2017-2018 hart interactive honors geometry module 5 ... · module 5+: conics this file derived from...

2017-2018

Hart Interactive

Honors Geometry

Module 5+

Student Workbook

Conics

Lessons 1- 8

Name: _____________________________

Period: ______

Teacher: ___________________________

GEOMETRY • MODULE 5+

Hart Interactive –Geometry

Mathematics Curriculum Table of Contents1 Student Workbook

Conics Unit 5+: Conics

Unit 5+ Vocabulary ................................................................................................................................ S.1

Lesson 1: Writing the Equation for a Circle .......................................................................................... S.5

Lesson 2: Recognizing Equations of Circles ....................................................................................... S.13

Lesson 3: Tangent Lines to Circles ..................................................................................................... S.23

Lesson 4: Equations for Tangent Lines to Circles................................................................................ S.31

Lesson 5: Curves in Geometry – The Ellipse ....................................................................................... S.39

Lesson 6: The Definition of a Parabola .............................................................................................. S.51

Lesson 7: Connecting to the Vertex Form of a Parabola ................................................................... S.61

Lesson 8: Curves from Geometry – Hyperbolas ................................................................................ S.69

1Each lesson is ONE day, and ONE day is considered a 45-minute period.

Module 5+: Conics

1

This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from GEO-M2-TE-1.3.0-08.2015

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.

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Hart Interactive – Geometry Plus Standard M5+ Unit Table of Contents HONORS GEOMETRY

Module 5+: Conics

2







HONORS GEOMETRY • MODULE 5+

Hart Interactive – Geometry Plus Standard

Conics Vocabulary

Knowledge Rating*

Lesson Vocabulary

Terms Definition Picture/Example/Notation

1 Circle A circle is the set of _____ points in a plane that are a given __________ from a single point.

1 Center The center of a circle is the given __________ from which all points on a circle are the same __________.

1 Radius The radius is the __________ from the center of a circle to a __________on the circle.

1 Equations of a Circle

(𝑥𝑥 − 𝑎𝑎)2 + (𝑦𝑦 − 𝑏𝑏)2 = 𝑟𝑟2 is the center-radius form of the general equation for any circle with radius _____ and center (______, ____).

1 Tangent A tangent is a line which __________ a circle at just one point.

2 Completing the Square

Completing the square is a technique used to solve quadratic equations, graph quadratic functions, and rewrite equations of circles.

2 Perfect Square Trinomials

Perfect square trinomials are algebraic expressions with __________terms that are created by multiplying a binomial to __________.

Knowledge Rating: N = I have no knowledge of the word. S = I’ve seen the word, but I’m not sure what it means. U = I understand this word and can use it correctly.

Module 5+: Conics

S.1







HONORS GEOMETRY

M5+ Conics Vocabulary Hart Interactive – Geometry Plus Standard

Knowledge Rating*

Lesson Vocabulary


3 Secant of a Circle

A secant is a line that intersects a circle at _____ points.

5 Ellipse

An ellipse is the set of points such that the _____ of the distances to two fixed points (the foci) is __________.

5 Focus (Foci) The foci of an ellipse are two __________ points on the interior of an ellipse

5 Equation of an Ellipse

6 Parabola

For a given point, called the focus, and a given line not through the focus, called the directrix, a parabola is the ____ of points such that the __________ to the focus equals the distance to the directrix.

6 Directrix A directrix is a straight __________ used in constructing a curve such as a parabola.


Module 5+: Conics S.2







HONORS GEOMETRY


Knowledge Rating*

Lesson Vocabulary


6 Vertex of a Parabola

The vertex of a parabola is the point where the parabola crosses its _____ of symmetry. If the coefficient of the x2 term is __________, the vertex will be the lowest point on the graph, the point at the bottom of the “ U ”-shape. If the coefficient of the x2 term is __________, the vertex will be the highest point on the graph, the point at the top of the “ U ”-shape.

6

Axis of Symmetry of a Parabola

The axis of symmetry of a parabola is a vertical __________that divides the parabola into two congruent halves.

6 Equation of a Parabola

Vertical Parabola Horizontal Parabola

2( )y a x h k= − + 2( )x a y k h= − +

8 Hyperbola

A hyperbola can be defined as follows: For two given points, the foci, a hyperbola is the _____ of points such that the difference between the distances to each focus is __________.

8 Asymptotes

An asymptote is a value that you get closer and closer to, but never quite __________. It is a horizontal, vertical, or slanted line that a graph approaches but never __________.

8 Equation of a Hyperbola or

2 2 2 2

2 2 2 21 1x y y xa b a b

− = − =









HONORS GEOMETRY


Writing about Math

Use at least five words from the Unit 5 vocabulary list to write a one paragraph “news” story. You may want to use some of the sentence starters listed below to help you. Remember this is a fake news story so have fun with it but be sure to be accurate in your mathematics.

It is amazing to think about_________ Experts agree that ________________

Let me explain ___________________ Historically, ______________________

You’ll be excited to learn that________ It’s hard to believe, but _____________

It is interesting to note that_________ You will find that __________________

Experts agree that _______________ You’ll soon discover why ____________

It’s incredible that _______________ You’ll soon discover that ____________

Most often, _____________________ No one will argue that ______________

Many people believe that ___________ Without a doubt, __________________

In my experience, ________________ What’s so great about, _____________?

There are many reasons why ________ You may be surprised to learn that _____

Positively, ______________________ Strangely enough, __________________

Try to visualize __________________ Most people are unfamiliar with ________

Surprisingly, ____________________ Experience shows that ______________








HONORS GEOMETRY

Hart Interactive – Geometry Plus Standard M5+ Lesson 1

Lesson 1: Writing the Equation for a Circle

Exploratory Activity

1. A. If we graph all of the points whose distance from the origin is equal to 5, what shape will be formed? Try it in the grid at the right.

B. Our goal now is to find the coordinates of eight of those points that comprise the circle. Four are very easy to find. What are they?

If we need to locate more points on the circle, how could we do that? We need the distance from the origin (i.e., the center of the circle) to be 5. Graphically, we are looking for the coordinates (𝑥𝑥,𝑦𝑦) that are exactly 5 units from the center of the circle.

C. Identify one other point on the circle that appears to be 5 units from the center of the circle. What can we do to be sure that the distance between the center of the circle and the identified point is in fact 5?

D. Use the Pythagorean Theorem to create an equation for this circle.

Lesson 1: Writing the Equation for a Circle Module 5+: Conics

S.5







HONORS GEOMETRY


2. Let’s look at another circle, one whose center is not at the origin. Shown at the right is a circle with center (2, 3) and radius 5.

A. Again, there are four points that are easy to

locate and others that can be verified using the Pythagorean Theorem or distance formula. List all 12 points that can easily be read from the graph.

B. What are the differences between this circle

and the one we just looked at in Exercise 1? C. Are the circles congruent? Is there a sequence of basic rigid motions that would take this circle to the

origin? Explain. D. What effect does the translation have on all of the points from the circle above?

The effect that translation has on the points can be expressed as the following. Let (𝑥𝑥,𝑦𝑦) be any point on the circle with center (2,3). Then, the coordinates of all of the points (𝑥𝑥,𝑦𝑦) after the translation are �(𝑥𝑥 − 2), (𝑦𝑦 − 3)�.

E. Since the radius is equal to 5, we can locate any point (𝑥𝑥,𝑦𝑦) on the circle using the Pythagorean Theorem as we did before.

(𝑥𝑥 − 2)2 + (𝑦𝑦 − 3)2 = 52 The solutions to this equation are all the points of a circle whose radius is 5 and center is at (2,3). What do the numbers 2, 3, and 5 represent in the equation above?


S.6







HONORS GEOMETRY


3. Assume we have a circle with radius 5 whose center is at (𝑎𝑎, 𝑏𝑏). What is an equation whose graph is that circle?

4. Assume we have a circle with radius 𝑟𝑟 whose center is at (𝑎𝑎, 𝑏𝑏). What is an equation whose graph is that

circle?

5. Write an equation for the circle whose center is at (9, 0) and has radius 7.

6. Write an equation whose graph is the circle at the right.

7. What is the radius and center of the circle given by the equation (𝑥𝑥 + 12)2 + (𝑦𝑦 − 4)2 = 81?

(𝑥𝑥 − ______)2 + (𝑦𝑦 − _____)2 = ______2


S.7







HONORS GEOMETRY


8. Petra is given the equation (𝑥𝑥 − 15)2 + (𝑦𝑦 + 4)2 = 100 and identifies its graph as a circle whose center is (−15, 4) and radius is 10. Has Petra made a mistake? Explain.

9. A. What is the radius of the circle with center (3, 10) that passes through (12, 12)?

B. What is the equation of this circle?

10. A circle with center (2,−5) is tangent to the 𝑥𝑥-axis. A. What is the radius of the circle?

B. What is the equation of the circle?


S.8







HONORS GEOMETRY


11. Two points in the plane, 𝐴𝐴(−3, 8) and 𝐵𝐵(17, 8), represent the endpoints of the diameter of a circle. A. What is the center of the circle? Explain how you found the answer.

B. What is the radius of the circle? Explain.

C. Write the equation of the circle.

12. Consider the circles with the following equations:

𝑥𝑥2 + 𝑦𝑦2 = 25 and

(𝑥𝑥 − 9)2 + (𝑦𝑦 − 12)2 = 100.

A. What are the radii of the circles?

B. What is the distance between the centers of the circles?

C. Make a rough sketch of the two circles to explain why the circles must be tangent to one another.


S.9







HONORS GEOMETRY


13. A circle is given by the equation (𝑥𝑥2 + 2𝑥𝑥 + 1) + (𝑦𝑦2 + 4𝑦𝑦 + 4) = 121.

A. What is the center of the circle?

B. What is the radius of the circle?

C. Describe what you had to do in order to determine the center and the radius of the circle.

14. Determine the center and radius of the circle B. 2(𝑥𝑥 + 1)2 + 2(𝑦𝑦 + 2)2 = 10

Lesson Summary

(𝑥𝑥 − 𝑎𝑎)2 + (𝑦𝑦 − 𝑏𝑏)2 = 𝑟𝑟2 is the center-radius form of the general equation for any circle with radius 𝑟𝑟 and center (𝑎𝑎, 𝑏𝑏).


S.10







HONORS GEOMETRY


Homework Problem Set

1. Write the equation for a circle with center �12

, 37� and radius √13.

2. What is the center and radius of the circle given by the equation 𝑥𝑥2 + (𝑦𝑦 − 11)2 = 144? 3. A circle is given by the equation 𝑥𝑥2 + 𝑦𝑦2 = 100. Which of the following points are on the circle?

A. (0, 10) B. (−8, 6) C. (−10,−10) D. (45, 55) E. (−10, 0) 4. Determine the center and radius of each circle.

A. 3𝑥𝑥2 + 3𝑦𝑦2 = 7 B. 4(𝑥𝑥 − 2)2 + 4(𝑦𝑦 − 9)2 − 64 = 0

5. A circle has center (−13,𝜋𝜋) and passes through the point (2,𝜋𝜋).

A. What is the radius of the circle? B. Write the equation of the circle.

6. Two points in the plane, 𝐴𝐴(19, 4) and 𝐵𝐵(19,−6), represent the endpoints of the diameter of a circle.

A. What is the center of the circle? B. What is the radius of the circle? C. Write the equation of the circle.


S.11







HONORS GEOMETRY


7. Write the equation of the circle shown below. 8. Write the equation of the circle shown below.

9. Consider the circles with the following equations:

𝑥𝑥2 + 𝑦𝑦2 = 2 and

(𝑥𝑥 − 3)2 + (𝑦𝑦 − 3)2 = 32.

A. What are the radii of the two circles?

B. What is the distance between their centers? C. Make a rough sketch of the two circles to explain why the circles must be tangent to one another.


S.12







HONORS GEOMETRY


Lesson 2: Recognizing Equations of Circles

Opening Exercise

1. Justin says the following is the equation of a circle with radius 5 and center (1, 2).

𝑥𝑥2 − 2𝑥𝑥 + 1 + 𝑦𝑦2 − 4𝑦𝑦 + 4 = 25

A. How can you verify Justin’s statement?

B. Work with your group to determine if Justin is correct.

Lesson 2: Recognizing Equations of Circles Module 5+: Conics

S.13







HONORS GEOMETRY


From Exercise 1, you can see how important it is to be able to factor trinomial squares and complete the square in order to recognize equations of circles. Before we do that, let’s remember the patterns that are formed from squaring a binomial. Squaring a Binomial 2. In Algebra 1, you learned to multiply binomials using an area model or double distribution. Use the

models below to help you rewrite each expression. Use one method while your partner uses the other method. Then switch for the second problem.

A. Use an area model to express (𝑥𝑥 − 5)2 as a trinomial.

A. Use the distributive property to express (𝑥𝑥 − 5)2 as a trinomial.

( 5)( 5)x x− − =

2( 5) _____________________x − = 2( 5) _____________________x − =

B. Use the distributive property to express (𝑥𝑥 + 4)2 as a trinomial.

( 4)( 4)x x+ + =

B. Use an area model to express (𝑥𝑥 + 4)2 as a trinomial.

2( 4) _____________________x + = 2( 4) _____________________x + =


S.14







HONORS GEOMETRY


3. Use an area model to square each binomial below.

A. Use an area model to express (𝑎𝑎 + 𝑏𝑏)2 as a trinomial.

B. Use an area model to express (𝑎𝑎 − 𝑏𝑏)2 as a trinomial.

2( ) _____________________a b+ = 2( ) _____________________a b− =

Trinomials in this form are called Perfect Square Trinomials. They are important for to us when we want to complete the square. 4. For each trinomial below, decide if it is a perfect square trinomial. If it is, then factor it.

A. 𝑥𝑥2 + 12𝑥𝑥 + 36 B. 2 14 49x x− + C. 2 2y y+ −

D. 2 20 100y y− + E. 2 2 3x x+ − F. 2 2 1y y+ +

5. Rewrite the following equations in the form (𝑥𝑥 − 𝑎𝑎)2 + (𝑦𝑦 − 𝑏𝑏)2 = 𝑟𝑟2.

A. 𝑥𝑥2 + 4𝑥𝑥 + 4 + 𝑦𝑦2 − 6𝑥𝑥 + 9 = 36 B. 𝑥𝑥2 − 10𝑥𝑥 + 25 + 𝑦𝑦2 + 14𝑦𝑦 + 49 = 4

6. What is the center and radius of each circle in Exercise 5?


S.15







HONORS GEOMETRY


7. For each circle below, rewrite the equation and then give the center and radius.

A. 2 28 16 6 9 25x x y y+ + + − + = B. 2 214 49 2 1 1x x y y− + + + + =

Completing the Square Sometimes an equation will not be easy to put in the form (𝑥𝑥 − 𝑎𝑎)2 + (𝑦𝑦 − 𝑏𝑏)2 = 𝑟𝑟2. Equations like

2 6 40x x+ = are not perfect square trinomials, but we can manipulate them to get them into that form. This is called completing the square.

8. The diagram below shows how this was done using an area model. Discuss with your group what is going on in the diagram. Then complete the square to solve the following equation: 𝑥𝑥2 + 6𝑥𝑥 = 40.

𝑥𝑥 𝑥𝑥

3

3

𝑥𝑥2 3𝑥𝑥

3𝑥𝑥 ? 40

=

𝑥𝑥 𝑥𝑥

3

3

𝑥𝑥2 3𝑥𝑥

3𝑥𝑥 9 49

=


S.16







HONORS GEOMETRY


9. Explain the steps that Traci used to rewrite the equation 𝑥𝑥2 + 4𝑥𝑥 + 𝑦𝑦2 − 12𝑦𝑦 = 41 so that she could find the center and radius of the circle it defines.

𝑥𝑥2 + 4𝑥𝑥 + 𝑦𝑦2 − 12𝑦𝑦 = 41 Given equation.

(𝑥𝑥2 + 4𝑥𝑥) + (𝑦𝑦2 − 12𝑦𝑦) = 41

(𝑥𝑥2 + 4𝑥𝑥 + _____) + (𝑦𝑦2 − 12𝑦𝑦 + _____) = 41 + _____ + _____

(𝑥𝑥2 + 4𝑥𝑥 + 4) + (𝑦𝑦2 − 12𝑦𝑦 + 36) = 41 + 4 + 36

(𝑥𝑥 + 2)2 + (𝑦𝑦 − 6)2 = 81

Center: (-2, 6) Radius: 9

10. Identify the center and radius for each of the following circles. A. 𝑥𝑥2 − 20𝑥𝑥 + 𝑦𝑦2 + 6𝑦𝑦 = 35

B. 𝑥𝑥2 − 3𝑥𝑥 + 𝑦𝑦2 − 5𝑦𝑦 = 192


S.17







HONORS GEOMETRY


11. Could the circle with equation 𝑥𝑥2 − 6𝑥𝑥 + 𝑦𝑦2 − 7 = 0 have a radius of 4? Why or why not?

12. Stella says the equation 𝑥𝑥2 − 8𝑥𝑥 + 𝑦𝑦2 + 2y = 5 has a center of (4,−1) and a radius of 5. Is she correct? Why or why not?

13. Could 𝑥𝑥2 + 𝑦𝑦2 + 𝐴𝐴𝑥𝑥 + 𝐵𝐵𝑦𝑦 + 𝐶𝐶 = 0 represent a circle? Explain your thinking.


S.18







HONORS GEOMETRY


Circle, Point or Empty Set?

Once we have completed the square and have the equation in the form (𝒙𝒙 − 𝒂𝒂)𝟐𝟐 + (𝒚𝒚 − 𝒃𝒃)𝟐𝟐 = 𝒓𝒓𝟐𝟐, we can use the constant on the right side to determine if we really do have a circle, or if this is the equation of a point, or if there is no possible solution.

14. Identify the graphs of the following equations as a circle, a point, or an empty set. A. 𝑥𝑥2 + 𝑦𝑦2 + 4𝑥𝑥 = 0 B. 𝑥𝑥2 + 𝑦𝑦2 + 6𝑥𝑥 − 4𝑦𝑦 + 15 = 0 C. 𝑥𝑥2 + 𝑦𝑦2 − 6𝑥𝑥 + 4𝑦𝑦 + 13 = 0

15. Summarize with your group, the conditions to create a circle, point or the empty set from an equation of the form (𝑥𝑥 − 𝑎𝑎)2 + (𝑦𝑦 − 𝑏𝑏)2 = 𝑟𝑟2.


S.19







HONORS GEOMETRY


Lesson Summary

[source: http://www.regentsprep.org/regents/math/algtrig/ate12/completesqlesson.htm]


S.20







http://www.regentsprep.org/regents/math/algtrig/ate12/completesqlesson.htm

HONORS GEOMETRY


Homework Problem Set 1. Identify the centers and radii of the following circles.

A. (𝑥𝑥 + 25)2 + 𝑦𝑦2 = 1 B. 𝑥𝑥2 + 2𝑥𝑥 + 𝑦𝑦2 − 8𝑦𝑦 = 8

C. 𝑥𝑥2 − 20𝑥𝑥 + 𝑦𝑦2 − 10𝑦𝑦 + 25 = 0 D. 𝑥𝑥2 + 𝑦𝑦2 = 19

2. Sketch graphs of the following equations. A. 𝑥𝑥2 + 𝑦𝑦2 + 10𝑥𝑥 − 4𝑦𝑦 + 33 = 0 B. 𝑥𝑥2 + 𝑦𝑦2 + 14𝑥𝑥 − 16𝑦𝑦 + 104 = 0

C. 𝑥𝑥2 + 𝑦𝑦2 + 4𝑥𝑥 − 10𝑦𝑦 + 29 = 0


S.21







HONORS GEOMETRY


3. Chante claims that two circles given by (𝑥𝑥 + 2)2 + (𝑦𝑦 − 4)2 = 49 and 𝑥𝑥2 + 𝑦𝑦2 − 6𝑥𝑥 + 16𝑦𝑦 + 37 = 0 are

externally tangent. She is right. Show that she is.

4. Draw a circle. Randomly select a point in the interior of the circle; label the point 𝐴𝐴. Construct the greatest radius circle with center 𝐴𝐴 that lies within the circular region defined by the original circle. Hint: Draw a line through the center, the circle, and point 𝐴𝐴.


S.22







Hart Interactive – Geometry Plus Standard M5+ Lesson 3 HONORS GEOMETRY

Lesson 3: Tangent Lines to Circles

Exploration Activity

You will need: a straightedge, compass and sheet of unlined paper or patty paper

1. Which line is the tangent? Which is the secant? How do you know?

2. Can you visualize another line through point 𝐶𝐶 that is also tangent to the circle? Try to draw a second tangent line through point 𝐶𝐶, and label its point of intersection with the circle as point 𝐹𝐹.

3. How can you map CD

onto CF

?

Lesson 3: Tangent Lines to Circles Module 5+: Conics

S.23

© 2015 Common Core, Inc. Some rights reserved. commoncore.org This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.





Paper Folding to Get a Tangent

4. Use a sheet of unlined white paper or patty paper and choose a point in the middle of your paper and label it O. Use your compass to draw a circle with center O. Choose a point on Circle O and label it D. Use your straightedge to draw the line through O and D, extending it beyond the circle.

5. How can you fold the paper so that the crease is tangent to the circle at point D? Describe the process below and then fold your paper to create the tangent.

6. What do you notice about the tangent and the line DO? 7. Choose a point on the crease, and label it C. Fold the paper in such a way that you get a second tangent to

the circle through point C. Describe your method below.

8. With your partner, discuss how you could construct tangent CD instead of using paper folding. Write or sketch your ideas in the space below.


S.24






9. Use the steps you and your partner wrote to construct a tangent to the circle below at point D.

Extending our Focus 10. To construct a tangent to a circle from a point not on the circle we need to think back to the work you did

in geometry. Use the circle below to determine the angle measure of ∠BCD.

•

• D


S.25






11. What changes will you need to make if point C is not on the circle? Construct the tangent to the circle from the point not on the circle. Write your steps below.

12. Analyze the construction shown below. Argue that the dashed lines shown are tangent to the circle with

center 𝐵𝐵.


S.26






13. Use a compass to construct a line that is tangent to the circle at point 𝐹𝐹. Then choose a point 𝐺𝐺 on the tangent line, and construct another tangent to the circle through 𝐺𝐺.

F

O


S.27






Lesson Summary

Constructing a tangent to a circle from a point outside the circle.

Steps Construction

Given circle with center C and a point P outside of the circle.

Connect point C and point P and then find midpoint M by bisecting the segment.

Using M as the center, draw the circle passing through points C and P. Let N be the point of intersection of the two circles

Construct a line through P and N. This is the desired tangent line.

[source: http://pdfs.cpm.org/state_supplements/Construction_of_a_Tangent_Line.pdf]


S.28





http://pdfs.cpm.org/state_supplements/Construction_of_a_Tangent_Line.pdf



1. Prove Thales’ Theorem: If B, C and D are points on a circle A where BD is a diameter of the circle, then ∠BCD is a right angle. (Hint: Draw AC and use the sum of the angles in a triangle.)

2. Construct the tangent lines from point 𝑃𝑃 to the circle given below.


S.29






3. Prove that if segments from a point 𝑃𝑃 are tangent to a circle O at points 𝐴𝐴 and 𝐵𝐵, then 𝑃𝑃𝐴𝐴�� = 𝑃𝑃𝐵𝐵��.

4. Given points 𝐴𝐴, 𝐵𝐵, and 𝐶𝐶 so that 𝐴𝐴𝐵𝐵 = 𝐴𝐴𝐶𝐶, construct a circle so that 𝐴𝐴𝐵𝐵�� is tangent to the circle at 𝐵𝐵 and 𝐴𝐴𝐶𝐶�� is tangent to the circle at 𝐶𝐶.


S.30






Lesson 4: Equations for Tangent Lines to Circles

Exploratory Exercise

1. Consider the circle with equation (𝑥𝑥 − 3)2 + (𝑦𝑦 − 5)2 = 20. Find the equations of two tangent lines to the circle that each has slope −1

2.

Lesson 4: Equations for Tangent Lines to Circles Module 5+: Conics

S.31








Approaching this Type of Problem Analytically

2. Consider the circle with equation (𝑥𝑥 − 4)2 + (𝑦𝑦 − 5)2 = 20. Find the equations of two tangent lines to the circle that each has slope 2.

A. What is the center of the circle? Label the center 𝑂𝑂. The radius of the circle?

B. If the tangent lines are to have

a slope of 2, what must be the slope of the radii to those tangent lines? Why?

C. Write the slope formula using the point 𝐴𝐴(𝑥𝑥,𝑦𝑦) on the circle.

D. We have y−5x−4

= −12

and (𝑥𝑥 − 4)2 + (𝑦𝑦 − 5)2 = 20. Use these equation to solve for x.


S.32








E. As expected, there are two possible values for 𝑥𝑥. Why are two values expected?

F. What are the coordinates of the points of tangency? How can you determine the 𝑦𝑦-coordinates? G. How can we verify that these two points lie on the circle? H. How can we find the equations of these tangent lines? What are the equations?


S.33








Looking at the General Case

3. Refer to the diagram below.

Let 𝑝𝑝 > 1. What is the equation of the tangent line to the circle 𝑥𝑥2 + 𝑦𝑦2 = 1 through the point (𝑝𝑝, 0) on the 𝑥𝑥-axis with a point of tangency in the upper half-plane?

A. Use 𝑄𝑄(𝑥𝑥,𝑦𝑦)as the point of tangency, as shown in the diagram provided. Label the center as 𝑂𝑂(0,0).

B. What do we know about segments 𝑂𝑂𝑄𝑄 and 𝑃𝑃𝑄𝑄? C. Write an equation that uses this idea. D. Combine the two equations to find an expression for 𝑥𝑥. E. Use the expression for 𝑥𝑥 to find an expression for 𝑦𝑦. F. What are the coordinates of the point 𝑄𝑄 (the point of tangency)?

G. What is the slope of 𝑂𝑂𝑄𝑄�� in terms of 𝑝𝑝? H. What is the slope of 𝑄𝑄𝑃𝑃�� in terms of 𝑝𝑝? I. What is the equation of line 𝑄𝑄𝑃𝑃?


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4. Using the same diagram from Exercise 3, label the point of tangency in the lower half-plane as 𝑄𝑄′. A. What are the coordinates of 𝑄𝑄′?

B. What is the slope of 𝑂𝑂𝑄𝑄′��?

C. What is the slope of 𝑄𝑄′𝑃𝑃��?

D. Find the equation of the second tangent line to the circle through (𝑝𝑝, 0).

5. Show that a circle with equation (𝑥𝑥 − 2)2 + (𝑦𝑦 + 3)2 = 160 has two tangent lines with equations

𝑦𝑦 + 15 = 13

(𝑥𝑥 − 6) and 𝑦𝑦 − 9 = 13

(𝑥𝑥 + 2).


S.35








6. Kamal gives the following proof that 𝑦𝑦 − 1 = 89

(𝑥𝑥 + 10) is the equation of a line that is tangent to a circle given by (𝑥𝑥 + 1)2 + (𝑦𝑦 − 9)2 = 145. The circle has center (−1, 9) and radius 12.04. The point (−10, 1) is on the circle because (−10 + 1)2 + (1 − 9)2 = (−9)2 + (−8)2 = 145.

The slope of the radius is 9 − 1−1 + 10

= 89;

therefore, the equation of the tangent line is 𝑦𝑦 − 1 = 89

(𝑥𝑥 + 10).

A. Kerry said that Kamal has made an error. What was Kamal’s error? Explain what he did wrong.

B. What should the equation for the tangent line be?

Lesson Summary

THEOREM:

A tangent line to a circle is perpendicular to the radius of the circle drawn to the point of tangency.

TANGENT TO A CIRCLE: A tangent line to a circle is a line in the same plane that intersects the circle in one and only one point. This point is called the point of tangency.


S.36








Homework Problem Set 1. Consider the circle (𝑥𝑥 − 1)2 + (𝑦𝑦 − 2)2 = 16. There are two lines tangent to this circle having a slope of 0.

A. Find the coordinates of the points of tangency.

B. Find the equations of the two tangent lines.

2. Consider the circle 𝑥𝑥2 − 4𝑥𝑥 + 𝑦𝑦2 + 10𝑦𝑦 + 13 = 0. There are two lines tangent to this circle having a slope of 2

3.

A. Find the coordinates of the two points of tangency.

B. Find the equations of the two tangent lines.

3. What are the coordinates of the points of tangency of the two tangent lines through the point (1, 1) each tangent to the circle 𝑥𝑥2 + 𝑦𝑦2 = 1?


S.37








4. What are the coordinates of the points of tangency of the two tangent lines through the point (−1,−1) each tangent to the circle 𝑥𝑥2 + 𝑦𝑦2 = 1?

5. What is the equation of the tangent line to the circle 𝑥𝑥2 + 𝑦𝑦2 = 1 through the point (6, 0)?

6. D’Andre said that a circle with equation (𝑥𝑥 − 2)2 + (𝑦𝑦 − 7)2 = 13 has a tangent line represented by the equation 𝑦𝑦 − 5 = −3

2(𝑥𝑥 + 1). Is he correct? Explain.

7. Describe a similarity transformation that maps a circle given by 𝑥𝑥2 + 6𝑥𝑥 + 𝑦𝑦2 − 2𝑦𝑦 = 71 to a circle of radius 3 that is tangent to both axes in the first quadrant.

8. Could a circle given by the equation (𝑥𝑥 − 5)2 + (𝑦𝑦 − 1)2 = 25 have tangent lines given by the equations 𝑦𝑦 − 4 = 4

3(𝑥𝑥 − 1) and 𝑦𝑦 − 5 = 3

4(𝑥𝑥 − 8)? Explain how you know.


S.38








Lesson 5: Curves in Geometry – The Ellipse

The ellipse has some interesting reflective properties. Imagine that you and a friend are standing at the focal points of an elliptical room, as shown in the diagram below. Though your friend may be 100 feet away, you would be able to hear what she is saying, even if she were facing away from you and speaking at the level of a whisper! How can this be? This phenomenon is based on the reflective property of ellipses: Every ray emanating from one focus of the ellipse is reflected off the curve in such a way that it travels to the other focal point of the ellipse.

There are many examples of rooms with this curious property. One is “The Whispering Gallery” at Grand Central Station in New York City. The brief video clip below serves to illustrate this phenomenon.

https://www.youtube.com/watch?v=7nbxSCxiq2o

Lesson 5: Curves in Geometry – The Ellipse Module 5+: Conics

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This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M3-TE-1.3.0-08.2015






https://www.youtube.com/watch?v=7nbxSCxiq2o


Opening Activity 1. Now, let’s turn our attention to the mathematical properties of elliptical rooms. If every sound wave

emanating from one focus bounces off the walls of the room and reaches the other focus at exactly the same instant, what can we say about the lengths of the segments shown in the following two diagrams if the ellipses are congruent and the fixed points are in identical locations?

2. Draw as many segments as you can from one focus, to the ellipse, and back to the other focus. Share what you notice with your partner and then write your observations below. This leads us to the distance property of ellipses:

For any point on an ellipse, the sum of the distances to the foci is constant.

3. Let’s look at a concrete demonstration of the distance property for an ellipse. Take a length of string, and attach it to the two fixed points below. Then, place your pencil on the string, and pull it taut. Move the pencil around the paper, keeping the string taut. Because the string has a fixed length, the sum of the distances to the foci remains the same. Don’t forget to draw the curve on the other side of the two fixed points. This drawing technique generates an ellipse.

20 ft50 ft yx

•

fixed point

•

fixed point


S.40








An ellipse centered at the origin is a set of points that satisfy an equation of the form 𝑥𝑥2

𝑎𝑎2+ 𝑦𝑦2

𝑏𝑏2= 1. Do you

think that a room with the reflection property discussed above can be described by this kind of equation?

4. Since we are going to describe an ellipse using an equation, let’s bring coordinate axes into the picture. Let’s place one axis along the line that contains the two focal points, and let’s place the other axis at the midpoint of the two foci. Where is the center of the ellipse?

5. Suppose that the focal points are located 8 feet apart, and a sound wave traveling from one focus to the other moves a total distance of 10 feet. Exactly which points in the plane are on this elliptical curve? Perhaps it helps at this stage to add in a few labels. Write an equation that represents the distance condition that defines this ellipse.

My equation: ___________________________________________

6. Use the fact that the foci are 8 feet apart to determine the coordinates of F and G.

F ( ________, _________ ) and G ( ________, _________ )


S.41








7. We are trying to get an equation of an ellipse that is in the form 2 2

2 2 1x ya b

+ = . We’ll use the annotated

picture below to help us do this.

A. Use the Pythagorean Theorem to write an equation with PF.

B. Use the Pythagorean Theorem to

write an equation with PG. C. Use the fact that PF + PG = 10 and

your equations from Parts A and B to write a new equation in terms of x and y.

D. Now that it’s done, we can set about verifying that a curve with this special distance property can be

written in the very simple form 𝑥𝑥2


𝑏𝑏2= 1. Work with your partner to simplify this equation until it

is in the standard form of an ellipse.

The distance from points F and G to the 𝑦𝑦-axis is 4. If point P is a distance of 𝑥𝑥 from the 𝑦𝑦-axis, that means the horizontal distance from P to F is 4 + 𝑥𝑥 and from P to G is 4 − 𝑥𝑥.


S.42








Here is a recap of our work up to this point:

An ellipse is a set of points where the sum of the distances to two fixed points (called foci) is constant. We studied a particular example of such a curve,

showing that it can be written in the form 𝑥𝑥2


𝑏𝑏2= 1.

8. Let’s see what else we can learn about the graph of this ellipse 2 2

125 9x y

+ = by examining the equation

more closely. Where do you think the curve intersects the axes?

9. Now, let’s verify that these features are consistent with the facts we started with, that is, that the foci were 8 feet apart and that the distance each sound wave travels between the foci is 10 feet. Do the 𝑥𝑥-intercepts make sense in light of these facts? Then check to see if the 𝑦𝑦-intercepts are consistent with the initial conditions of this problem. Write your conclusions below.


S.43








10. Tammy takes an 8-inch length of string and tapes the ends to a sheet of paper in such a way that the ends of the string are 6 inches apart. Then, she pulls the string taut and traces the curve below using a pencil.

A. Where does this curve intersect the 𝑦𝑦-axis? Where does the curve intersect the 𝑥𝑥-axis? Use the diagram to help you.

B. The equation for Tammy’s ellipse is 2 2

17 16x y

+ = .

How can we use the information from Part A to write this equation without deriving it as we did in Exercise 7?

Closing

11. Let 𝐹𝐹 and 𝐺𝐺 be the foci of an ellipse. If 𝑃𝑃 and 𝑄𝑄 are points on the ellipse, what conclusion can you draw about distances from 𝐹𝐹 and 𝐺𝐺 to 𝑃𝑃 and 𝑄𝑄?

12. What information do you need in order to derive the equation of an ellipse? 13. What is fundamentally true about every ellipse? 14. What is the standard form of an ellipse centered at the origin?

15. Summarize what you know about equations of ellipses centered at the origin with vertices (a, 0), (―a, 0), (0, b) and (0, ―b).


S.44








Lesson Summary

[source: http://www.slideshare.net/roneick/ellipses-32078423]


S.45







http://www.slideshare.net/roneick/ellipses-32078423



1. Points 𝐹𝐹 and 𝐺𝐺 are located at (0, 3) and (0,−3). Let 𝑃𝑃(𝑥𝑥,𝑦𝑦) be a point such that 𝑃𝑃𝐹𝐹 + 𝑃𝑃𝐺𝐺 = 8. Use this information to show that the equation of the ellipse is 𝑥𝑥2

7+ 𝑦𝑦2

16= 1.

2. Derive the equation of the ellipse with the given foci 𝐹𝐹 and 𝐺𝐺 that passes through point 𝑃𝑃. Write your

answer in standard form: 𝑥𝑥2


𝑏𝑏2= 1.

A. The foci are 𝐹𝐹(−2,0) and 𝐺𝐺(2,0), and point 𝑃𝑃(𝑥𝑥,𝑦𝑦) satisfies the condition 𝑃𝑃𝐹𝐹 + 𝑃𝑃𝐺𝐺 = 5. B. The foci are 𝐹𝐹(0,−1) and 𝐺𝐺(0,1), and point 𝑃𝑃(𝑥𝑥,𝑦𝑦) satisfies the condition 𝑃𝑃𝐹𝐹 + 𝑃𝑃𝐺𝐺 = 4.


S.46








3. The semi-major axes of an ellipse are the segments from the center to the farthest vertices, and the semi-minor axes are the segments from the center to the closest vertices. For each of the ellipses in Problem 2, find the lengths 𝑎𝑎 and 𝑏𝑏 of the semi-major axes.

4. Without deriving the equation, use what you learned in the lesson (Exercise 10) to write the equation of each ellipse described below.

A. An ellipse centered at the origin with 𝑥𝑥-intercepts (−2,0), (2,0) and 𝑦𝑦-intercepts (0,8), (0,−8)

B. An ellipse centered at the origin with 𝑥𝑥-intercepts (−√5, 0), (√5, 0) and 𝑦𝑦-intercepts (0,3), (0,−3)

5. Examine the ellipses and the equations of the ellipses you have worked with, and describe the ellipses with

equation 𝑥𝑥2


𝑏𝑏2= 1 in the three cases 𝑎𝑎 > 𝑏𝑏, 𝑎𝑎 = 𝑏𝑏, and 𝑏𝑏 > 𝑎𝑎.

6. Is it possible for 𝑥𝑥2

4+ 𝑦𝑦2

9= 1 to have foci at (−𝑐𝑐, 0) and (𝑐𝑐, 0) for some real number 𝑐𝑐?


S.47








7. For each value of 𝑘𝑘 specified in Parts A – E, plot the set of points in the plane that satisfy the equation 𝑥𝑥2

4+ 𝑦𝑦2 = 𝑘𝑘.

A. 𝑘𝑘 = 1

B. 𝑘𝑘 = 14

C. 𝑘𝑘 = 19

D. 𝑘𝑘 = 116

E. Describe what

happens to the graph of

𝑥𝑥2

4+ 𝑦𝑦2 = 𝑘𝑘 as

𝑘𝑘 → 0.

8. A. Make a conjecture: Which points in the plane satisfy the equation 𝑥𝑥2

4+ 𝑦𝑦2 = 0?

B. Explain why your conjecture in part (g) makes sense algebraically.

C. Which points in the plane satisfy the equation 𝑥𝑥2

4+ 𝑦𝑦2 = −1?

-3

-2

-1

0

1

2

3

-3 -2 -1 0 1 2 3


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9. For each value of 𝑘𝑘 specified in Parts A – E, plot the set of points in the plane that satisfy the equation 𝑥𝑥2

𝑘𝑘+ 𝑦𝑦2 = 1.

A. 𝑘𝑘 = 1

B. 𝑘𝑘 = 2

C. 𝑘𝑘 = 4

D. 𝑘𝑘 = 10

E. 𝑘𝑘 = 25

F. Describe what happens to the graph of

𝑥𝑥2

𝑘𝑘+ 𝑦𝑦2 = 1 as

𝑘𝑘 → ∞.

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6


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10. For each value of 𝑘𝑘 specified in Parts A – E, plot the set of points in the plane that satisfy the equation

𝑥𝑥2 + 𝑦𝑦2

𝑘𝑘= 1.

A. 𝑘𝑘 = 1

B. 𝑘𝑘 = 2

C. 𝑘𝑘 = 4

D. 𝑘𝑘 = 10

E. 𝑘𝑘 = 25

F. Describe what happens to the graph of

𝑥𝑥2 + 𝑦𝑦2

𝑘𝑘= 1 as

𝑘𝑘 → ∞.

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6


S.50







HONORS GEOMETRY


Lesson 6: The Definition of a Parabola

Opening Exercise

1. Suppose you are viewing the cross-section of a mirror. Where would the incoming light be reflected in each type of design? Sketch your ideas below.

Discussion: Telescope Design

When Newton designed his reflector telescope (Figure 1), he understood two important ideas.

The curved mirror needs to focus all the light to a single point that we will call the focus. An angled flat mirror is placed near this point and reflects the light to the eyepiece of the telescope.

The reflected light needs to arrive at the focus at the same time. Otherwise, the image is distorted.

Mirror 1

Incoming Light

Mirror 3

Incoming Light

Mirror 2

Incoming Light

Szőcs Tamás Figure 1

Lesson 6: The Definition of a Parabola Module 5+: Conics

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This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015






HONORS GEOMETRY


In Algebra 1, you studied parabolas and their transformations. In this lesson and the next one, you’ll look at parabolas more analytically and see how their shape is defined by the focus and directrix. You’ll create a parabola by folding paper that involved the focus and directrix.

You will need: one sheet of lined paper and a ruler

2. Follow the steps below to create your parabola.

Mark a spot near the middle of the piece of paper on one of the lines.

This mark will be the focus of the parabola. Label it.

Label the edge on your paper as a number line. For ease, the marks can be labeled, with 0 directly under the focus. This edge line is the directrix of our parabola. Label it.

Fold the paper so that the point 0 is directly on the focus. Fold the paper, and unfold it. This fold line is the perpendicular bisector of a line from the focus to the directrix (line below the parabola) and represents the minimum of the parabola, the vertex.

Repeat the process, this time using the marks labelled -1 and 1. These fold lines are also perpendicular bisectors of the lines joining the focus points on the directrix. They’re tangents to the parabola, and start showing the edge of the curve.

By the time you’ve repeated this process a number of times, the parabola becomes very clear.


S.52







HONORS GEOMETRY


3. The definition of a parabola states:

Definition: A parabola with directrix 𝐿𝐿 and focus point 𝐹𝐹 is the set of all points in the plane that are equidistant from the point 𝐹𝐹 and line 𝐿𝐿.

Use your paper folded parabola from Exercise 2 to see if this true. With your ruler, draw segments from the directrix to the parabola and then from that point on the parabold to the focus. Measure both segments. Are they equal?

4. Figure 2 to the right illustrates this definition of a parabola. Use tick marks to show which segments are equal in length. (You do NOT need to measure them.)

Further Discussion on Newton’s Telescope Design

All parabolas have the reflective property illustrated in Figure 3. Rays parallel to the axis reflect off the parabola and through the focus point, 𝐹𝐹.

Thus, a mirror shaped like a rotated parabola would satisfy Newton’s requirements for his telescope design.

Figure 4 at the right shows several different line segments representing the reflected light with one endpoint on the curved mirror that is a parabola and the other endpoint at the focus. Anywhere the light hits this type of parabolic surface, it always reflects to the focus, 𝐹𝐹, at exactly the same time.

Figure 5 shows the same image with a directrix. Imagine for a minute that the mirror was not there. Then, the light would arrive at the directrix all at the same time. Since the distance from each point on the parabolic mirror to the directrix is the same as the distance from the point on the mirror to the focus, and the speed of light is constant, it takes the light the same amount of time to travel to the focus as it would have taken it to travel to the directrix. In the diagram, this means that 𝐴𝐴𝐹𝐹 = 𝐴𝐴𝐹𝐹𝐴𝐴, 𝐵𝐵𝐹𝐹 = 𝐵𝐵𝐹𝐹𝐵𝐵, and so on. Thus, the light rays arrive at the focus at the same time, and the image is not distorted.

Figure 3

Figure 2

Figure 4

Figure 5


S.53







HONORS GEOMETRY


x

y

𝑭𝑭(𝟎𝟎,𝟐𝟐)

𝑨𝑨(𝒙𝒙,𝒚𝒚)

𝑭𝑭′(𝒙𝒙,𝟎𝟎)

Finding an Analytic Equation for a Parabola

5. If we are given a focus and a directrix, we should be able to create an equation for a parabola.

Focus: 𝐹𝐹(0,2)

Directrix: 𝑥𝑥-axis

Parabola: 𝑃𝑃 = {(𝑥𝑥,𝑦𝑦)| (𝑥𝑥,𝑦𝑦) is equidistant from 𝐹𝐹 and the 𝑥𝑥-axis.}

Let 𝐴𝐴 be any point (𝑥𝑥,𝑦𝑦) on the parabola 𝑃𝑃. Let 𝐹𝐹′ be a point on the directrix with the same 𝑥𝑥-coordinate as point 𝐴𝐴.

A. Find the length 𝐴𝐴𝐹𝐹′.

B. Use the distance formula to create an expression that represents the length 𝐴𝐴𝐹𝐹.

C. Create an equation that relates the two lengths, and solve it for 𝑦𝑦.

D. Use your equation to determine the vertex of the parabola. Does this make sense from your picture?


S.54







HONORS GEOMETRY


6. Derive the equation of a parabola given the focus of (0,4) and the directrix 𝑦𝑦 = 2. Use the diagram to help you work this problem.

A. Label the focus, directrix and vertex of the parabola on the

diagram. B. Label a point (𝑥𝑥,𝑦𝑦) anywhere on the parabola.

C. Write an expression for the distance from the point (𝑥𝑥,𝑦𝑦) to

the directrix.

D. Write an expression for the distance from the point (𝑥𝑥,𝑦𝑦) to the focus.

E. Apply the definition of a parabola to create an equation in terms of 𝑥𝑥 and 𝑦𝑦. Solve this equation for 𝑦𝑦.

F. What is the translation that takes the graph of this parabola to the graph of the equation derived in Exercise 5?


S.55



-2 2

2

4

x

y





HONORS GEOMETRY


Lesson Summary

PARABOLA: A parabola with directrix line 𝐿𝐿 and focus point 𝐹𝐹 is the set of all points in the plane that are equidistant from the point 𝐹𝐹 and line 𝐿𝐿

AXIS OF SYMMETRY: The axis of symmetry of a parabola given by a focus point and a directrix is the perpendicular line to the directrix that passes through the focus

VERTEX OF A PARABOLA: The vertex of a parabola is the point where the axis of symmetry intersects the parabola.

In the Cartesian plane, the distance formula can help in deriving an analytic equation for a parabola.


S.56







HONORS GEOMETRY


Homework Problem Set 1. Demonstrate your understanding of the definition of a parabola by drawing three pairs of congruent

segments given each parabola, its focus, and directrix. Measure the segments that you drew in either inches or centimeters to confirm the accuracy of your sketches.

A. B.

C. D.

2. Find the distance from the point (4,2) to the point (0,1).

3. Find the distance from the point (4,2) to the line 𝑦𝑦 = −2.


S.57







HONORS GEOMETRY


4. Find the distance from the point (𝑥𝑥, 4) to the line 𝑦𝑦 = −1.

5. Find the distance from the point (𝑥𝑥,−3) to the line 𝑦𝑦 = 2.

6. Find the values of 𝑥𝑥 for which the point (𝑥𝑥, 4) is equidistant from (0,1), and the line 𝑦𝑦 = −1.

7. Find the values of 𝑥𝑥 for which the point (𝑥𝑥,−3) is equidistant from (1,−2), and the line 𝑦𝑦 = 2.


S.58







HONORS GEOMETRY


8. Consider the equation 𝑦𝑦 = 𝑥𝑥2. A. Find the coordinates of the three points on the graph of 𝑦𝑦 = 𝑥𝑥2 whose 𝑥𝑥-values are 1, 2, and 3.

B. Show that each of the three points in part (A) is equidistant from the point �0, 14�, and the line 𝑦𝑦 = −1

4.

C. Show that if the point with coordinates (𝑥𝑥,𝑦𝑦) is equidistant from the point �0, 14�, and the line 𝑦𝑦 = −1

4,

then 𝑦𝑦 = 𝑥𝑥2.

9. Consider the equation 𝑦𝑦 = 12𝑥𝑥2 − 2𝑥𝑥.

A. Find the coordinates of the three points on the graph of 𝑦𝑦 = 12𝑥𝑥2 − 2𝑥𝑥 whose 𝑥𝑥-values are −2, 0, and 4.

B. Show that each of the three points in part (A) is equidistant from the point �2,−32� and the line 𝑦𝑦 = −5

2.

C. Show that if the point with coordinates (𝑥𝑥,𝑦𝑦) is equidistant from the point �2,−32�, and the line 𝑦𝑦 = −5

2 ,

then 𝑦𝑦 = 12𝑥𝑥2 − 2𝑥𝑥.


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HONORS GEOMETRY


10. Derive the analytic equation of a parabola with focus (1,3) and directrix 𝑦𝑦 = 1. Use the diagram to help you work this problem. A. Label a point (𝑥𝑥,𝑦𝑦) anywhere on the parabola.

B. Write an expression for the distance from the point (𝑥𝑥, 𝑦𝑦) to the

directrix.

C. Write an expression for the distance from the point (𝑥𝑥, 𝑦𝑦) to the focus (1,3).

D. Apply the definition of a parabola to create an equation in terms of 𝑥𝑥 and 𝑦𝑦. Solve this equation for 𝑦𝑦.

E. Describe a sequence of transformations that would take this parabola to the parabola with equation 𝑦𝑦 = 1

4𝑥𝑥2 + 1 derived in Exercise 5.

11. Consider a parabola with focus (0,−2) and directrix on the 𝑥𝑥-axis. A. Derive the analytic equation for this parabola.

B. Describe a sequence of transformations that would take the parabola with equation 𝑦𝑦 = 14𝑥𝑥2 + 1

derived in Exercise 5 to the graph of the parabola in part (A).


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-2 2 4

2

4

x

y





HONORS GEOMETRY


Lesson 7: Connecting to the Vertex Form of a Parabola

Opening Exercise

1. Draw the parabola for each focus and directrix given below. The directrix is the dashed line. 2. How can you use the definition of a parabola to quickly locate at least three points on the graph of the

parabola with a focus (0,1) and directrix 𝑦𝑦 = −1? 3. How can you use the definition of a parabola to quickly locate at least three points on the graph of the

parabola with a focus (0,2) and directrix 𝑦𝑦 = −2? 4. Generalize this process of finding three points on a parabola with a given focus and directrix.

Lesson 7: Connecting to the Vertex Form of a Parabola Module 5+: Conics

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-4 -2 2 4

-4

-2

2

4

x

y

-4 -2 2 4

-4

-2

2

4

x

y





HONORS GEOMETRY


𝑝𝑝 12𝑝𝑝

12𝑝𝑝

5. Draw the parabola with the given focus and directrix.

6. Draw the parabola with the given focus and directrix.

Discussion 7. What can you conclude about the relationship between the parabolas in Exercises 5 and 6?

8. Let 𝑝𝑝 be the number of units between the focus and the directrix, as shown. As the value of 𝑝𝑝 increases, what happens to the shape of the resulting parabola?


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HONORS GEOMETRY


Derive an Equation for a Parabola

Consider a parabola 𝑃𝑃 with distance 𝑝𝑝 > 0 between the focus with coordinates �0, 1

2𝑝𝑝�, and directrix 𝑦𝑦 = −1

2𝑝𝑝.

9. What is the equation that represents this parabola?

We have shown that any parabola with a distance 𝑝𝑝 > 0 between the focus �0, 12𝑝𝑝� and directrix 𝑦𝑦 = −1

2𝑝𝑝

has a vertex at the origin and is represented by a quadratic equation of the form 𝑦𝑦 = 12𝑝𝑝𝑥𝑥2. As shown in the

exercise above.

10. Suppose that the vertex of a parabola with a horizontal directrix that opens upward is (ℎ,𝑘𝑘), and the distance from the focus to directrix is 𝑝𝑝 > 0. Then, the focus has coordinates �_____, _____ + 1

2𝑝𝑝�, and the

directrix has equation 𝑦𝑦 = _______ − 12𝑝𝑝.

If we go through the above derivation with our new focus and directrix, we should not be surprised to get a quadratic equation. In fact, if we complete the square on that equation, we can write it in the form 𝑦𝑦 = 1

2𝑝𝑝(𝑥𝑥 − ℎ)2 + 𝑘𝑘.

In Algebra I, Module 4, we saw that any quadratic function can be put into vertex form:

𝑓𝑓(𝑥𝑥) = 𝑎𝑎(𝑥𝑥 − ℎ)2 + 𝑘𝑘.

Now we see that any parabola that opens upward can be described by a quadratic function in vertex form,

where 𝑎𝑎 = 12𝑝𝑝

.

11. Write the new equation without a.

12. If the parabola opens downward, then the equation is __________________________________ and the graph of any quadratic equation of this form is a parabola with vertex at (ℎ,𝑘𝑘), distance 𝑝𝑝 between focus and directrix, and opening downward.

Likewise, we can derive analogous equations for parabolas that open to the left and right.


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HONORS GEOMETRY


Lesson Summary

PARABOLAS

[image source: https://www.cliffsnotes.com/study-guides/algebra/algebra-ii/conic-sections/parabola]

Parabola Opens Up or Down Parabola Opens Left or Right

Given a parabola 𝑃𝑃 with vertex (ℎ,𝑘𝑘), horizontal directrix, and distance 𝑝𝑝 > 0 between focus and directrix, the analytic equation that describes the parabola 𝑃𝑃 is

• 𝑦𝑦 = 12𝑝𝑝

(𝑥𝑥 − ℎ)2 + 𝑘𝑘 if the parabola opens

upward, and

• 𝑦𝑦 = − 12𝑝𝑝

(𝑥𝑥 − ℎ)2 + 𝑘𝑘 if the parabola opens

downward.

Given a parabola 𝑃𝑃 with vertex (ℎ,𝑘𝑘), vertical directrix, and distance 𝑝𝑝 > 0 between focus and directrix, the analytic equation that describes the parabola 𝑃𝑃 is

• 𝑥𝑥 = 12𝑝𝑝

(𝑦𝑦 − 𝑘𝑘)2 + ℎ if the parabola opens to

the right, and

• 𝑥𝑥 = − 12𝑝𝑝

(𝑦𝑦 − 𝑘𝑘)2 + ℎ if the parabola opens

to the left.

Conversely, if 𝑝𝑝 > 0, then • The graph of the quadratic equation 𝑦𝑦 =

12𝑝𝑝

(𝑥𝑥 − ℎ)2 + 𝑘𝑘 is a parabola that opens

upward with vertex at (ℎ,𝑘𝑘) and distance 𝑝𝑝 from focus to directrix, and

• The graph of the quadratic equation 𝑦𝑦 =− 1

2𝑝𝑝(𝑥𝑥 − ℎ)2 + 𝑘𝑘 is a parabola that opens

downward with vertex at (ℎ,𝑘𝑘) and distance 𝑝𝑝 from focus to directrix.

Conversely, if 𝑝𝑝 > 0, then • The graph of the quadratic equation 𝑥𝑥 =

12𝑝𝑝

(𝑦𝑦 − 𝑘𝑘)2 + ℎ is a parabola that opens to

the right with vertex at (ℎ,𝑘𝑘) and distance 𝑝𝑝 from focus to directrix, and

• The graph of the quadratic equation 𝑥𝑥 =− 1

2𝑝𝑝(𝑦𝑦 − 𝑘𝑘)2 + ℎ is a parabola that opens to

the left with vertex at (ℎ,𝑘𝑘) and distance 𝑝𝑝 from focus to directrix.


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https://www.cliffsnotes.com/study-guides/algebra/algebra-ii/conic-sections/parabola

HONORS GEOMETRY



1. Show that if the point with coordinates (𝑥𝑥, 𝑦𝑦) is equidistant from (4,3), and the line 𝑦𝑦 = 5, then 𝑦𝑦 = −14𝑥𝑥2 + 2𝑥𝑥.

2. Show that if the point with coordinates (𝑥𝑥, 𝑦𝑦) is equidistant from the point (2,0) and the line 𝑦𝑦 = −4, then 𝑦𝑦 = 1

8(𝑥𝑥 − 2)2 − 2.

3. Find the equation of the set of points which are equidistant from (0,2) and the 𝑥𝑥-axis. Sketch this set of points.

4. Find the equation of the set of points which are equidistant from the origin and the line 𝑦𝑦 = 6. Sketch this set of points.


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HONORS GEOMETRY


5. Find the equation of the set of points which are equidistant from (4,−2) and the line 𝑦𝑦 = 4. Sketch this

set of points.

6. Find the equation of the set of points which are equidistant from (4,0) and the 𝑦𝑦-axis. Sketch this set of points.

7. Find the equation of the set of points which are equidistant from the origin and the line 𝑥𝑥 = −2. Sketch this set of points.


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HONORS GEOMETRY


8. Use the definition of a parabola to sketch the parabola defined by the given focus and directrix.

A. Focus: (0,5) ; Directrix: 𝑦𝑦 = −1 B. Focus: (−2,0); Directrix: 𝑦𝑦-axis

C. Focus: (4,−4); Directrix: 𝑥𝑥-axis

D. Focus: (2,4) ; Directrix: 𝑦𝑦 = −2

9. Find an analytic equation for each parabola described in Problem 8.


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HONORS GEOMETRY


10. Sketch each parabola, labeling its focus and directrix.

A. 𝑦𝑦 = 12𝑥𝑥2 + 2

B. 𝑦𝑦 = −14𝑥𝑥2 + 1

C. 𝑥𝑥 = 18𝑦𝑦2

D. 𝑥𝑥 = 12𝑦𝑦2 + 2

E. 𝑦𝑦 = 110

(𝑥𝑥 − 1)2 − 2


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Lesson 8: Curves in Geometry - Hyperbolas

Classwork Opening Discussion

In the previous lessons, you learned how to describe an ellipse, how to graph an ellipse, and how to derive the standard equation of an ellipse knowing its foci. In this lesson, you’ll learn to perform these tasks for a hyperbola.

When a satellite moves in a closed orbit around a planet, it follows an elliptical path. However, if the satellite is moving fast enough, it overcomes the gravitational attraction of the planet and breaks out of its closed orbit. The minimum velocity required for a satellite to escape the closed orbit is called the escape velocity. The velocity of the satellite determines the shape of its orbit. To see this visualized, check out the YouTube video https://www.youtube.com/watch?v=pRvVK2m_wGE.

1. If the velocity of the satellite is less than the escape velocity, it follows a path that looks like 𝐸𝐸2 in the diagram. Describe the path, and give the mathematical term for this curve.

2. If the velocity of the satellite is exactly equal to the escape velocity, it follows a path that looks like 𝑃𝑃 in the diagram. Describe the path, and give the mathematical term for this curve.

3. If the velocity of the satellite exceeds the escape velocity, it follows a path that looks like 𝐻𝐻 in the diagram. Describe the path, and give the mathematical term for this curve.

Escape velocity – the lowest speed that an object must have in order to escape the gravitational pull of a planet.

Lesson 8: Curves in Geometry - Hyperbolas Module 5+: Conics

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https://www.youtube.com/watch?v=pRvVK2m_wGE


Analysis of a Hyperbola

4. Consider the equation 𝑥𝑥2 + 𝑦𝑦2 = 1. When we graph the set of points (𝑥𝑥,𝑦𝑦) that satisfy this equation, what sort of curve do we get? What are the properties of this curve?

Let’s make one small change to this equation: Consider 𝑥𝑥2 − 𝑦𝑦2 = 1.

5. What sort of curve does this produce? Develop an argument, and share it with your group. Let’s explore this question together. We focus on three features that are of general interest when studying curves: intercepts, symmetries, and end behavior.

6. Does this curve intersect the 𝑥𝑥-axis? Does it intersect the 𝑦𝑦-axis? If so, where? 7. What sort of symmetries do you expect this graph to have? Now, let’s try to get a feel for what this curve looks like. It never hurts to plot a few points, so let’s start with that approach. Solving for 𝑦𝑦 makes this process a bit easier, so we’ll isolate the 𝑦𝑦-variable and get the following.

𝑦𝑦2 = 𝑥𝑥2 − 1, which means 𝑦𝑦 = ±√𝑥𝑥2 − 1.

We are only going to deal with points in the first quadrant, so we can simply consider 𝑦𝑦 = √𝑥𝑥2 − 1.

8. Use this relation to find the 𝑦𝑦-values that correspond to 𝑥𝑥 = 1, 2, 3, 4, and 5.

𝑥𝑥 1 2 3 4 5

𝑦𝑦

Approximate value of y

To get a better understanding of these numbers, use your calculator to obtain a decimal approximation for each square root.


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9. Plot the points from Exercise 8.

10. Perhaps you are beginning to get a sense of what the graph looks like, but more data is surely useful. Let’s use a spreadsheet or a graphing calculator to generate some more data. In particular, what happens to the 𝑦𝑦-values when the 𝑥𝑥-values get larger and larger? Give the approximate value with three decimal places.

𝑥𝑥 10 20 50 100 500

𝑦𝑦

11. The data are telling us a particular story, but the power of mathematics lies in its ability to explain that story. A. If (𝑥𝑥,𝑦𝑦) satisfies 𝑥𝑥2 − 𝑦𝑦2 = 1, is it true that 𝑦𝑦 < 𝑥𝑥? How can we be sure?

B. What does this tell us about the curve we are studying? In particular, where is the graph in relation to the line 𝑦𝑦 = 𝑥𝑥?

C. The line 𝑦𝑦 = 𝑥𝑥 acts as an upper boundary for the curve. Let’s draw this boundary line for visual reference.

0

1

2

3

4

5

6

0 1 2 3 4 5 6


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Since the points on the curve are getting closer and closer to the line 𝑦𝑦 = 𝑥𝑥, this graph provides visual confirmation of our observation that the 𝑦𝑦-value gets closer and closer to the 𝑥𝑥-value as 𝑥𝑥 gets larger and larger.

12. To sum up, we have four important facts about the graph of 𝑥𝑥2 − 𝑦𝑦2 = 1: A. The graph contains the point (_____,0).

B. The graph lies below the line y = _______.

C. As we move along to the right, the points on the curve get extremely _________ to the line 𝑦𝑦 = ______.

D. The graph is symmetric with respect to both the _______________ and the ________________.

Let’s bring symmetry to bear on this discussion. Since the graph is symmetric with respect to the 𝑥𝑥-axis, we can infer the location of points in the fourth quadrant:

Does this remind you of anything from the start of today’s lesson? That’s right! It’s the trajectory followed by a satellite that has exceeded its escape velocity. This curve is called a hyperbola; therefore, we say that the satellite is following a hyperbolic path.


S.72








The boundary lines, which for this hyperbola have equations 𝑦𝑦 = 𝑥𝑥 and 𝑦𝑦 = −𝑥𝑥, are known as oblique or slant asymptotes. The location of the dot is called a focus of the hyperbola. In fact, since the curve is also symmetrical with respect to the 𝑦𝑦-axis, a hyperbola actually has two foci.

13. Use the points from Exercise 8 and your knowledge of symmetry to sketch the complete hyperbolic

curve. Don’t forget to use the slant asymptotes to help you further define the curve.

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6


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Analysis of 𝒙𝒙𝟐𝟐

𝟑𝟑𝟐𝟐− 𝒚𝒚𝟐𝟐

𝟐𝟐𝟐𝟐= 𝟏𝟏

Now that we have a sense for what a hyperbola looks like, let’s analyze the basic equation. We began with 𝑥𝑥2 − 𝑦𝑦2 = 1. What do you suppose would happen if we took the curve 𝑥𝑥2 − 𝑦𝑦2 = 1 and replaced 𝑥𝑥 and 𝑦𝑦 with 𝑥𝑥

3 and 𝑦𝑦

2?

14. What would the 𝑥𝑥-intercepts be? How do the 𝑥𝑥-intercepts compare to those of the graph of 𝑥𝑥2 − 𝑦𝑦2 = 1?

15. Fill in the table of values for the first quadrant and use it to sketch the complete hyperbola in all four quadrants.

𝑥𝑥 1 2 3 4 5 6 7 8

𝑦𝑦

-8

-7

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

Notice that this is really the same shape as before; it is just that the plane has been stretched horizontally by a factor of 3 and vertically by a factor of 2.


S.74








16. What was the boundary line for points on the curve generated by the original equation 𝑥𝑥2 − 𝑦𝑦2 = 1?

17. What is the boundary line for points on the new curve 𝑥𝑥2

32− 𝑦𝑦2

22= 1? How can you easily find the

boundary lines (slant asymptotes) for any hyperbola?

Where do you suppose the foci of this curve are located? This is not an easy question. Recall that in the last lesson, we used the foci of an ellipse to generate an equation for the ellipse. We can use a similar procedure to find the relationship between the foci of a hyperbola and the equation that generates the hyperbola.

Formal Properties of Hyperbolas Like the ellipse, the formal definition of the hyperbola involves distances. In the figure below, the dotted segments from the foci to point P1 differ in length by a certain amount. Let’s call this difference k. The dashed segments from the foci to point P2 differ in length by exactly the same amount, k.

18. Write an equation that would describe this idea.

• P1

• P2

• G

• F


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The standard equation of the hyperbola is given by 𝒚𝒚𝟐𝟐

𝒂𝒂𝟐𝟐− 𝒙𝒙𝟐𝟐

𝒃𝒃𝟐𝟐= 𝟏𝟏.

Deriving the Equation of a Hyperbola Let’s take 𝐹𝐹(−1, 0) and 𝐺𝐺(1, 0) to be the foci of a hyperbola, with each point 𝑃𝑃 on the hyperbola satisfying either 𝑃𝑃𝐹𝐹 − 𝑃𝑃𝐺𝐺 = 1 or 𝑃𝑃𝐺𝐺 − 𝑃𝑃𝐹𝐹 = 1. What is the equation of such a hyperbola? In particular, can we

express the equation in the standard form 𝑥𝑥2

𝑎𝑎2− 𝑦𝑦2

𝑏𝑏2= 1?

19. A. Use the diagram below to write an equation for PG in terms of x and y. (Hint: The Pythagorean

Theorem is very useful.)

B. Write an equation for PF in terms of x and y.

C. Use your equation from Exercise 18 that would connect these two distances.


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Now we have 𝑃𝑃𝐹𝐹 − 𝑃𝑃𝐺𝐺 = �(𝑥𝑥 + 1)2 + 𝑦𝑦2 − �(𝑥𝑥 − 1)2 + 𝑦𝑦2 = 1.

20. In the last lesson, we learned a technique that can be used to deal with equations that contain two radical expressions. Apply that technique to this equation, and see where you end up! Be sure to rewrite

the equation in the standard form 𝑥𝑥2


𝑏𝑏2= 1.


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Closing Activity/Lesson Summary

21. Complete the graphic organizer comparing ellipses and hyperbolas.

Ellipse

Equation:

Center:

Asymptotes:

Symmetry:

Hyperbola – Opening Up and Down

Equation:

Center:

Asymptotes:

Symmetry:

Hyperbola – Opening Left and Right

Equation:

Center:

Asymptotes:

Symmetry:


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Homework Problem Set For each hyperbola described below:

(A) Derive an equation of the form 𝑥𝑥2


𝑏𝑏2= 1 or 𝑦𝑦

2

𝑏𝑏2− 𝑥𝑥2

𝑎𝑎2= 1.

(B) State any 𝑥𝑥- or 𝑦𝑦-intercepts. (C) Find the equations for the asymptotes of the hyperbola. (D) Graph the hyperbola.

1. Let the foci be 𝐴𝐴(−2,0) and 𝐵𝐵(2,0), and let 𝑃𝑃 be a point for which either 𝑃𝑃𝐴𝐴 − 𝑃𝑃𝐵𝐵 = 2 or 𝑃𝑃𝐵𝐵 − 𝑃𝑃𝐴𝐴 = 2.

2. Consider 𝐴𝐴(0,−3) and 𝐵𝐵(0,3), and let 𝑃𝑃 be a point for which either 𝑃𝑃𝐴𝐴 − 𝑃𝑃𝐵𝐵 = 2.5 or 𝑃𝑃𝐵𝐵 − 𝑃𝑃𝐴𝐴 = 2.5.

A.

A.

B. B.

C. C.

D. D.


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For each value of 𝑘𝑘 specified in Problems 3 – 9, plot the set of points in the plane that satisfy the equation 𝑥𝑥2 − 𝑦𝑦2 = 𝑘𝑘.

3. 𝑘𝑘 = 4

4. 𝑘𝑘 = 1

5. 𝑘𝑘 = 14

6. 𝑘𝑘 = 0

7. 𝑘𝑘 = −4

8. 𝑘𝑘 = −1

9. 𝑘𝑘 = −14

10. Describe the hyperbolas 𝑥𝑥2 − 𝑦𝑦2 = 𝑘𝑘 for different values of 𝑘𝑘. Consider both positive and negative

values of 𝑘𝑘, and consider values of 𝑘𝑘 close to zero and far from zero. 11. Are there any values of 𝑘𝑘 so that the equation 𝑥𝑥2 − 𝑦𝑦2 = 𝑘𝑘 has no solution? Explain your thinking.


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For each value of 𝑘𝑘 specified in Problems 12 – 17, plot the set of points in the plane that satisfy the equation 𝑥𝑥2

𝑘𝑘− 𝑦𝑦2 = 1.

12. 𝑘𝑘 = −1

13. 𝑘𝑘 = 1

14. 𝑘𝑘 = 2

15. 𝑘𝑘 = 4

16. 𝑘𝑘 = 10

17. 𝑘𝑘 = 25

18. Describe what happens to the graph of 𝑥𝑥2

𝑘𝑘− 𝑦𝑦2 = 1 as 𝑘𝑘 → ∞.

19. Suppose work similar to that done in Exercises 12 – 17 was completed for the equation 𝑥𝑥2 − 𝑦𝑦2

𝑘𝑘= 1.

What would happen to the graphs of 𝑥𝑥2 − 𝑦𝑦2

𝑘𝑘= 1 as 𝑘𝑘 → ∞.


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