
x
The point in rectangular form is
93Converting from Polar to Rectanglar
On your own, convert (3, 5p/3) to rectangular
coordinates.
We are given the radius of 3 and angle of 5p/3 or
300o. Find the values of x and y.
-60o
The point in rectangular form is
94Rectangular and Polar Equations
Equations in rectangular form use variables (x,
y), whileequations in polar form use variables
(r, ß) where ß is an angle.
Converting from one form to another involves
changing the variables from one form to the other.
We have already used all of the conversions which
are necessary.
Converting Polar to Rectangular
Converting Rectanglar to Polar
cos ß x/rsin ß y/rtan ß y/xr2 x2 y2
x r cos ß y r sin ß x2 y2 r2
95Convert Rectangular Equationsto Polar Equations
The goal is to change all xs and ys to rs and
ßs.When possible, solve for r.
Example 1 Convert x2 y2 16 to polar form.
Since x2 y2 r2, substitute into the equation.
r2 16
Simplify.
r 4
r 4 is the equivalent polar equation to x2 y2
16
96Convert Rectangular Equationsto Polar Equations
Example 2 Convert y 3 to polar form.
Since y r sin ß, substitute into the equation.
r sin ß 3
Solve for r when possible.
r 3 / sin ß
r 3 csc ß is the equivalent polar equation.
97Convert Rectangular Equationsto Polar Equations
Example 3 Convert (x - 3)2 (y 3)2 18 to
polar form.
Square each binomial.
x2 6x 9 y2 6y 9 18
Since x2 y2 r2, re-write and simplify by
combining like terms.
x2 y2 6x 6y 0
Substitute r2 for x2 y2, r cos ß for x and r
sin ß for y.
r2 6rcos ß 6rsin ß 0
Factor r as a common factor.
r(r 6cos ß 6sin ß) 0
r 0 or r 6cos ß 6sin ß 0
Solve for r r 0 or r 6cos ß 6sin ß
98Convert Polar Equationsto Rectangular Equations
The goal is to change all rs and ßs to xs and
ys.
Example 1 Convert r 4 to rectangular form.
Since r2 x2 y2, square both sides to get r2.
r2 16
Substitute.
x2 y2 16
x2 y2 16 is the equivalent polar equation to
r 4
99Convert Polar Equationsto Rectangular Equations
Example 2 Convert r 5 cos ß to rectangular
form.
Since cos ß x/r, substitute for cos ß.
Multiply both sides by r.
r2 5x
Substitute for r2.
x2 y2 5x is rectangular form.
100Convert Polar Equationsto Rectangular Equations
Example 3 Convert r 2 csc ß to rectangular
form.
Since csc ß r/y, substitute for csc ß.
Multiply both sides by y/r.
Simplify
y 3 is rectangular form.