Algebra II Notes – Unit Ten: Conic Sections

Page 1 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

Syllabus Objective: 10.1 – The student will sketch the graph of a conic section with centers either at or not at the origin. (PARABOLAS)

( )1 1,x yReview: The Midpoint Formula

The midpoint M of the line segment connecting the points and ( )2 2,x y is

1 2 1 2,2 2

x x y yM + + =

.

Ex: Find the midpoint of the line segment joining ( )7,1− and ( )2,5− .

Let ( ) ( )1 1, 7,1x y = − and ( ) ( )2 2, 2,5x y = − . Substitute the values into the formula and simplify.

1 2 1 2 7 2 1 5 9, , ,32 2 2 2 2

x x y yM + + − + − + = = = −

Conic Sections

: curves that are formed by the intersection of a plane and a double-napped cone

Parabola:

( ) ( )2 4y k p x h− = −

the conic formed by connecting all points equidistant from a point (the focus)

and a line (the directrix) with the equation of the form or

( ) ( )2 4x h p y k− = −

Vertex ( ),h k: the point, , that lies on the axis of symmetry halfway between the focus and directrix

Axis of Symmetry

( ) ( )2 4y k p x h− = −

: the line of symmetry of a parabola that passes through the vertex.

For , the axis of symmetry is horizontal, and has the equation x h= . For

( ) ( )2 4x h p y k− = − , the axis of symmetry is vertical, and has the equation y k= .

p : the distance the focus and directrix are from the vertex

Focus: a point on the axis of symmetry of a parabola equidistant from the vertex as the directrix

Directrix: a line perpendicular to the axis of symmetry equidistant from the vertex as the focus

Algebra II Notes – Unit Ten: Conic Sections

Page 2 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

Parabola 2 4 , 0x py p= > Parabola 2 4 , 0x py p= <

Parabola 2 4 , 0y px p= > Parabola 2 4 , 0y px p= >

Ex: Sketch the graph of the parabola 213

x y= − .

Vertex: ( )0,0 Because y is squared, the axis of symmetry is horizontal: 0y =

Find p: 2 3y x= − 4 3

34

p

p

= −

= − Because p is negative,

the parabola will open to the left.

Directrix: 34

x = Focus: 3 ,04

−

Directrix y p= −

Focus

( )0, p

Vertex ( )0,0

Directrix y p= −

Focus

( )0, p

Vertex ( )0,0

Directrix x p= − Directrix

x p= −

Focus

( ),0p

Focus

( ),0p

Vertex

( )0,0

Vertex

( )0,0

Algebra II Notes – Unit Ten: Conic Sections

Page 3 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

Standard Form of the Equation of a Parabola:

( ) ( )2 4y k p x h− = −

Horizontal Axis (opens left or right):

Vertex: ( ),h k Axis of Symmetry: y k= Focus: ( ),h p k+ Directrix: x h p= −

Vertical Axis (opens up or down): ( ) ( )2 4x h p y k− = −

Vertex: ( ),h k Axis of Symmetry: x h= Focus: ( ),h k p+ Directrix: y k p= −

Ex: Sketch the graph of the parabola ( ) ( )27 8 3x y+ = − . Identify the vertex, focus, and

directrix.

Because it is in the form ( ) ( )2 4x h p y k− = − , the parabola has a

vertical axis.

Find p: 4 8

2p

p==

p is positive, so the parabola opens up.

Vertex: ( )7,3−

Focus: p units up from the vertex ( ) ( )7,3 2 7,5− + = −

Directrix: horizontal line p units down from the vertex

3 2 1y y= − ⇒ =

Ex: Write an equation of the parabola whose vertex is at ( )3,2 and whose focus

is at ( )4,2 .

Begin with a sketch. The parabola opens toward the focus, so it opens right.

Find p: The distance from the focus to the vertex is 1. The parabola opens right, so p = 1.

Because the parabola has a horizontal axis, we will use the equation ( ) ( )2 4y k p x h− = − .

The vertex ( ) ( )3,2 ,h k= , and p = 1. ( ) ( )22 4 3y x− = −

Ex: A store uses a parabolic mirror to see all of the aisles in the store. A cross section of the mirror is shown. Write an equation for the cross section of the mirror and identify the focus.

Algebra II Notes – Unit Ten: Conic Sections

Page 4 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

Because the parabola has a vertical axis, we will use the equation ( ) ( )2 4x h p y k− = − . The vertex is

( ) ( )0,0 ,h k= , so we now have 2 4x py= .

The parabola passes through the point ( ) ( )8,2 ,x y= . Use this to find p.

( ) ( )22 4 8 4 2 64 8 8x py p p p= ⇒ = ⇒ = ⇒ =

The equation is ( )2 24 8 32x y x y= ⇒ = . The focus is p units up from the vertex: ( )0,8

You Try: ( ) ( )26 4 8y x− = − − Graph the equation . Identify the vertex, focus, and directrix.

QOD: Parabolas can be found many places in real life. Find at least three real-life examples of parabolas. What is the significance of the focus in these real-life examples?

Algebra II Notes – Unit Ten: Conic Sections

Page 5 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

Syllabus Objective: 10.1 – The student will sketch the graph of a conic section with centers either at or not at the origin. (CIRCLES)

( )1 1,x y

Review: The Distance Formula

The distance d between the points and ( )2 2,x y is ( ) ( )2 22 1 2 1d x x y y= − + − .

Use the Pythagorean Theorem to show this on the coordinate plane by forming a right triangle with the points ( )1 1,x y , ( )2 2,x y , and ( )2 1,x y . Ex: Find the distance between the points ( )2, 4− and ( )5, 1− − . Let ( ) ( )2 2, 5, 1x y = − − and ( ) ( )1 1, 2, 4x y = − . Substitute the values into the formula and simplify.

( ) ( ) ( ) ( )( ) ( ) ( )22 2 2 2 22 1 2 1 5 2 1 4 7 3 49 9 58d x x y y= − + − = − − + − − − = − + = + =

Circle:

( ) ( )22 2x h y k r− + − =

the conic formed by connecting all points equidistant from a point with the equation of

the form

Center of a Circle ( ),h k: the point, , that is equidistant from all points on the circle

Radius

Ex: Show that the equation of a circle with center

: the distance r between the center and any point on the circle

( ),h k and radius r is

( ) ( )22 2x h y k r− + − = .

Let one point on the circle be ( ),x y and the center be ( ),h k . The radius, r, is the distance between the

center and a point on the circle.

( ) ( )2 22 1 2 1d x x y y= − + − ⇒ ( ) ( )2 2r x h y h= − + −

Squaring both sides, we have ( ) ( )22 2r x h y k= − + − .

Ex: Draw the circle given by the equation 2 24y x= − .

Write in standard form. 2 2 4x y+ =

Identify the center and the radius. Center: ( )0,0 Radius: 2 4 2r r= ⇒ =

Sketch the graph by plotting the center then plotting points on the circle 2 units

Algebra II Notes – Unit Ten: Conic Sections

Page 6 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

above, below, and to the left and right of the center.

Ex: Write an equation of the circle with center ( )1,4− and radius 3.

Center: ( ) ( )1,4 ,h k− = Radius: 3r =

Equation: ( ) ( ) ( )( ) ( ) ( ) ( )22 2 2 22 2 21 4 3 1 4 9x h y k r x y x y− + − = ⇒ − − + − = ⇒ + + − =

Ex: Write an equation of the circle shown.

The center of the circle is ( ) ( )0,0 ,h k= .

Equation: 2 2 2x y r+ =

To find r, we will use the point given on the circle ( )5,1 .

( ) ( )2 22 2 2 25 1 26x y r r r+ = ⇒ + = ⇒ =

The equation of the circle is 2 2 26x y+ = .

Ex: A street light can be seen on the ground within 30 yd of its center. You are driving and are 10 yd east and 25 yd south of the light. Write an inequality to describe the region on the ground that is lit by the light. Is the street light visible?

Write the equation of the circle (use the center ( )0,0 ). 2 2 2 2 230 900x y x y+ = ⇒ + =

The region lit by the light is the region inside the circle, so we want to include all distances less than or

equal to the radius. 2 2 900x y+ ≤

To check if the street light is visible, substitute the point ( )10,25 into the inequality.

( ) ( )2 210 25 900100 625 900 true

+ ≤

+ ≤ YES, the street light is visible.

You Try ( ) ( )2 24 5 36x y+ + − =: Graph the circle.

QOD: Describe how to graph a circle on the graphing calculator.

Algebra II Notes – Unit Ten: Conic Sections

Page 7 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

Syllabus Objective: 10.1 – The student will sketch the graph of a conic section with centers either at or not at the origin. (ELLIPSES) Ellipse: the set of all points P such that the sum of the distances between P and two distinct fixed points is a constant.

Foci: the two fixed points that create an ellipse

Vertices: the two points at which the line through the foci intersect the ellipse

Major Axis: the line segment joining the two vertices

Center of the Ellipse: the midpoint of the major axis

Co-Vertices: the two points at which the line perpendicular to the major axis at the center intersects the ellipse

Minor Axis: the line segment joining the co-vertices

Equation of an Ellipse

( ) ( )2 2

2 2 1x h y k

a b− −

+ =

:

: Horizontal Major Axis

Center: ( ),h k Vertices: ( ),h a k± Co-Vertices: ( ),h k b± Foci: ( ),h c k±

( ) ( )2 2

2 2 1x h y k

b a− −

+ = : Vertical Major Axis

Center: ( ),h k Vertices: ( ),h k a± Co-Vertices: ( ),h b k± Foci: ( ),h k c±

Note: 2 2 2c a b= − The foci of the ellipse lie on the major axis, c units from the center where .

Ellipse centered at the origin with a horizontal major axis:

Algebra II Notes – Unit Ten: Conic Sections

Page 8 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

Ex: Draw the ellipse given by 2 24 25 100x y+ = . Identify the foci.

Write the equation in standard form. (Must be set equal to 1.)

2 2

2 2

4 25 100100 100 100

125 4

x y

x y

+ =

+ = Center: ( )0,0

Length of Major Axis (horizontal) = 2a = 10

Length of Minor Axis (vertical) = 2b = 4

2 25 4 21

21

c

c

= − =

= Foci: ( ) ( )21,0 , 21,0−

Ex: Write an equation of the ellipse with center at ( )0,0 , a vertex at ( )0, 3− , and a co-vertex at

( )1,0− .

The vertex is at ( )0, 3− , so the ellipse has a vertical major axis. We will use ( ) ( )2 2

2 2 1x h y k

b a− −

+ = .

The center is ( ) ( )0,0 ,h k= . 2 2

2 2 1x yb a

+ =

The distance between the center and the vertex is 3, so 3a = . The distance between the center and the

co-vertex is 1, so 1b = . 2 2 2

22 2 1 1

1 3 9x y yx+ = ⇒ + =

Ex: Graph the ellipse ( ) ( )2 26 2

125 100

x y− −+ = . Identify the center, vertices, co-vertices, and

foci.

Center: ( )6,2 Length of Major Axis (vertical) = 2a = 20

Length of Minor Axis (horizontal) = 2b = 10

Vertices: ( ) ( ) ( )6,2 10 6, 8 and 6,12± = − Co-Vertices: ( ) ( ) ( )6 5,2 1,2 and 11,2± =

Algebra II Notes – Unit Ten: Conic Sections

Page 9 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

2 2 2 100 25 75c a b= − = − = 75 5 3c = =

Foci: ( ) ( ) ( )6,2 5 3 6, 6.66 and 6,10.66± ≈ −

Ex: The planet Jupiter ranges from 460.2 million miles away to 507.0 million miles away from the sun. The center of the sun is a focus of the orbit. If Jupiter’s orbit is elliptical, write an equation for its orbit in millions of miles.

460.2507.0

a ca c− =+ =

2 967.2

483.6a

a==

507.0

483.6 507.023.4

a cc

c

+ =+ =

= ( ) ( )

2 2 2

2 2 2

2

23.4 483.6

233321.4

c a b

b

b

= −

= −

=

2 2

1233868.96 233321.4

x y+ =

You Try ( )0,0: Write an equation of the ellipse with the center at , vertex at ( )4,0 , and focus at ( )2,0 .

QOD: How can you tell from the equation of an ellipse whether the major axis is horizontal or vertical?

Algebra II Notes – Unit Ten: Conic Sections

Page 10 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

Syllabus Objective: 10.1 – The student will sketch the graph of a conic section with centers either at or not at the origin. (HYPERBOLAS)

Hyperbola: the set of all points P such that the difference of the distances from P to two fixed points, called the foci is constant

Vertices: the two points at which the line through the foci intersects the hyperbola

Transverse Axis: the line segment joining the vertices

Center: the midpoint of the transverse axis

( ) ( )2 2

2 2 1x h y k

a b− −

− =

Equation of a Hyperbola

: Horizontal Transverse Axis

Center: ( ),h k Vertices: ( ),h a k±

( ) ( )2 2

2 2 1y k x h

a b− −

− = : Vertical Transverse Axis

Center: ( ),h k Vertices: ( ),h k a±

Note: 2 2 2c a b= + The foci of the hyperbola lie on the transverse axis, c units from the center where .

Hyperbolas have slant asymptotes. Draw the rectangle formed by the vertices and the points ( ),h k b±

for horizontal and ( ),h b k± for vertical. The lines that pass through the corners of this rectangle are the

slant asymptotes of the hyperbola.

Ex: Draw the hyperbola given by the equation 2 29 16 144x y− = . Find the vertices, foci, and asymptotes.

Write the equation in standard form (set equal to 1).

2 2

2 2

9 16 144144 144 144

116 9

x y

x y

− =

− =

Center: ( )0,0 Vertices: ( ) ( )

2 16 44,0 and 4,0

a a= ⇒ =

− Foci:

( ) ( )

2 2 2 2 16 9 255 5,0 and 5,0

c a b cc

= + ⇒ = + =

= ⇒ −

Algebra II Notes – Unit Ten: Conic Sections

Page 11 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

Asymptotes: 3 3and 4 4

y x y x= − =

Note: The hyperbola itself is only the curve.

(Dash the asymptotes.)

Ex: Write an equation of the hyperbola with foci at ( )0, 2− and ( )0,2 and vertices at ( )0, 1−

and ( )0,1 .

The foci and vertices lie on the y-axis, so the transverse axis is vertical. We will use the equation

( ) ( )2 2

2 2 1y k x h

a b− −

− = . Center ( ) ( ), 0,0h k = . This is the midpoint of the vertices.

2 2

2 2 1y xa b

− =

The foci are 2 units from the center, so 2c = . The vertices are 1 unit from the center, so 1a = .

2 2 2 2 2

2

2 13

c a b bb

= + ⇒ = +

=

2 2 2

22 1 1

1 3 3y x xy− = ⇒ − =

Ex: Graph the hyperbola ( ) ( )22 2

5 116

yx

+− − = .

Center ( ) ( ), 5, 2h k = − 1, 4a b= =

Vertices: ( ) ( ) ( )5 1, 2 4, 2 an d6, 2± − = − −

Note: The hyperbola itself is only the curve.

(Dash the asymptotes.)

Algebra II Notes – Unit Ten: Conic Sections

Page 12 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

Ex: The diagram shows the hyperbolic cross section of a large hourglass. Write an equation that models the curved sides.

Center ( ) ( ), 0,0h k = 2a =

Vertices: ( ) ( )2,0 and 2,0−

The transverse axis is horizontal, so we will use the equation

( ) ( )2 2

2 2 1x h y k

a b− −

− = .

2 2

2 2 12x y

b− = Substitute in a point on the hyperbola ( ) ( )4,6 ,x y= and solve for b.

2 2

2 2 2

2 2

4 6 36 361 4 1 34

3 36 12b b b

b b

− = ⇒ − = ⇒ − = −

= ⇒ =

2 2

14 12x y

− =

You Try ( )2 21 16 16x y+ − =: Graph the hyperbola . Identify the vertices and foci.

QOD2 2

2 2 1x ya b

− =: What are the asymptotes of the hyperbola ? What are the asymptotes of the hyperbola

2 2

2 2 1y xa b

− = ?

Algebra II Notes – Unit Ten: Conic Sections

Page 13 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

Syllabus Objective: 10.2 – The student will classify a conic section given its equation with its center either at or not at the origin.

General Form of a Second-Degree Equation: 2 2 0Ax Bxy Cy Dx Ey F+ + + + + =

Discriminant 2 4B AC−:

Classifying a Conic from Its Equation 2 2 0Ax Bxy Cy Dx Ey F+ + + + + =

If 2 4 0B AC− < and 0B = and A C= : CIRCLE

If 2 4 0B AC− < and 0B ≠ or A C≠ : ELLIPSE

If 2 4 0B AC− = : PARABOLA

If 2 4 0B AC− > : HYPERBOLA

2 22 8 6 0x y y+ + + =

Graphing from the General Form

To graph from general form, complete the square for both variables to write in standard form.

Ex: Classify the conic given by . Then write in standard form and graph.

( ) ( )2 24 0 4 2 1 8B AC− = − = − Because A C≠ , this is an ellipse.

Complete the square: ( )( )

2 2

22

2 8 16 6 0 16

2 4 10

x y y

x y

+ + + + = +

+ + =

Rewrite in standard form (set equal to 1):

( )

( )

22

22

42 1010 10 10

41

5 10

yx

yx

++ =

++ =

Center: ( )0, 4− 5 2.236a = ≈

10 3.162b = ≈

Vertices: ( ) ( ) ( )0, 4 3.162 0, 0.838 , 0, 7.162− ± = − −

Co-Vertices: ( ) ( ) ( )0 2.236, 4 2.236, 4 , 2.236, 4± − = − − −

Algebra II Notes – Unit Ten: Conic Sections

Page 14 of 14 McDougal Littell: 10.1 – 10.6 Alg II Notes Unit 10: Conic Sections

Ex: Classify the conic given by 2 24 16 4 4 0x y x y− − − − = . Then write in standard form and graph.

( ) ( )2 24 0 4 4 1 16 0B AC− = − − = > This is a hyperbola.

Complete the square: ( ) ( )( ) ( )

2 2

2 2

4 4 4 4 4 4 0 16 4

4 2 2 16

x x y y

x y

− + − + + − = + −

− − + =

Rewrite in standard form (set equal to 1):

( ) ( )

( ) ( )

2 2

2 2

4 2 2 1616 16 16

2 21

4 16

x y

x y

− +− =

− +− =

Center: ( )2, 2− 4, 2a b= =

Transverse Axis is horizontal.

Vertices: ( ) ( ) ( )2 2, 2 0, 2 , 4, 2± − = − −

You Try 2 2 20 94 0y x y− − + =: Classify the conic given by . Then write in standard form and graph.

QOD: Describe how to determine which conic section an equation represents in general form.