Transcript chapter8_Sec1

College Algebra
Fifth Edition
James Stewart  Lothar Redlin

Saleem Watson
Conic
9 Sections
Chapter Overview
Conic sections are the curves we get
when we make a straight cut in a cone.
• For example, if a cone is cut
horizontally, the cross section
is a circle.
• So, a circle is a conic section.
Chapter Overview
Other ways of cutting a cone produce
parabolas, ellipses, and hyperbolas.
Chapter Overview
Our goal in this chapter is to find
equations whose graphs are the conic
sections.
• We already know from Section 2.2 that the graph
of the equation x2 + y2 = r2 is a circle.
• We will find equations for each of the other conic
sections by analyzing their geometric properties.
8.1
Parabolas
Parabolas
We saw in Section 4.1 that the graph of
the equation y = ax2 + bx + c is a U-shaped
curve called a parabola that opens either
upward or downward—depending on whether
the sign of a is positive or negative.
• Here, we study parabolas from a geometric
rather than an algebraic point of view.
Parabolas
We begin with the geometric definition
of a parabola and show how this leads to
the algebraic formula that we are already
familiar with.
Parabola—Geometric Definition
A parabola is the set of points in the plane
equidistant from a fixed point F (focus) and
a fixed line l (directrix).
• The vertex V lies halfway
between the focus and
the directrix.
• The axis of symmetry is
the line that runs through
the focus perpendicular
to the directrix.
Parabolas
In this section, we restrict our attention to
parabolas that:
• Are situated with the vertex at the origin.
• Have a vertical or horizontal axis of symmetry.
Parabolas in more general positions will be
considered in Section 8.4.
Parabolas
If the focus of such a parabola is the point
F(0, p), the axis of symmetry must be vertical
and the directrix has the equation y = –p.
• The figure
illustrates
the case p > 0.
Parabolas
If P(x, y) is any point on the parabola,
• The distance from P to the focus F
(using the Distance Formula)
is:
x 2  ( y  p )2
• The distance from
P to the directrix
is:
| y – (–p) | = | y + p |
Parabolas
By the definition of a parabola, these two
distances must be equal:
x  ( y  p)  y  p
2
2
2
x  ( y  p)  y  p  ( y  p)
2
2
x  y  2 py  p  y  2py  p
2
2
2
2
x  2 py  2py
2
x  4 py
2
2
2
Parabolas
If p > 0, then the parabola opens upward.
If p < 0, it opens downward.
When x is replaced by –x, the equation
remains unchanged.
• So, the graph is symmetric about the y-axis.
Equations and Graphs
of Parabola
Equations and Graphs of Parabola
We now summarize what we have just
proved about the equation and features
of a parabola with a vertical axis.
Parabola with Vertical Axis
The graph of the equation
x2 = 4py
is a parabola with these properties.
Vertex
V(0, 0)
Focus
F(0, p)
Directrix
y = –p
Parabola with Vertical Axis
The parabola opens:
• Upward if p > 0.
• Downward if p < 0.
E.g. 1—Finding the Equation of a Parabola
Find the equation of the parabola
with vertex V(0, 0) and focus F(0, 2),
and sketch its graph.
• Since the focus is F(0, 2), we conclude that p = 2
(and so the directrix is y = –2).
• Thus, the equation is:
x2 = 4(2)y
x2 = 8y
E.g. 1—Finding the Equation of a Parabola
Since p = 2 > 0, the parabola opens
upward.
E.g. 2—Finding Focus and Directrix from Equation
Find the focus and directrix of the
parabola y = –x2, and sketch the graph.
• To find the focus and directrix, we put the given
equation in the standard form x2 = –y.
• Comparing this to the general equation x2 = 4py,
we see that 4p = –1; so, p = –¼.
• The focus is F(0, –¼) and the directrix is y = ¼.
E.g. 2—Finding Focus and Directrix from Equation
Here’s the graph of the parabola,
together with the focus and the directrix.
E.g. 2—Finding Focus and Directrix from Equation
We can also draw the graph using
a graphing calculator.
Equations and Graphs of Parabola
Reflecting the graph in this figure about
the diagonal line y = x has the effect of
interchanging the roles of x and y.
• This results in
a parabola with
horizontal axis.
• By the same method
as before, we can
prove the following
properties.
Parabola with Horizontal Axis
The graph of the equation
y2 = 4px
is a parabola with these properties.
Vertex
V(0, 0)
Focus
F(p, 0)
Directrix
x = –p
Parabola with Horizontal Axis
The parabola opens:
• To the right if p > 0.
• To the left if p < 0.
E.g. 3—Parabola with Horizontal Axis
A parabola has the equation
6x + y2 = 0
(a) Find the focus and directrix of the parabola,
and sketch the graph.
(b) Use a graphing calculator to draw the graph.
E.g. 3—Parabola with Horiz. Axis
Example (a)
We put the given equation
in the standard form y2 = –6x.
• Comparing this to the general equation
y2 = 4px, we see that 4p = –6; so, p = (–3/2).
• The focus is F(–3/2, 0) and the directrix
is x = 3/2.
E.g. 3—Parabola with Horiz. Axis
Example (a)
Since p < 0, the parabola opens
to the left.
E.g. 3—Parabola with Horiz. Axis
Example (b)
To draw the graph using a graphing
calculator, we need to solve for y.
6x + y2 = 0
y2 = –6x
y   6x
E.g. 3—Parabola with Horiz. Axis
Example (b)
To obtain the graph of the parabola,
we graph both functions.
Note
The equation y2 = 4px does not define
y as a function of x.
• So, to use a graphing calculator to graph
a parabola with horizontal axis, we must first
solve for y.
• This leads to two functions
y  4 px and y   4 px
• We need to graph both y  6 x and y   6 x
to get the complete graph of the parabola.
Width of Parabola
We can use the coordinates of the focus
to estimate the “width” of a parabola when
sketching its graph.
• The line segment that runs through the focus
perpendicular to the axis—with endpoints on
the parabola—is called the latus rectum.
• Its length is the focal diameter of the parabola.
Width of Parabola
From the figure, we can see that the distance
from an endpoint Q of the latus rectum to
the directrix is | 2p |.
• So, the distance from Q
to the focus must be | 2p |
as well (by the definition
of a parabola).
• Hence, the focal diameter
is | 4p |.
Width of Parabola
In the next example, we use the focal
diameter to determine the “width”
of a parabola when graphing it.
E.g. 4—Focal Diameter of a Parabola
Find the focus, directrix, and focal
diameter of the parabola y = ½x2,
and sketch its graph.
• We put the equation in the form x2 = 4py.
y = ½x2
x2 = 2y
• We see that 4p = 2.
• So, the focal diameter is 2.
E.g. 4—Focal Diameter of a Parabola
Solving for p gives p = ½.
• So, the focus is (0, ½) and the directrix
is y = –½ .
Since the focal diameter is 2, the latus rectum
extends 1 unit to the left and 1 unit to the right
of the focus.
E.g. 4—Focal Diameter of a Parabola
Here’s the graph.
Family of Parabolas
In the next example, we graph a family
of parabolas—to show how changing
the distance between the focus and
the vertex affects the “width” of a parabola.
E.g. 5—Family of Parabolas
(a) Find equations for the parabolas
with vertex at the origin and foci
F1 0,
1
8
, F2 0,
1
2
, F3 0, 1, F4 (0, 4)
(b) Draw the graphs of the parabolas in (a).
• What do you conclude?
E.g. 5—Family of Parabolas
Example (a)
Since the foci are on the positive y-axis,
the parabolas open upward and have
equations of the form x2 = 4py.
• This leads to the following equations.
E.g. 5—Family of Parabolas
We see that
the closer
the focus to
the vertex,
the narrower
the parabola.
Example (b)
Applications
Applications
Parabolas have an important property
that makes them useful as reflectors
for lamps and telescopes.
Applications
Light from a source placed at the focus of
a surface with parabolic cross section will be
reflected in such a way that it travels parallel
to the axis of the parabola.
• Thus, a parabolic mirror
reflects the light into
a beam of parallel rays.
Reflection Property
Conversely, light approaching the reflector
in rays parallel to its axis of symmetry is
concentrated to the focus.
• This reflection property—which can
be proved using calculus—is used in
the construction of reflecting telescopes.
E.g. 6—Focal Point of a Searchlight Reflector
A searchlight has
a parabolic reflector
that forms a “bowl,”
12 in. wide from rim
to rim and 8 in. deep.
• If the filament of the light bulb is located
at the focus, how far from the vertex
of the reflector is it?
E.g. 6—Focal Point of a Searchlight Reflector
We introduce a coordinate system and
place a parabolic cross section of the reflector
so that:
• Its vertex is at the origin.
• Its axis is vertical.
E.g. 6—Focal Point of a Searchlight Reflector
Then, the equation of this parabola has
the form x2 = 4py.
• We see that
the point (6, 8) lies on
the parabola.
• We use this to find p.
E.g. 6—Focal Point of a Searchlight Reflector
62 = 4p(8)
36 = 32p
p = (9/8)
• The focus is F(0, (9/8)).
• So, the distance between the vertex
and the focus is 98  1 81 in.
• Since the filament is positioned at the focus,
it is located 1 81 in. from the vertex of the reflector.