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Conics

- Chapter 12

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

Distance and Midpoint Formulas

- Section 12.1

The Distance Formula

The Distance Formula The distance between two

points P1 (x1, y1) and P2 (x2, y2), is

The Distance Formula

Example Determine the distance between (?3, ?5)

and (3, 3).

The Distance Formula

Example Find the distance between ( 6, 6) and

( 5, 2).

The Midpoint Formula

The midpoint of a line segment is the point

located exactly halfway between the two endpoints

of the line segment.

The Midpoint Formula

Example Find the midpoint of the line segment

joining P1 (0, 8) and P2 (4, 6).

M (2, 1)

Circles

- Section 12.2

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.

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.

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

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.

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.

Parabolas

- Section 12.3

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.

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).

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.

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.

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 )

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.

A Parabola Whose Vertex is Not the Origin

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.

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.

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)

Ellipses

- Section 12.4

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

Ellipses with Center at Origin

y

P (x, y)

x

F2 (c, 0)

F1 ( c, 0)

Ellipses with Center at Origin

Example

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.

An Ellipse Whose Center is Not the Origin

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).

Hyperbolas

- Section 12.5

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

Equations of a Hyperbola

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.

Hyperbolas with Center at Origin

Example continued

Locate points above and below the foci.

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.

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.

Asymptotes

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.

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.

Nonlinear Systems of Equations

- Section 12.6

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.

Solving Using Substitution

Example continued

Solve the resulting quadratic equation.

Continued.

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.

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.

Solving Using Substitution

Example continued

The graph of the system also verifies the

solution.

x 3y 5

x2 y2 25

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.

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.

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

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.

Solving Using Elimination

Example continued

Substitute the values for y into either equation.

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.

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).