Conics

1
Conics

• Chapter 12

2
Chapter Sections
12.1 Distance and Midpoint Formulas 12.2
Circles 12.3 Parabolas 12.4 Ellipses 12.5
Hyperbolas 12.6 Nonlinear Systems of Equations
3
Distance and Midpoint Formulas

• Section 12.1

4
The Distance Formula
The Distance Formula The distance between two
points P1 (x1, y1) and P2 (x2, y2), is
5
The Distance Formula
Example Determine the distance between (?3, ?5)
and (3, 3).
6
The Distance Formula
Example Find the distance between ( 6, 6) and
( 5, 2).
7
The Midpoint Formula
The midpoint of a line segment is the point
located exactly halfway between the two endpoints
of the line segment.
8
The Midpoint Formula
Example Find the midpoint of the line segment
joining P1 (0, 8) and P2 (4, 6).
M (2, 1)


9
Circles

• Section 12.2

10
Conic Sections
Conics, an abbreviation for conic sections are
curves that result from the intersection of a
right circular cone and a plane. The four conics
are shown below.
11
Radius and Center

• A circle is a set of all points in the Cartesian
  plain that are a fixed distance r from a fixed
  point (h, k). The fixed distance r is called the
  radius, and the fixed point (h, k) is called the
  center of the circle.

12
Standard Form of a Circle
The standard form of an equation of a circle with
radius r and center (h, k) is (x h)2 (y
k)2 r2.
Example Determine the equation of the circle
with radius 4 and center ( 5, 2).
(x h)2 (y k)2 r2
center ( 5, 2)
r 4
(x ( 5))2 (y (2))2 42
(x 5)2 (y 2)2 16
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Graphing a Circle
Example Graph the equation (x 5)2 (y 2)2
16.
h 5, k 2
r 4
The center is ( 5, 2).
The radius is 4.
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General Form of a Circle
The general form of the equation of a circle is
given by the equation x2 y2 ax by c
0 when the graph exists.
Example Determine the equation of the circle
x2 y2 2x 8y 8 0
Regroup the terms.
(x2 2x) (y2 8y) 8
(x2 2x 1) (y2 8y 16) 8 1 16
Complete the square in both x and y.
(x 1)2 (y 4)2 9
Factor.
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Parabolas

• Section 12.3

16
The Parabola
Parabolas that Open Up or Down The graph of y
a(x h)2 k or y ax2 bx c is a parabola
that 1. opens up if a gt 0 and opens down if a lt
0. 2. has vertex (h, k) if the equation is of
the form y a(x h)2 k. 3. has a vertex whose
x-coordinate is The y-coordinate
is found by evaluating the equation at the
x-coordinate of the vertex.
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The Parabola
A parabola is the collection of all points P in
the plane that are the same distance from a fixed
point F as they are from a fixed line D. The
point F is called the focus of the parabola, and
the line D is its directrix. In other words, a
parabola is a set of points P for which d(F,P)
d(P,D).
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Equations of a Parabola
From the distance formula we can obtain the fact
that the equation of the parabola whose vertex is
at the origin and opens to the right is y2
4ax.
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Graphing a Parabola
Example Graph the equation x2 16y.
The equation is of the form x2 4ay, where
4a 16, so that a 4.
The graph of the equation is a parabola with
vertex (0, 0) and focus at (0, a) (0, 4) so
the parabola opens down.
The directrix is the line y 4.
x2 16y
x2 16( 4)
Substitute y 4 into the equation.
x 8
Take the square root of both sides.
Continued.
20
Graphing a Parabola
Example continued Graph the equation x2 16y.
The points on the parabola to the left and right
of the focus are ( 8, 4) and (8, 4).
V(0, 0)
( 8, 4 )
(8, 4 )
F(0, 4 )
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Finding the Equation of a Parabola
Example Find an equation of the parabola with
vertex at (0, 0) and focus at ( 8, 0).
The distance from the vertex (0, 0) to the focus
at ( 8, 0) is a 8.
The focus lies on the negative x-axis, so the
parabola will open to the left.
The equation of the parabola is of the form y2
4ax with a 8 .
y2 4(8)x
The equation of the parabola is y2 32x.
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A Parabola Whose Vertex is Not the Origin
23
A Parabola Whose Vertex is Not the Origin
Example Graph the parabola x2 2x 8y 25
0.
Complete the square in x to write the equation in
standard form.
x2 2x 8y 25
Isolate the terms involving x.
x2 2x 1 8y 25 1
Complete the square.
x2 2x 1 8y 24
Simplify.
(x 1)2 8(y 3)
Factor.
The equation is of the form (x h)2 4a(y k).
This is a parabola that opens up with vertex (
1, 3).
Continued.
24
A Parabola Whose Vertex is Not the Origin
Example continued Graph the parabola x2 2x
8y 25 0.
Since 4a 8, a 2, and since the parabola opens
up, the focus will be a 2 units above the
vertex at ( 1, 5).
Find two additional points on the graph.
(x 1)2 8(5 3)
Let y 5.
(x 1)2 16
Simplify.
Take the square root of both sides.
x 1 ?4
Subtract 1 from both sides.
x 1 ?4
x 5 or x 3
Continued.
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A Parabola Whose Vertex is Not the Origin
Example continued Graph the parabola x2 2x
8y 25 0.
The points ( 5, 5) and (3, 5) are on the graph
of the parabola.
( 5, 5 )
(3, 5)
F( 1, 5 )
V( 1, 3)
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Ellipses

• Section 12.4

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The Ellipse
An ellipse is the collection of points in the
plane such that the sum of the distances from two
fixed points, called the foci, is a constant.
The line containing the foci is called the major
axis.
The midpoint of the line segment joining the foci
is called the center.
The line through the center and perpendicular to
the major axis is called the minor axis.
The two points of intersection of the ellipse and
the major axis are the vertices.
P (x, y)
y
x
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Ellipses with Center at Origin
y
P (x, y)
x
F2 (c, 0)
F1 ( c, 0)
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Ellipses with Center at Origin
Example
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Finding the Equation of an Ellipse
Example Find the equation of an ellipse whose
center is at (0, 0) a focus at (0, 4), and
vertex at (0, 7).
Since the focus and vertex lie on the y-axis, it
is the major axis.
The distance from the center of the ellipse to
the vertex is b 7 units.
The distance from the center of the ellipse to
the focus is c 4 units.
Because a2 b2 c2, we have a2 72 42 49
16 33.
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An Ellipse Whose Center is Not the Origin
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Ellipses with Center at (h, k)
Example
This is the equation of an ellipse with center at
(h, k) (4, ?3), a2 9, b2 4.
The major axis is parallel to the x-axis.
The vertices are a 3 units to the left and
right of the center.
Plot points b 2 units above and below the
center at (4, 1) and (4, 5).
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Hyperbolas

• Section 12.5

34
The Hyperbola
A hyperbola is the collection of all points in
the plane the difference of whose distances from
two fixed points, called the foci, is a constant.
The line containing the foci is called the
transverse axis.
The line through the center and perpendicular to
the transverse axis is called the conjugate axis.
Conjugate axis
The vertices are the two points of intersection
of the hyperbola and the transverse axis.
Transverse axis
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Equations of a Hyperbola
36
Hyperbolas with Center at Origin
Example
a 3 and b 6, so the vertices are at (a, 0),
(?a,0) or (3, 0), (?3,0).
Continued.
37
Hyperbolas with Center at Origin
Example continued
Locate points above and below the foci.
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Finding the Equation of a Hyperbola
Example Find an equation of the hyperbola with
center at the origin, one focus at (3, 0) and
vertex at (2, 0).
Since all of the points are on the x-axis, the
transverse axis is the x-axis and the hyperbola
will open left and right.
The vertex is (2, 0) so a 2, and the focus is
(3, 0) so c 3.
b2 c2 a2 9 4 5.
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Asymptotes
As x and y get larger in both the positive and
negative direction, the branches of the hyperbola
approach two lines, called asymptotes of the
hyperbola.
Both vertices are a units from the origin.
Both vertices are a units from the origin.
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Asymptotes
41
Graphing with Asymptotes
Example Graph the following equation using the
asymptotes as a guide.
The center is at (?3, 1), and the hyperbola is
horizontal.
Since a 4, the vertices of the hyperbola are at
(?3 4, 1), or (?7, 1) and (1, 1).
Continued.
42
Graphing with Asymptotes
Example continued
Plot points b 2 units above and below the
center at ( 3, 3) and ( 3, 1).
Draw the rectangle and use it to draw the
hyperbola.
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Nonlinear Systems of Equations

• Section 12.6

44
Nonlinear Systems of Equations
A system of nonlinear equations in two variables
is a system of equations in which at least one of
the equations is not linear.
Example Solve the following system of equations
by using substitution
Solve equation (2) for x.
Substitute (3y 5) into equation (1).
Continued.
45
Solving Using Substitution
Example continued
Solve the resulting quadratic equation.
Continued.
46
Solving Using Substitution
Example continued
Substitute the values for y into either equation.
The solutions of the system are (?5, 0) and (4,
3).
Continued.
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Solving Using Substitution
Example continued
The solutions of the system are (?5, 0) and (4,
3).
Check
x 3y 5
x2 y2 25
x 3y 5
x2 y2 25
42 (3)2 25
4 3(3) 5
( 5)2 0 25
5 3(0) 5
16 9 25
4 9 5
25 25
5 5
?
?
25 25
5 5
?
?
Continued.
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Solving Using Substitution
Example continued
The graph of the system also verifies the
solution.
x 3y 5
x2 y2 25
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Solving Using Substitution
Example Solve the following system of equations
by using substitution
x y 2
Equation (2)
(y 2)2 y 4
Substitute (y 2) into equation (1).
y2 4y 4 y 4
Simplify.
y2 5y 0
y(y 5) 0
Factor.
Continued.
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Solving Using Substitution
Example continued
y(y 5) 0
y 0 or y 5
Substitute the values for y into either equation.
y 2 x
y 2 x
0 2 x
5 2 x
x 2
x 3
Continued.
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Solving Using Substitution
Example continued
The solutions are (2, 0) and ( 3, 5). Check
both solutions in both equations.
y 2 x
x2 y 4
y 2 x
x2 y 4
( 3)2 ( 5) 4
5 2 3
22 0 4
0 2 2
?
4 4
3 3
?
4 4
2 2
?
?
The graph of the system also verifies the
solution.
y 2 x
x2 y 4
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Solving Using Elimination
Example Solve the following system using
elimination.
Multiply equation (2) by ?3.
13y2 26
Add equation (1) and (2).
y2 2
Solve for y.
Continued.
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Solving Using Elimination
Example continued
Substitute the values for y into either equation.
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Solving Using Elimination
Example Solve the following system using
elimination.
Multiply equation (1) by 1.
Equation (2)
Add equations (1) and (2).
3y2 0
y 0
Solve for y.
Continued.
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Solving Using Elimination
Example continued
y 0
Substitute the values for y into either equation.
x2 2y2 4
x2 2(0)2 4
x2 4
x 2 or x 2
The solutions are ( 2, 0) and (2, 0).