Transcript Circles, Parabolas, Ellipses, and Hyperbolas

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Circles, Parabolas, Ellipses, and
Hyperbolas
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Table of Contents
Title Slide
Table of Contents
Equations of the Curves centered at the origin (0,0)
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Equations for Curves not centered at the origin
Circles
Parabolas
Ellipse
Hyperbolas
Example 1 Circle
How to Solve Example 1 Circle
Example 2 Parabola
Solution to Example 2 Parabola
Example 3 Ellipse
How to Solve Example 3 Ellipse
Example 4 Hyperbola
How to Solve Example 4 Hyperbola
Problem 1
Problem 2
Problem 3
Problem 4
More Practice
References
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Equations of the Curves
centered at the origin (0,0)
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• Circle
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x y r
2
2
2
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• Parabola
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y  ax
2
x  ay
2
or
• Ellipse
x2 y 2
 2 1
2
a
b
where a>b>0
• Hyperbola
x2 y2
 2  1 or
2
a
b
y2 x2
 2 1
2
a
b
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Equations for Curves
not centered at the origin
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• Circle
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( x  h)  ( y  k )  r
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2
2
2
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Centered at (h, k)
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• Parabola
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y  a( x  h)  k
2
or
x  a( y  h)  k
2
• Ellipse
( x  h) 2 ( y  k ) 2

 1 where a>b>0
2
2
a
b
Hyperbola
or
( x  h) 2 ( y  k ) 2

1
2
2
a
b
centered at (h, k)
( y  k ) 2 ( x  h) 2

1
2
2
b
a
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Circles
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( x  h)  ( y  k )  r
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2
2
2
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x y r
2
2
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2
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Circles are a special type of ellipses. There is a center that is the
same distance from every point on the diameter. In the equation
the center is at (h, k). The distance from the center to any point
on the line is called the radius of the circle. From the equation to
find the radius you take the square root of r2.
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y  a( x  h)  k
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Parabolas
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x  a( y  h)  k
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A parabola is a curve that is oriented either up, down, left, or right. The vertex of
the parabola is at (h, k). In the equation the h value added or subtracted to x
moves the parabola left and right. If you subtract the value of h the parabola
moves to the right. If you add the value of h the parabola moves to the left. The
parabola can be made skinnier and wider by changing the value of a in the
equation. If a is a whole number the parabola will become skinnier; if it’s a
fraction the parabola will become wider. When you add or subtract a value of k
the parabola moves up or down by the value. Parabolas are symmetrical across
the line through the vertex of the parabola.
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Ellipse
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( x  h) 2 ( y  k ) 2

1
a2
b2
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x2 y 2
 2 1
2
a
b
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The center of the ellipse is at (h, k). The radius of ellipses are not a constant
distance from the center. To find the distance to the curve from the center you have
to find the distance from the center to the curve for the x and y separately, these
points are called vertices. The vertices are on the major axis and minor axis. The
major axis is the longer axis and the minor axis is the shorter axis through the
center of the ellipse. To find the distance from the center in the x direction you take
the square root of a2. To find the distance from the center in the y direction you
take the square root of b2. You then will have two points on the x direction and two
points in the y direction and you use these four points to draw your ellipse. Ellipses
are symmetrical across both of there axis's.
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Hyperbola
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( x  h) 2 ( y  k ) 2

1
2
2
a
b
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( y  k ) 2 ( x  h) 2

1
2
2
b
a
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Hyperbolas look like two parabolas opening in opposite directions. The equations of the
asymptotes are y= k + (b/a)(x-h). The asymptotes help you to graph the hyperbola. The
center of the hyperbola is also at (h, k). The vertices of the hyperbola depend on whether
the hyperbolas open left and right or up and down. You can determine which way the
hyperbola opens by looking to see if the x or y term has a negative sign. In the equation in
the upper left corner the y term has the negative and since the y has the negative the
hyperbolas open left and right. When opening left and right the vertices are (h+a, k). For
the equation in the upper right corner the x value has the negative sign which means that
the hyperbola opens up and down. The vertices for an equation that opens up and down
are (h, k+b).
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Example 1
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Circle
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Graph the following equation of a circle
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(x-3)2 +(y-3)2=16
Also find without graphing
•the center for the circle
and
•the radius for the circle.
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How to Solve Example 1
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Circle
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1. First we start by looking at the equation and comparing the given equation to
the standard equation. The equation is (x-3)2 +(y-3)2=16. We know the form
of the equation of a circle is (x-h)2 +(y-k)2=r2.From the two equations we
know that h=3 and k=3, therefore our center is (3,3). We have 16= r2 so r=4.
2. Since we know our center and radius we can start to graph our function. We
start by graphing the center of the circle by plotting (3,3)
3. Next we use the radius to find the four points on the circle that are to the
direct right, left, up, and down from the center. You find these points by
adding/ subtracting the radius to the value of h, and then adding/ subtracting
the radius from the value of k. So you have (3-4, 3) = (-1,3), (3+4, 3) = (7,3),
(3, 3-4) = (3, -1), and (3, 3+4) = (3, 7).
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How to Solve Example 1
Continued
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A
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4.Next you plot the four
points that you just
found.
5.Then connect the four
points around the
center to create your
circle.
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C: (3.00, 3.00)
A: (3.00, 7.00)
B: (7.00, 3.00)
D: (3.00, -1.00)
E: (-1.00, 3.00)
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r
E
B
C
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D
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Example 2
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Parabola
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Graph the following equation of the parabola.
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Y= 2(x+2)2 + 1
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For the above equation find the following before graphing the
equation
•The vertex of the parabola.
•Is the parabola skinnier or wider then a standard parabola.
•What way does the parabola open?
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Solution to Example 2
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Parabola
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1. First we look at the equation compared to the standard parabola equation. Y=2(x+2)2 +1
Y=a (x + h)2 + k. So a = 2, h = 2, and k = 1.
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2. By looking at the equation the vertex of this equation is (-2, 1).
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3. We know that a = 2 which is a whole number and whole numbers make the parabola
skinner. We also know that since a is positive and that the equation equals y that the
parabola will open up.
4. To graph the equation the first thing that we do is plot the vertex of the parabola.
5. Next we need to pick a value of x that is an equal distance from the center to solve for y.
these two y values should be the same since the parabola is symmetrical through the
center. Lets use x=-3 and x=-1. So for x=-3, y=2(-3+2)2+1 = 3 and for x=-1, y= 2(1+2)2+1= 3
So we have the points (-3,3) and (-1,3).
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Solution to Example 2 Continued
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6. Next we pick at least two more values of x
to find values of y’s for to be able to graph
the parabola accurately. Lets use x = -4 and
x= 0.
So for x = -4, y = 2(-4+2)2+1 = 2(-2)2+1= 9
and for x = 0, y = 2(0+2)2+1= 9.
So we have the points (-4,9) and (0,9).
7. Now we can graph the center of the
parabola (-2,1) and the points that we have
found to lie on the parabola (-1,3), (-3,3),
(-4,9), and (0,9).
8. Next we can connect the pints and continue
the ends of the curve up-15 to create the
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parabola.
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fx  = 2x+22 +1
V: (-2.00, 1.00)
D
B
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A: (-1.00, 3.00)
B: (0.00, 9.00)
C: (-3.00, 3.00)
D: (-4.00, 9.00)
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C
A
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V
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Example 3
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Ellipse
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Graph the Following equation
( x  1) 2 ( y  3) 2

1
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9
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For the above equation find without graphing
•The center of the ellipse
•State the major and minor axis
•Find the vertices of the ellipse.
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How to Solve Example 3
Ellipse
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1. First we compare the equation of the ellipse to the standard equation of the ellipse to
find the values of h, k, a, and b.
2
2
2
2
(
x

1
)
(
y

3
)
( x  h)
( y  k)

1

1
2
2
4
9
a
b
By comparing the equations we know that k=3, h=1, a=2, and b=3.
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2. We know that the center is at (h, k) so our center is at (1,3).
3. We can also determine our major axis and minor axis by looking at our values of a and
b. The larger value of a and b is associated with the major axis and the smaller value is
associated with the minor axis. In this case b is larger then a, so b is associated with the
major axis. This means that the major axis is the diameter of the ellipse that is parallel
to the y axis. The value of a is smaller, so it is associated with the minor axis and this
axis is the diameter parallel to the x axis.
4. We find our vertices by adding and subtracting our value of a to the x value of the
center and then adding and subtracting our value of b to the y value of the center. So in
this case (1+2, 3) = (3,3), (1-2, 3) =( -1,3), (1, 3+3) = (1,6), and (1, 3-3) =(1, 0).
5. Now we can graph the ellipse.
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How to Solve Example 3 Continued
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A
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6. First you can plot the center
of the ellipse.
7. Now plot the four vertices.
8. Then connect the four vertices
with a curved line in the
shape of an ellipse
9. You can now draw in and
label your major and minor
axis if you need to.
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C: (1.00, 3.00)
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Minor
D
C
2
B
A: (1.00, 6.00)
B: (3.00, 3.00)
E: (1.00, 0.00)
D: (-1.00, 3.00)
Major
E
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Example 4
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Hyperbola
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Graph the following equations
( x  2) 2 ( y  3) 2

1
4
9
Before graphing the equation find
•The center
•The asymptotes and their equations
•The vertices of the hyperbola
•What way does the hyperbola open?
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How to Solve Example 4
Hyperbola
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1. First we compare the given equation of the hyperbola with the standard equation to find
the value of h, k, a, and b.
2
2
2
2
(
x

2
)
(
y

3
)
( x  h) ( y  k )

1

1
2
2
4
9
a
b
In this equation we can see that h=2, k=3, a=2, and b=3.
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2. Next we know that the center of a hyperbola is (h, k). So for this hyperbola the center is at
(2,3).
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3. Next we can find the values of the asymptotes which are +/- b/a. We can also find the
equations of the asymptotes which are in the form k+(b/a)(x-h). So for this hyperbola the
values of the asymptotes are +3/2. The equations for the asymptotes are 3+(3/2)(x-2).
4. Next we need to determine what direction the hyperbola opens. We know that the way the
hyperbola opens depends on which value x or y has the negative sign associated with it. In
this equation the y value has the negative sign associated with it. This tells us that the
hyperbolas open left and right.
5. Since we know which direction that the hyperbola opens we can find our vertices. You
find the vertices by adding and subtracting the value of a or b to the center. Since the
hyperbola opens left and right we know that we need to add and subtract a value of a to
the center. So we have (2+2, 3) =(4,3) and (2-2, 3) = (0,3).
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How to Solve Example 4 Continued
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6. To graph the hyperbolas we need to
graph our asymptotes.
7. Now we need to plot the two vertices.
8. We can now draw in our hyperbola by
drawing two parabola looking curves
through the vertices that get close to the
asymptotes but never touch the
asymptotes and so that the parabola
shaped curves open left and right.
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C: (2.00, 3.00)
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V1: (4.00, 3.00)
V2: (0.00, 3.00)
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V2
C
V1
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Problem 1
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Click on the correct answer to move to the next question.
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What type of object/curve is produced by the equation below
curve? Also find the center from the equation.
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A. Circle (6,4)
B. Ellipse (6,2)
C. Ellipse (6,4)
D. Hyperbola (6,4)
E. Circle (6,2)
( x  6) 2 ( y  4) 2

1
36
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Problem 2
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Click on the correct answer to continue.
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What is the center of this hyperbola? What are the equations for
the asymptotes from this hyperbolic equation?
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( x  7) 2 ( y  9) 2

1
36
25
A. Center (7,9) Asymptotes 9+(36/25)(x-7) and 9-(36/25)(x-7)
B. Center (6,5) Asymptotes 5+(36/25)(x-6) and 5-(36/25)(x-6)
C. Center (5,6) Asymptotes 25+(7/5)(x-36) and 25-(7/5)(x-36)
D. Center (7,9) Asymptotes 9+(6/5)(x-7) and 9-(6/5)(x-7)
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Problem 3
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Click on the correct answer to move to the next problem.
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What type of object/ curve is given by the equation below? What is
the center of the equation?
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A. Circle Center (3,2)
B. Ellipse Center (2,10)
C. Parabola Center (2,10)
D. Parabola Center (3,2)
y  3( x  2)2  10
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Problem 4
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Click the correct answer to continue.
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What is the center and the radius of the following circle equation?
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( x 1)  y  100
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A. Center (1,10) Radius = 10
B. Center (10,1) Radius = 1
C. Center (0,1) Radius = 10
D. Center (1,0) Radius = 10
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More Practice
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Print and complete the following worksheets.
Writing Equations from Graphs
Graphing Curves
Equations of Curves
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References
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Barnett, Raymond A., Michael R. Ziegler, and Karl E. Byleen. PreCalculus graphs and
models. Boston: McGraw Hill, 2005.
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Calvert, J. B. “Ellipse”. 2005. 3 Dec.2009 http://mysite.du.edu/~jcalvert/math/
ellipse.htm
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Dawkins, Paul. “Algebra.” 2009. 3 Dec. 2009 <http://tutorial
.math.lamar.edu/classes/alg/hyperbolas.aspx>.
Picture from google images:
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