Chapter 8 Conic Sections

Worksheet 44 (8.1)

Chapter 8 Conic Sections

8.1 Graphing Parabolas

Summary 1:

Parabolas

The graph of any equation of the form y = ax2 + bx + c where a,b, and c are real numbers and a  0 is a parabola.

Parabolas of this type will always be symmetrical about a vertical line called the line of symmetry. This line will pass through the vertex of the parabola and divide it into 2 symmetrical halves. The vertex of the parabola is the lowest point or minimum value if the parabola opens upward. The vertex is the highest point or maximum value if the parabola opens downward.

Many parabolas will be exhibit y-axis symmetry. The graph of an equation is symmetric with respect to the y-axis if replacing x with -x results in an equivalent equation.

Summary 2:

Graphing Parabolas of the Form y = x2+ k

The basic parabola is produced by the equation y = x2. Its vertex is at the origin (0,0) and its line of symmetry is x = 0.

Adding a number to x2, for example, y = x2 + 2, results in the graph movingup two units.

Subtracting a number from x2, for example, y = x2 - 1, results in the graph moving down one unit.

In general, the graph of a quadratic equation of the form y = x2 + k is the same as the graph of y = x2 except moved up or down k units depending on whether k is positive or negative. The graph of y = x2 + k is a vertical translationof the graph of y = x2. All equations of this type exhibit y- axis symmetry.

Worksheet 44 (8.1)

Warm-up 1.a) Graph: y = x2 + 2

x / y = x2 + 2
0 / 2
1
2
-1
-2

Note: The graph moved up two units because 2 was added to x2.

Problem

1. Graph: y = x2 - 1y

Summary 3:

Graphing Parabolas of the Form y = ax2

Multiplying x2 by a number whose absolute value is greater than 1, for example, y = 3x2, results in the graph becoming narrower. Multiplying x2 by a number whose absolute value is less than 1, for example, y = ¼x2, results in the graph becoming wider.

In general, the graph of a quadratic equation of the form y = ax2 is narrowerthan the basic parabola if a > 1 and wider if a < 1.

If a is a positive number the parabola will open upward. If a is a negative number the parabola will open downward.

Worksheet 44 (8.1)

Warm-up 2.a) Graph: y = 3x2

x / y = 3x2
0 / 0
1 / 3
2 / 12
-1
-2

Note: The graph became narrower due to the coefficient of 3 in front of x2.

Problem

2. Graph: y = ¼x2y

Warm-up 3.a) Graph y = -½x2

x / y= -½x2
0 / 0
1 / -½
2 / -2
-1
-2

Note: The graph became wider and opens downward.

Worksheet 44 (8.1)

Problem

3. Graph: y = -2x2 y

Summary 4:

Graphing Parabolas in the Form y = (x - h)2

Adding a number to x before it is squared, for example, y = (x + 3)2, results in the graph moving left 3 units.

Subtracting a number from x before it is squared, for example, y = (x - 1)2, results in the graph moving right 1 unit.

In general, the graph of a quadratic equation of the form y = (x - h)2is the same as the graph of the basic parabola, y = x2 but moved to the right

h units if h is positive or moved to the left h units if h is negative.

Warm-up 4.a) Graph: y = (x + 3)2

x / y = (x + 3)2
-5 / 4
-4
-3 / 0
-2
-1

Note: The graph moved left 3 units.

Problem

4. Graph: y = (x - 1)2y

Worksheet 44 (8.1)

Warm-up 5.a) Graph: y = -2(x - 3)2 + 1

To predict the changes in the graph, note what is happening to the equation:

1. The negative in front of 2 will make the graph open downward.

2. The 2 in front of the squared term will make the graph become narrower.

3. The 3 subtracted from x will move the graph to the right 3 units.

4. The 1 added to the squared term will move the graph up 1 unit.

x / y = -2(x - 3)2 + 1
1 / -7
2
3 / 1
4
5

Problem

5. Graph: y = -(x + 2)2 - 4y

Note: Equations of parabolas in the form illustrated in warm-up 5, and problem 5 are in standard form: y = a(x - h)2 + k, where (h,k) is the vertex.

Worksheet 45 (8.2)

8.2 More Parabolas and Some Circles

Summary 1:

Standard Form

The general form of a quadratic equation, y = ax2 + bx + c, can be changed to standard form, y = a(x - h)2 + k, by the process of completingthe square.

In standard form, facts about the graph of the equation are more easily interpreted.

Summary 2:

Completing the Square

1. If the coefficient of the squared term is not 1, factor the coefficient from the 2 terms containing an x.

2. Leave the constant term on the right. (Do not factor the coefficient from the constant term.)

3. Take one-half of the coefficient of the x term, square it and add this quantity beside the two terms containing an x.

4. To the right of the constant, perform the opposite operation to compensate for the change made in the equation in step 3.

5. Write the equation in standard form, by factoring the perfect square trinomial created and simplifying the constants.

Warm-up 1.a) Change the equation to standard form by completing the square,

and graph the equation.

y = x2 + 2x - 3Take ½ of 2, square it andadd

y = (x2 + 2x + ) - 3 - beside 2x. Subtract the same quantity from the constant.

y = (x + 1)2 - Write in standard form.

The vertex = ( , ). The parabola will move left and down .

Worksheet 45 (8.2)

Warm-up 1.a) continued

x / y = (x + 1)2 - 4
-1 / -4 (vertex)
-2
0
-3
1

In many applications of parabolas we are interested in the x-intercepts. The last two values in the table are the x-intercepts of this parabola. These values can easily be found by replacing y with 0 and solving the resulting equation:

0 = x2 + 2x - 3

0 = (x + 3)(x - 1)

x + 3 = 0 or x - 1 = 0

x = -3 or x = 1

The x-intercepts are (-3, 0) and (1, 0).

b) Change the equation to standard form by completing the square and graph:

y = -2x2 + 8x - 7

y = -2(x2 - 4x ) - 7Factor -2 from the first two terms,

move -7 to the right. y = -2(x2 - 4x + ) - 7 + Take ½ of the coefficient of x, square it, and add inside the parentheses. To the right of -7 add 8 to compensate for the change made of -2(4) = -8.

y = -2( )2 + Write in standard form.

Vertex = (2, 1)

The parabola will: a) reflect over the x axis

b) become narrower

c) move 2 units right

d) move 1 unit up

x / y= -2(x - 2)2 + 1
2 / 1
1
3
0
4

Note: The first ordered pair in the table is the vertex. Choose two points on either side of the vertex.

Worksheet 45 (8.2)

Problems - Change the equation to standard form by completing the square and graph.

1. y = x2 + 6x + 4y

2. y = -2x2 - 4x + 2y

Summary 3:

Circles

A circle is the set of all points in a plane equidistant from a given fixed point called the center.

The radius of the circle is any line segment connecting the center with any point on the circle.

The general form of the equation of a circle is:

x2 + y2 + Dx + Ey + F = 0 where D, E and F are integers

The standard form of the equation of a circle is:

(x - h)2 + (y - k)2 = r2 where (h,k) is the center, and r is the radius.

The standard form of the equation of a circle with a center at the origin is:

x2 + y2 = r2

Standard form can be used to:

1. Find the equation of a circle given the center and radius.

2. Find the center and the radius of a circle given the equation.

Worksheet 45 (8.2)

Warm-up 2.a) Write the equation of a circle that has a center at (-2, 3) and a radius of 4. Express your final equation in the general form:

x2 + y2 + Dx + Ey + F = 0

1) Begin with standard form:

(x - h)2 + (y - k)2 = r2

2) Substitute in values: (h,k) = (-2, 3) and r = 4

(x - )2 + (y - )2 = ( )2

(x + )2 + (y - )2 = ( )2

3) Multiply and write in general form.

x2 + 4x + 4 + - + =

x2 + y2 + 4x - - = 0

b) Find the center and radius of the circle and graph.

x2 + y2 - 2x + 6y + 6 = 0

To find the center and radius of the circle, the equation needs to be written in standard form. Complete the square on x and y to write the equation in standard form.

1) Group the terms with x together and the terms with y together, move the constant to the right side of the equation.

(x2 - 2x ) + (y2 + 6y ) = -6

2) Take ½ the coefficient of the x term, square it and add inside the first parentheses. Do the same procedure to complete the square on y. Add the same two quantities to the right side of the equation.

(x2 - 2x + ) + (y2 + 6y + ) = -6 + +

(x - )2 + (y + )2 =

3) Rewrite in standard form:

(x - )2 + (y + )2 = ( )2

Center = ( , ) and radius =

Worksheet 45 (8.2)

Warm-up 2b) continued

4) Graph the equation obtained in step 3:

Problems

3. Write the equation of a circle with a center at (3,-1) and a radius of 3. Express your final equation in general form.

4. Rewrite the equation in standard form to find the center and radius of the circle: x2 + y2 - 8x + 4y - 5 = 0

Worksheet 46 (8.3)

8.3 Graphing Ellipses

Summary 1:

Ellipses

Ax2 + By2 = C is an equation for an ellipse, if A, B, and C are of the same sign, but AB. To graph an ellipse, find the x and y intercepts and join with a smooth curve.

Warm-up 1.a) Graph: 25x2 + 9y2 = 225

1) Let y = 0 and find the x-intercepts:

25x2 + 9(0)2= 225

25x2= 225

x2= 9

x = ±3

x-intercepts: (3, 0) and (-3, 0)

2) Let x = 0 and find the y-intercepts:

25(0)2 + 9y2= 225

9y2= 225

y2=

y-intercepts: (0, ) and (0, )

Note: The line segment containing the y-intercepts of (0, 5) and (0, -5) is calledthe major axis of the ellipse. The shorter line segment containing the x-intercepts of (3, 0) and (-3, 0) is called the minor axis. Both the x and y axes serve as axes of symmetry.

3) Graph the intercepts and join with a smooth curve:

Worksheet 46 (8.3)

Problem

1. Graph: 4x2 + 16y2 = 64y

Summary 2:

Other Ellipses

Ellipses whose centers are not at the origin can be graphed by completing the square as done with circles.

Warm-up 2.a) Graph x2 - 6x + 4y2 + 16y + 9 = 0

Complete the square on both x and y.

(x2 - 6x + ) + 4(y2 + 4y + )= -9 + +

(x - )2 + 4(y + 2)2=

The graph is an ellipse because the 1 and 4 are the same sign, but not equal. The center of the ellipse is at (3, -2).

To find the endpoints of the major axis let y = -2 and solve for x.

(x - 3)2 + 4(-2 + 2)2= 16

(x - 3)2 + 4( )2= 16

(x - 3)2= 16

x - 3= ±

x= 3 ±

x= or x =

The endpoints of the major axis are (7, -2) and (-1, -2).

To find the endpoints of the minor axis let x = 3 and solve for y.

(3 - 3)2 + 4(y + 2)2= 16

( )2 + 4(y + 2)2= 16

Worksheet 46 (8.3)

Warm-up 2. continued

4(y + 2)2= 16

(y + 2)2=

y + 2= ±

y = -2 ±

y= or y =

The endpoints of the minor axis are (3, ) and (3, ).

Graph the 2 sets of endpoints and draw a smooth curve.

Problem

2. Graph: 4x2 + 32x + y2 - 2y + 29 = 0y

Worksheet 47 (8.4)

8.4 Graphing Hyperbolas

Summary 1:

Hyperbolas

Ax2 + By2 = C is an equation for a hyperbola if A, B, and C are nonzero constants and A and B are unlike signs. A hyperbola will have either

x-intercepts or y-intercepts, but never both.

Warm-up 1.a) Graph: x2 - 4y2 = 16

1) Let y = 0 and find the x-intercepts:

x2 - 4(0)2= 16

x2= 16

x-intercepts are ( , 0) and ( ,0)

2) Let x = 0, solve for y, there are no y-intercepts.

02 - 4y2= 16

-4y2= 16

y2= -4

The square root of -4 is not a real number, therefore there are no y-intercepts.

3) Solve the equation for y to find other points.

x2 - 4y2= 16

-4y2= -x2 + 16

y2=

Worksheet 47 (8.4)

Note: We must choose values where x  4 or x  -4 so the radicand remains positive. Complete the remainder of the table.

x / y
4 / 0
-4
5 / ±3/2
-5
6 / ±5±2.2
-6

Note: To help us sketch the hyperbola we can draw in two dashed lines called asymptotes. Each branch of the hyperbola approaches these lines, but does not intersect them. To find the equations of the asymptotes, let the constant term be = 0, and solve for y.

4) Equations of asymptotes:

x2 - 4y2= 0

-4y2= -x2

y2=

y= ±½x

The lines y = ½x and y = -½x are asymptotes.

5) Graph the hyperbola by sketching in the asymptotes as dashed lines, graphing the points in the table, and connecting the points with a smooth curve for each branch, never intersecting the asymptotes.

Worksheet 47 (8.4)

Problem

1. Graph: y2 - 9x2 = 9

Summary 2:

Other Hyperbolas

Hyperbolas whose centers are not at the origin can be graphed by completing the square as done with ellipses.

Warm-up 2.a) Graph 9x2 + 36x -25y2 + 50y - 214 = 0

1) Complete the square on both x and y.

9(x2 + 4x + ) - 25(y2 - 2y + )= 214 + -

9(x + )2 - 25(y - 1)2=

The graph is a hyperbola because the 9 and -25 have different signs. The center of the hyperbola is at (-2, 1).

2) Let y = 1 and solve for x:

9(x + 2)2 - 25(1 - 1)2= 225

9 (x + 2)2 - 25( )2= 225

(x + 2)2= 225

x + 2= ±

x= 2 ±

x= or x =

The hyperbola intersects the horizontal line y = 1 at (7, 1) and (-1, 1).

3) Let x = -2 and solve for y:

9(-2 + 2)2 - 25(y - 1)2= 225

9 ( )2 - 25(y - 1)2= 225

-25(y - 1)2= 225

(y - 1)2= -9

Worksheet 47 (8.4)

Warm-up 2. continued

The square root of -9 is not a real number, therefore the hyperbola does not intersect the vertical line x = -2.

3) Solve the equation for y to find other points.

9(x + 2)2 - 25(y - 1)2= 225

-25(y - 1)2= 225 - 9(x + 2)2

y - 1 =

Note: We must choose values where x  3 or x  -7 so the radicand remains positive. Complete the remainder of the table.

x / y
3 / 0
4
5 / 5.9 and -3.9
-7
-8 / 4.6 and -2.6
-9

4) Equations of asymptotes:

9(x + 2)2 - 25(y - 1)2= 0

(3(x + 2) + 5(y - 1))(3(x + 2) + 5(y - 1)) = 0

(3(x + 2) + 5(y - 1)) = 0 (3(x + 2) - 5(y - 1)) = 0

5y - 5 = 3x + 6 5y - 5 = -3x - 6

y = y =

Worksheet 47 (8.4)

Warm-up 2. continued

The lines y =and y =are asymptotes.

Problem

2. Graph: 9x2 - 18x - 4y2 - 8y - 31 = 0y