Circles, Parabolas, Ellipses, and Hyperbolas

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Table of ContentsTitle Slide

Equations of the Curves centered at the origin (0,0)

Equations for Curves not centered at the originCircles

Parabolas

Ellipse

HyperbolasExample 1 Circle

How to Solve Example 1 CircleExample 2 Parabola

Solution to Example 2 ParabolaExample 3 Ellipse

How to Solve Example 3 Ellipse

Example 4 Hyperbola

How to Solve Example 4 HyperbolaProblem 1Problem 2Problem 3 Problem 4More PracticeReferences

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Equations of the Curves centered at the origin (0,0)

• Circle

• Parabola or

• Ellipse

where a>b>0

• Hyperbola

or

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2

2

2

b

y

a

x

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2

b

y

a

x1

2

2

2

2

b

x

a

y

2axy 2ayx

222 ryx

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Equations for Curves not centered at the origin

• Circle Centered at (h, k)

• Parabola

or • Ellipse where a>b>0

Hyperbola or centered at (h, k)

222 )()( rkyhx

1)()(

2

2

2

2

b

ky

a

hx1

)()(2

2

2

2

a

hx

b

ky

1)()(

2

2

2

2

b

ky

a

hx

khxay 2)( khyax 2)(

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Circles

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.

222 )()( rkyhx 222 ryx

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Parabolas

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.

khxay 2)( khyax 2)(

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Ellipse1

)()(2

2

2

2

b

ky

a

hx1

2

2

2

2

b

y

a

x

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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Hyperbola1

)()(2

2

2

2

b

ky

a

hx1

)()(2

2

2

2

a

hx

b

ky

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 1Circle

Graph the following equation of a circle

(x-3)2 +(y-3)2=16Also find without graphing

•the center for the circle and •the radius for the circle.

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How to Solve Example 1Circle

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

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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-2

-5 5

r

C: (3.00, 3.00)

E: (-1.00, 3.00)

D: (3.00, -1.00)

B: (7.00, 3.00)

A: (3.00, 7.00)

BE

A

D

C

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Example 2Parabola

Graph the following equation of the parabola.

Y= 2(x+2)2 + 1

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 2Parabola

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.

2. By looking at the equation the vertex of this equation is (-2, 1).

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

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 to create the parabola.

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2

-2

-5

D: (-4.00, 9.00)

C: (-3.00, 3.00)

B: (0.00, 9.00)

A: (-1.00, 3.00)

V: (-2.00, 1.00)

f x = 2x+2 2+1D B

C A

V

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Example 3Ellipse

Graph the Following equation

For the above equation find without graphing

•The center of the ellipse

•State the major and minor axis

•Find the vertices of the ellipse.

19

)3(

4

)1( 22

yx

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How to Solve Example 3Ellipse

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. By comparing the equations we know that k=3, h=1, a=2, and b=3.

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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)1( 22

yx

1)()(

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How to Solve Example 3 Continued

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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-2

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Minor

Major

D: (-1.00, 3.00)E: (1.00, 0.00)

B: (3.00, 3.00)

A: (1.00, 6.00)

C: (1.00, 3.00)

D B

E

A

C

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Example 4Hyperbola

Graph the following equations

Before graphing the equation find

•The center

•The asymptotes and their equations

•The vertices of the hyperbola

•What way does the hyperbola open?

19

)3(

4

)2( 22

yx

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How to Solve Example 4Hyperbola

1. First we compare the given equation of the hyperbola with the standard equation to find the value of h, k, a, and b. In this equation we can see that h=2, k=3, a=2, and b=3.

2. Next we know that the center of a hyperbola is (h, k). So for this hyperbola the center is at (2,3).

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).

19

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)2( 22

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1)()(

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How to Solve Example 4 Continued

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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-2

-4

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C: (2.00, 3.00)

V1: (4.00, 3.00)

V2: (0.00, 3.00)

V2 V1C

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Problem 1

Click on the correct answer to move to the next question.

What type of object/curve is produced by the equation below curve? Also find the center from the equation.

19

)4(

36

)6( 22

yx

A. Circle (6,4)

B. Ellipse (6,2)

C. Ellipse (6,4)

D. Hyperbola (6,4)

E. Circle (6,2)

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Problem 2Click on the correct answer to continue.

What is the center of this hyperbola? What are the equations for the asymptotes from this hyperbolic equation?

125

)9(

36

)7( 22

yx

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 Click on the correct answer to move to the next problem.

What type of object/ curve is given by the equation below? What is the center of the equation?

10)2(3 2 xyA. Circle Center (3,2)

B. Ellipse Center (2,10)

C. Parabola Center (2,10)

D. Parabola Center (3,2)

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Problem 4 Click the correct answer to continue.

What is the center and the radius of the following circle equation?

100)1( 22 yx

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

Print and complete the following worksheets.

Writing Equations from Graphs

Graphing Curves

Equations of Curves

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References

Barnett, Raymond A., Michael R. Ziegler, and Karl E. Byleen. PreCalculus graphs and models. Boston: McGraw Hill, 2005.

Calvert, J. B. “Ellipse”. 2005. 3 Dec.2009 http://mysite.du.edu/~jcalvert/math/ ellipse.htm

Dawkins, Paul. “Algebra.” 2009. 3 Dec. 2009 <http://tutorial .math.lamar.edu/classes/alg/hyperbolas.aspx>.

Picture from google images: <http://www.stickergiant.com/Merchant2/imgs/450/ss35_450.jpeg>

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