9.2: Parabolas Assignment: P. 623-625: 1-4, 21, 22,

9.2: Parabolas
Objectives:
1. To define a parabola as
a conic section and as a
locus and find its parts
(vertex, focus, directrix)
2. To write the equation of
a parabola in standard
form
3. To graph the equation of
a parabola in standard
form
Assignment:
• P. 623-625: 1-4, 21, 22,
28, 32, 36, 39, 43, 47,
52, 53, 55, 56, 58
• P. 655-656: 3, 9, 15,
16, 23, 26, 39, 42, 49
• Challenge Problems
Warm-Up
When a solid is cut by a plane, the resulting
plane figure is called a section. A section
that is parallel to the base is a crosssection.
Warm-Up
Below is what is known as a double-napped
cone (aka double cone).
Each cone is
called a nappe.
Upper nappe
Vertex
The point of
intersection is
the vertex.
Lower nappe
Warm-Up
Describe each of the following sections of
the double cone, none of which touch the
vertex:
1. Plane is parallel to the “base”
2. Same plane as above except
tilted a few degrees
3. Plane is parallel to one of the
lateral sides
4. Plane is perpendicular to the
“base”
Objective 1
You will be able to define a
parabola as a conic section and as
a locus and find its parts (vertex,
focus, and directrix)
Conic Section
Here are the four basic conic sections,
which are formed by the intersection of a
plane and a double cone.
Plane does not pass through the vertex
Conic Section
Conics were first described by the Greeks and
were later instrumental in the development of
calculus in the 1500-somethings.
Plane does not pass through the vertex
Conic Section
Conics can be described by the general
quadratic equation: 𝐴𝑥 2 + 𝐵𝑦 2 + 𝐶𝑥 + 𝐷𝑦 + 𝐸 = 0
Conic Section
And here are some degenerate conic sections.
What’s the difference? They’re the simpletons
of the conic family of sections.
Plane passes through the vertex
Applications
Like all conic
sections, there are
plenty of parabolic
applications:
Galileo
discovered that
trajectories of
launched objects
are parabolas
Applications
Like all conic
sections, there are
plenty of parabolic
applications:
Galileo
discovered that
trajectories of
launched objects
are parabolas
Applications
Like all conic
sections, there are
plenty of parabolic
applications:
While the orbits
of some celestial
bodies are
elliptical, others
are parabolic
Applications
Like all conic
sections, there are
plenty of parabolic
applications:
Parabolic
reflectors have
parabolic crosssections
Objective 1
You will be able to define a
parabola as a conic section and as
a locus and find its parts (vertex,
focus, and directrix)
Definition: Locus
A locus is a set of points that share a
common geometric property.
A circle is
the locus of
coplanar
points (𝑥, 𝑦)
that are
equidistant
from a given
point called
the center.
Equation:
𝑥−ℎ
𝑥−ℎ
2
2
+ 𝑦−𝑘
+ 𝑦−𝑘
2
2
=𝑟
= 𝑟2
Center
Radius
Definition: Parabola
Parabolas can be defined as a function of x:
𝑓 𝑥 =𝑎 𝑥−ℎ 2+𝑘
• This parabola either
opens up or down
depending on the value
of a
• What about all of those
parabolas that open to
the left or right?
Definition: Parabola
Parabolas can also be defined as a locus:
A parabola is the set of
coplanar points (𝑥, 𝑦)
that are equidistant
from a fixed line called
a directrix and a fixed
point not on the line
called the focus.
Click to watch me move!
Parts of a Parabola
The vertex (h, k) is the
midpoint between the
focus and the directrix.
• The vertex and focus
lie on the axis of
symmetry
• The distance from the
vertex to the focus is p,
the focal length
Vertical Directrix
If the directrix is vertical,
then the parabola opens
to the left or right.
Vertical Directrix
Vertex
ℎ, 𝑘
Focus
ℎ + 𝑝, 𝑘
Directrix
𝑥 = ℎ– 𝑝
Axis of
Symmetry
𝑦=𝑘
Exercise 1
The vertex of a parabola has coordinates
(2, 5) while the focus is at (4, 5).
1. What is the focal length?
2. Is the axis of symmetry vertical or
horizontal?
3. Is the directrix vertical or horizontal?
4. Does the parabola open up/down or
right/left?
Objective 2
You will be able
to write the
equation of a
parabola in
standard form
Exercise 2
Use the locus
definition of a
parabola to derive
the equation of a
parabola with a
vertical directrix, a
vertex at (h, k), and
a focal length of p.
Exercise 2
(h – p, y)
According to the definition:
𝑃𝐴 = 𝑃𝐹
𝑥− ℎ−𝑝
2
+ 𝑦−𝑦
2
=
𝑥− ℎ+𝑝
2
+ 𝑦−𝑘
2
Exercise 2
According to the definition:
𝑃𝐴 = 𝑃𝐹
𝑥− ℎ−𝑝 2+ 𝑦−𝑦 2 = 𝑥− ℎ+𝑝 2+ 𝑦−𝑘 2
𝑥− ℎ−𝑝 2 = 𝑥− ℎ+𝑝 2+ 𝑦−𝑘 2
𝑥 2 − 2𝑥 ℎ − 𝑝 + ℎ − 𝑝 2 = 𝑥 2 − 2𝑥 ℎ + 𝑝 + ℎ + 𝑝 2 + 𝑦 − 𝑘
−2𝑥ℎ + 2𝑥𝑝 + ℎ − 𝑝 2 = −2𝑥ℎ − 2𝑥𝑝 + ℎ + 𝑝 2 + 𝑦 − 𝑘 2
4𝑥𝑝 + ℎ − 𝑝 2 = ℎ + 𝑝 2 + 𝑦 − 𝑘 2
4𝑥𝑝 + ℎ2 − 2ℎ𝑝 + 𝑝2 = ℎ2 + 2ℎ𝑝 + 𝑝2 + 𝑦 − 𝑘 2
4𝑥𝑝 − 4ℎ𝑝 = 𝑦 − 𝑘 2
4𝑝 𝑥 − ℎ = 𝑦 − 𝑘 2
Focal length
Coordinates of the Vertex
2
Vertical Directrix
If the directrix is vertical,
then the parabola opens
to the left or right.
𝒚−𝒌
𝟐
= 𝟒𝒑 𝒙 − 𝒉
Vertex
ℎ, 𝑘
Focus
ℎ + 𝑝, 𝑘
Directrix
𝑥 = ℎ– 𝑝
Axis of
Symmetry
𝑦=𝑘
Exercise 3
Write the equation of the parabola whose
vertex is (0, 0) and whose focus is (𝑝, 0).
Horizontal Directrix
If the directrix is horizontal,
then the parabola
opens up or down.
𝒙−𝒉
𝟐
= 𝟒𝒑 𝒚 − 𝒌
Vertex
ℎ, 𝑘
Focus
ℎ, 𝑘 + 𝑝
Directrix
𝑦 = 𝑘– 𝑝
Axis of
Symmetry
𝑥=ℎ
Exercise 4
Write the equation of the parabola whose
vertex is (0, 0) and whose focus is (0, 𝑝).
Exercise 5
What is the relationship between 𝑎 and 𝑝?
𝑦 =𝑎 𝑥−ℎ
2
+𝑘
𝑥−ℎ
2
= 4𝑝 𝑦 − 𝑘
Objective 2
You will be able
to write the
equation of a
parabola in
standard form
Exercise 6
Find the equation of the parabola in
standard form with its vertex at the origin
and focus at (3, 0).
Exercise 7
Find the equation of the parabola in
standard form with its vertex at (3, 2) and
focus at (1, 2).
Protip #1a
Consider the relationship between the vertex
and focus (which lies on the axis of symmetry)
and the directrix (which is perpendicular to the
axis of symmetry).
𝑥 − ℎ 2 = 4𝑝 𝑦 − 𝑘
Vertex
Focus
(3, 2)
(3, 5)
Vertical axis of symmetry
Protip #1b
Consider the relationship between the vertex
and focus (which lies on the axis of symmetry)
and the directrix (which is perpendicular to the
axis of symmetry).
𝑦 − 𝑘 2 = 4𝑝 𝑥 − ℎ
Vertex
Focus
(3, 2)
(5, 2)
Horizontal axis of symmetry
Protip #2
Consider the two quadratic equations below.
𝑥 2 − 10𝑥 − 8𝑦 + 10 = 0
𝑦 2 − 2𝑥 + 6𝑦 + 7 = 0
1. How can we tell that these equations
represent parabolas?
2. How can we tell if the parabola will open
up/down or left/right?
Protip #2
Consider the two quadratic equations below.
𝑥 2 − 10𝑥 − 8𝑦 + 10 = 0
𝑦 2 − 2𝑥 + 6𝑦 + 7 = 0
Two zeros on the 𝑥-axis
Two zeros on the 𝑦-axis
Exercise 8
Find the vertex, focus, and directrix of the
parabola given by 𝑥 2 − 10𝑥 − 8𝑦 + 17 = 0
Exercise 9
Find the vertex, focus, and directrix of the
parabola given by 𝑦 2 − 2𝑥 + 6𝑦 + 7 = 0
Objective 3
You will be able to
graph the equation
of a parabola in
standard form
Latus Rectum
The latus rectum of a
parabola is the
segment that
passes through
the focus, is
parallel to the
directrix, and has
endpoints that are
on the parabola.
Exercise 10
Find the length of the latus rectum of
𝑥 2 = 4𝑝𝑦.
Protip #3
Let the latus rectum help you graph a
parabola. Since its length is 4𝑝, it will be
± 2𝑝 from the focus.
𝑥−ℎ
2
= 4𝑝 𝑦 − 𝑘
Length of latus rectum
1. Plot vertex and focus
2. Plot endpoints of latus rectum
3. Draw parabola
Exercise 11
Graph the equation 𝑦 2 + 4𝑥 − 4𝑦 − 4 = 0
Exercise 8
Find the vertex, focus, and directrix of the
parabola given by 𝑥 2 − 10𝑥 − 8𝑦 + 17 = 0
Exercise 9
Find the vertex, focus, and directrix of the
parabola given by 𝑦 2 − 2𝑥 + 6𝑦 + 7 = 0
Exercise 8 and 9 (Again!)
8. 𝑥 2 − 10𝑥 − 8𝑦 + 17 = 0
9. 𝑦 2 − 2𝑥 + 6𝑦 + 7 = 0
9.2: Parabolas
Objectives:
1. To define a parabola as
a conic section and as a
locus and find its parts
(vertex, focus, directrix)
2. To write the equation of
a parabola in standard
form
3. To graph the equation of
a parabola in standard
form
Assignment
• P. 623-625: 1-4, 21,
22, 28, 32, 36, 39,
43, 47, 52, 53, 55,
56, 58
• P. 655-656: 3, 9, 15,
16, 23, 26, 39, 42,
49
• Challenge Problems
“Is that a pawabowa?”