Chapter 7 Analyzing Conic Sections

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Chapter 7
Analyzing Conic Sections
Jennifer Huss
7-1 The Distance and Midpoint
Formulas
• To find the distance between any two points
b) and (c, d), use the distance formula:
Distance = (c – a)2 + (d – b)2
• The midpoint of a line is halfway between the
two endpoints of a line
• To find the midpoint between (a, b) and (c, d),
use the midpoint formula:
Midpoint = (a + c) , (b + d)
2
2
(a,
7-1 Example
Find the distance between (-4, 2) and (-8, 4). Then find the
midpoint between the points.
Distance = [(-8) – (-4)]2 + (4 – 2)2 Midpoint = (-4) + (-8) , 2 + 4
2
2
2
2
Distance = (-8 + 4) + (2)
Distance = (-4)2 + (2)2
Distance = 16 + 4
Distance= 20
Midpoint = -12 , 6
2
2
Midpoint = ( -6, 3)
7-1 Problems
1. Find the distance between (0, 1) and (1, 5).
2. Find the midpoint between (6, -5) and (-2, -7).
3. Find the value for x if the Distance = 53 and
the endpoints are (-3, 2) and (-10, x).
4. If you are given an endpoint (3, 2) and
midpoint (-1, 5), what are the coordinates of
the other endpoint?
1) 17
2) (2, -6)
3) x = 0 or x = 4
4) (-5, 8)
7-2 Parabolas
• A parabola is a set of points on a
plane that are the same distance
from a given point called the focus
and a given line called the directrix
• The axis of symmetry is
perpendicular to the directrix and
passes through the parabola at a
point called the vertex
• The latus rectum goes through the
focus and is perpendicular to the
axis of symmetry
• If the equation of the parabola
begins with x= then the parabola is
not a function (fails the vertical line
test)
Axis of
Symmetry
Parabola
Focus
Latus
Rectum
Vertex
Directrix
7-2 Parabolas (cont.)
Important Information About the
Parabolas
Form of the
equation
y = a (x – h)2 + k
x = a (y – k)2 + h
Axis of
Symmetry
x=h
y=k
Vertex
(h, k)
(h, k)
Focus
(h, k + 1/4a)
(h + 1/4a, k)
Directrix
y = k – (1/4)a
x = h – (1/4)a
Direction of
Opening
Opens upward when a > 0
and downward when a < 0
Opens to the right when a > 0 and to
the left when a < 0
Length of Latus
Rectum
1/a units
1/a units
7-2 Example
Write y = x2 + 4x + 1 in the form y = a (x – h)2 + k and name
the vertex, axis of symmetry, and the direction the
parabola opens.
y = x2 + 4x + 1
y = (x2 + 4x + o ) + 1 – o
y = (x2 + 4x + 4) + 1 – 4
y = (x + 2)2 – 3
Vertex: (-2, -3)
Axis of Symmetry: x = -2
The parabola opens up because a = 1
so a > 0.
You can always check your
answers by graphing.
7-2 Problems
3.
Graph the equation x2 = 8y.
For the parabola y2 = -16x name the vertex, focus,
length of latus rectum, and direction of opening. Also,
give the equations of the directrix and axis of
symmetry.
Given the vertex (4, 1) and a point on the parabola
(8, 3), find the equation of the parabola.
Graph for
#1
1.
2.
Graph for
#2
2) Vertex: (0,0) Focus: (-4,0) Latus rectum: 16 Direction: left Directrix: x = 4 Axis of symmetry: y = 0
3) y = (1/8)(x – 4)2 + 1
7-3 Circles
• A circle is a set of points
equidistant from a center point
• The radius is a line between
the center and any point on the
circle
• The equation of a circle is
(x – h)2 + (y – k)2 = r2 where
the radius is r and the vertex is
(h, k)
• Sometimes you need to
complete the square twice to
get the equation in this form
(once for x and once for y)
Radius (r)
Vertex
(k, h)
7-3 Examples
1.
Find the center and radius of x2 + y2 + 4x – 12y – 9 = 0
and then graph the circle.
x2 + 4x + o + y2 – 12y + o = 9 + o + o
x2 + 4x + 4 + y2 – 12y + 36 = 9 + 4 + 36
(x + 2)2 + (y – 6)2 = 49
Radius = 7 and Center is (-2, 6)
7-3 Examples (cont.)
2.
If a circle has a center (3, -2) and a point on the circle
(7, 1), write the equation of the circle.
Find the radius by the distance formula.
Radius = (7 – 3)2 + (1 – (-2))2
r = (4)2 + (3)2
r = 16 + 9
r = 25
r=5
The equation of the circle will be (x – 3)2 + (y + 2)2 = 25
7-3 Problems
1) x2 + (y + 2)2 = 4
Center (0, -2) and Radius 2
2.
Find the center and radius of x2 + y2 + 4y = 0. Then
graph the circle.
If a circle has a center (0, 0) and a point on the circle
(-2, -4) write the equation of the circle.
#1
1.
2) x2 + y2 = 20
7-4 Ellipses
• An ellipse is the set of all points
in a plane such that the sum of
the distances from the foci is
constant
• An ellipse has two axes of
symmetry
• The axis of the longer side of
the ellipse is called the major
axis and the axis of the shorter
side is the minor axis
• The focus points always lie on
the major axis
• The intersection of the two axes
is the center of the ellipse
Major Axis
Focus
Center
Focus
Minor
Axis
7-4 Ellipses (cont.)
Foci Points
Is the major axis
horizontal or vertical?
Center of the
Ellipse
(x – h)2 + (y – k)2
=1
2
2
a
b
( h + c, k) and
(h – c, k)
Horizontal
(h, k)
(x – h)2 + (y – k)2 = 1
b2
a2
(h, k + c) and
(h, k – c)
Vertical
(h, k)
Equation of the Ellipse
Important Notes:
• In the above chart, c = a2 – b2
• a2 > b2 always so a2 is always the larger number
•If the a2 is under the x term, the ellipse is horizontal, if the a2 is under the y
term the ellipse is vertical
•You can tell that you are looking at an ellipse because: x2 is added to y2 and
the x2 and y2 are divided by different numbers (if numbers were the same, it’s
a circle)
7-4 Example
1.
Given an equation of an ellipse 16y2 + 9x2 – 96y – 90x = -225 find the
coordinates of the center and foci as well as the lengths of the major and
minor axis. Then draw the graph.
16 (y2 – 6y + o) + 9 (x2 – 10x + o) = -225 + 16 (o) + 9(o)
16 (y2 – 6y + 9) + 9 (x2 – 10x + 25) = -225 + 16(9) + 9(25)
16 (y – 3)2 + 9 (x – 5)2 = 144
(y – 3)2 + (x – 5)2
=1
9
16
Center: (5, 3)
16 > 9 so the foci are on the vertical axis
c=
16 – 9
c=
7
Foci: ( 5 + 7, 3) and (5 – 7, 3)
Major Axis Length = 4 (2) = 8
Minor Axis Length = 3 (2) = 6
7-4 Problems
1) Foci: (0, - 33) (0, 33)
2.
For 49x2 + 16y2 = 784 find the center, the foci, and the
lengths of the major and minor axes. Then draw the
graph.
Write an equation for an ellipse with foci (4, 0) and
(-4, 0). The endpoints of the minor axis are (0, 2) and
(0, -2).
#1
1.
Center: (0, 0) Length of major= 14 Length of minor= 8
2) X2 + y2
=1
20
4
7-5 Hyperbolas
•
•
•
•
•
•
A hyperbola is a set of all points on a
plane such that the absolute value of
the difference (subtraction) of the
distances from a point to the two foci
is constant
The center is the midpoint of the
segment connecting the foci
The vertex is the point on the
hyperbola closest to the center
The asymptotes are lines the
hyperbola can approach but never
touch
The transverse axis goes through
the foci
The conjugate axis is perpendicular
to the transverse axis at the center
point
Hyperbola
Conjugate
Axis
Transverse
Axis
Center
Focus
Vertex
Asymptotes
7-5 Hyperbolas (cont.)
Equation of Hyperbola
(x – h)2 _ (y – k)2
a2
b2
= 1
(y – k)2 _ (x – h)2
a2
b2
= 1
Center
Foci Points
Equation of
Asymptote
Vertex
Transverse
Axis
(h, k)
(h – c, k) and (h
+ c, k)
y = +/- (b/a) x
(h +a, k)
and
(h – a, k)
Horizontal
(h, k)
(h, k – c) and
(h, k + c)
(c = a2 + b2 )
y = +/- (a/b) x
(h, k + a)
and
(h, k – a)
Vertical
You must be looking at a
hyperbola because the x2 and
y2 terms are subtracted
(x2 – y2) or (y2 – x2)
7-5 Example
Write the standard form of the equation of the hyperbola
144y2 – 25x2 – 576y – 150x = 3249. Then find the coordinates of the center,
the vertices, the foci, and the equation of the asymptotes. Graph the
hyperbola and the asymptotes.
144(y2 – 4y + o) – 25(x2 + 6x + o) = 3249 + 144(o) + 25(o)
144(y2 – 4y + 4) – 25(x2 + 6x + 9) = 3249 + 144(4) + 25(9)
144(y – 2)2 – 25(x + 3)2 = 3600
(y-2)2 _ (x + 3)2
=1
25
144
Center: (-3, 2)
a = 5 so the vertices are (-3, 7) and (-3, -3)
a2 + b 2 = c 2
25 + 144 = c2
c = 13
The foci are (-3, 15) and (-3, -11).
7-5 Example (cont.)
Asymptotes have the formula y = +/- a/b x and we have center (-3, 2) and
slopes +/- 5/12.
y – 2 = 5/12 (x + 3)
y – 2 = (5/12) x + 15/12
y = (5/12) x + 13/4
y – 2 = -5/12 (x + 3)
y – 2 = (-5/12) x + -15/12
y = (-5/12) x + 3/4
7-5 Problems
Find the coordinates of the vertices and the foci. Give the
asymptote slopes for each hyperbola. Then draw the
graph.
1)
x2 _ y2 = 1
9
49
2)
25x2 – 4y2 = 100
2) Vertices: (-2, 0) (2, 0) Foci: ( 29, 0) (- 29, 0) Slope = +/- 5/2
1)
2)
1) Vertices: (-3, 0) (3, 0) Foci: (- 58, 0) ( 58, 0)
Slope = +/- 7/3
7-6 Conic Sections
•
•
•
•
Circles, ellipses, parabolas, and hyperbolas are all formed when a
double cone is sliced by a plane
The general equation of any conic section is :
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
The standard equations for each specific conic section are listed in
previous sections
If B = 0 and you look at A and C in the equations:
Conic Section
Name
Relationship of A and C
Parabola
A = 0 or C= 0, but never
both equal to 0
Circle
A=C
Ellipse
A and C have the same
sign, but A = C
Hyperbola
A and C have opposite
signs
7-6 Example
Identify 9x2 + 16y2 – 54x + 64y + 1 = 0 as one of the four
conic sections. Then graph the conic section.
9x2 + 16y2 – 54x + 64y = -1
9 (x2 – 6x + o) + 16(y2 + 4y + o) = -1 + 9(o) + 16(o)
9 (x2 – 6x + 9) + 16(y2 + 4y + 4) = -1 + 9(9) + 16(4)
9(x – 3)2 + 16(y + 2)2 = 144
(x – 3)2
(y + 2)2
+
16
9
=1
This conic section is an ellipse.
7-6 Problems
Write the equation in standard form and decide if the conic
section is a parabola, a circle, an ellipse, or a
hyperbola. Then graph the equation.
x2 + y2 + 6x = 7
5x2 – 6y2 – 30x – 12y + 9 = 0
1)
2)
1) (x + 3)2 + (y)2 = 16 circle
2) (x – 3)2 _ (y + 1)2 = 1
6
5
1)
2)
hyperbola
7-7 Solving Quadratic Systems
• When you solve a system of quadratic equations the
method is almost the same as solving a system of linear
equations
• If the system has one equation of a conic section and
one equation of a straight line, you can get zero, one, or
two solutions to the system
• If both the equations are conic sections, the system
should have zero, one, two, three, or four solutions
7-7 Example
Solve this system of equations using algebraic methods and by graphing the
equations.
y = (x – 2)2 + 1
y = -4x + 5
Set the equations equal to each other to solve for x.
-4x + 5 = (x – 2)2 + 1
-4x + 5 = x2 – 4x + 4 + 1
-4x + 5 = x2 – 4x + 5
5 = x2 + 5
x2 = 0
x=0
Then put x = 0 back in to solve for
y.
y = -4(0) + 5
y=5
The system of equations has one
solution, (0, 5). The graphs of
these equations confirms this.
7-7 Problems
Solve these systems of equations by using algebra and
graphing the equations.
4x2 + y2 = 25
2x2 – y2 = -1
2)
x2 + y2 = 10
y = x2 – 4
1) Four solutions: { (2, 3) (2, -3) (-2, 3) (-2, -3) }
1)
2) Four solutions: { (1, -3) (-1, -3) ( 6, 2) (- 6, -2) }

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