Table of Contents. Unit 6: Modeling Geometry. Answer Key...AK-1. Introduction... v

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2 These materials ma not be reproduced for an purpose. The reproduction of an part for an entire school or school sstem is strictl prohibited. No part of this publication ma be transmitted, stored, or recorded in an form without written permission from the publisher ISBN Copright 013 J. Weston Walch, Publisher Portland, ME Printed in the United States of America WALCH EDUCATION

3 Table of Contents Introduction... v Unit 6: Modeling Geometr Lesson 1: Deriving Equations... U6-1 Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas... U6-43 Lesson 3: Solving Sstems of Linear Equations and Circles... U6-65 Answer Ke...AK-1 iii Table of Contents

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5 Introduction Welcome to the CCGPS Analtic Geometr Student Resource Book. This book will help ou learn how to use algebra, geometr, data analsis, and probabilit to solve problems. Each lesson builds on what ou have alread learned. As ou participate in classroom activities and use this book, ou will master important concepts that will help to prepare ou for the EOCT and for other mathematics assessments and courses. This book is our resource as ou work our wa through the Analtic Geometr course. It includes explanations of the concepts ou will learn in class; math vocabular and definitions; formulas and rules; and exercises so ou can practice the math ou are learning. Most of our assignments will come from our teacher, but this book will allow ou to review what was covered in class, including terms, formulas, and procedures. In Unit 1: Similarit, Congruence, and Proofs, ou will learn about dilations, and ou will construct lines, segments, angles, polgons, and triangles. You will explore congruence and then define, appl, and prove similarit. Finall, ou will prove theorems about lines, angles, triangles, and parallelograms. In Unit : Right Triangle Trigonometr, ou will begin b exploring trigonometric ratios. Then ou will go on to appl trigonometric ratios. In Unit 3: Circles and Volume, ou will be introduced to circles and their angles and tangents. Then ou will learn about inscribed polgons and circumscribed triangles b constructing them and proving properties of inscribed quadrilaterals. You will construct tangent lines and find arc lengths and areas of sectors. Finall, ou will explain and appl area and volume formulas. In Unit 4: Extending the Number Sstem, ou will start working with the number sstem and rational exponents. Then ou will perform operations with complex numbers and polnomials. In Unit 5: Quadratic Functions, ou will begin b identifing and interpreting structures in expressions. You will use this information as ou learn to create and solve quadratic equations in one variable, including taking the square root of both sides, factoring, completing the square, appling the quadratic formula, and solving quadratic inequalities. You will move on to solving quadratic equations in two or more variables, and solving sstems v Introduction

6 of equations. You will learn to analze quadratic functions and to build and transform them. Finall, ou will solve problems b fitting quadratic functions to data. In Unit 6: Modeling Geometr, ou will stud the links between the two math disciplines, geometr and algebra, as ou derive equations of a circle and a parabola. You will use coordinates to prove geometric theorems about circles and parabolas and solve sstems of linear equations and circles. In Unit 7: Applications of Probabilit, ou will explore the idea of events, including independent events, and conditional probabilit. Each lesson is made up of short sections that explain important concepts, including some completed examples. Each of these sections is followed b a few problems to help ou practice what ou have learned. The Words to Know section at the beginning of each lesson includes important terms introduced in that lesson. As ou move through our Analtic Geometr course, ou will become a more confident and skilled mathematician. We hope this book will serve as a useful resource as ou learn. vi Introduction

7 UNIT 6 MODELING GEOMETRY Lesson 1: Deriving Equations Common Core Georgia Performance Standards MCC9 1.G.GPE.1 MCC9 1.G.GPE. Essential Questions 1. How can a circle be described in different was?. How can a parabola be described in different was? 3. How are the definitions of a circle and a parabola similar? 4. How are the definitions of a circle and a parabola different? WORDS TO KNOW center of a circle circle directrix of a parabola the point in the plane of the circle from which all points on the circle are equidistant. The center is not part of the circle; it is in the interior of the circle. the set of all points in a plane that are equidistant from a reference point in that plane, called the center. The set of points forms a -dimensional curve that measures 360º. a line that is perpendicular to the axis of smmetr of a parabola and that is in the same plane as both the parabola and the focus of the parabola; the fixed line referenced in the definition of a parabola distance formula a formula that states the distance between points (x 1, 1 ) focus of a parabola general form of an equation of a circle ( ) ( 1) + 1 and (x, ) is equal to x x a fixed point on the interior of a parabola that is not on the directrix of the parabola but is on the same plane as both the parabola and the directrix; the fixed point referenced in the definition of a parabola Ax + B + Cx + D + E = 0, where A = B, A 0, and B 0 U6-1 Lesson 1: Deriving Equations

8 parabola the set of all points that are equidistant from a fixed line, called the directrix, and a fixed point not on that line, called the focus. The parabola, directrix, and focus are all in the same plane. The vertex of the parabola is the point on the parabola that is closest to the directrix. b perfect square trinomial a trinomial of the form x + bx + Pthagorean Theorem a theorem that relates the length of the hpotenuse of a right triangle (c) to the lengths of its legs (a and b). The theorem states that a + b = c. quadratic function a function that can be written in the form f (x) = ax + bx + c, where a 0. The graph of an quadratic function is a parabola. radius standard form of an equation of a circle standard form of an equation of a parabola vertex of a parabola the distance from the center to a point on the circle (x h) + ( k) = r, where (h, k) is the center and r is the radius (x h) = 4p( k) for parabolas that open up or down; ( k) = 4p(x h) for parabolas that open right or left. For all parabolas, p 0 and the vertex is (h, k). the point on a parabola that is closest to the directrix and lies on the axis of smmetr U6- Unit 6: Modeling Geometr

9 Recommended Resources MathIsFun.com. Circle Equations. This interactive site reviews the standard form of the equation of a circle and how it is derived. It reminds users that the equation is based on the Pthagorean Theorem, and reviews how to expand binomials and simplif the resulting polnomials to obtain the general form, given the standard form. The site also demonstrates how to use the technique of completing the square to obtain the standard form, given the general form. Questions are provided to assess understanding. MathIsFun.com. Definition of Parabola. An interactive soccer plaer kicks a ball that follows a parabolic path. Users can change the values of a, b, and c to obtain equations of the form = ax + bx + c, and see how the graph changes to match the equation. This site might be useful as a review of what has alread been learned about quadratic functions and their graphs, which are parabolas. MathIsFun.com. Parabola. This site offers a hands-on activit showing how to use the definition of a parabola to draw one. The onl required materials are paper, a pencil, and a ruler. The site explains that ras parallel to the axis of smmetr are reflected from the parabola to its focus, and it lists some devices that use that propert, such as a satellite dish. Users can assess understanding b answering quiz questions. U6-3 Lesson 1: Deriving Equations

10 Lesson 6.1.1: Deriving the Equation of a Circle Introduction The graph of an equation in x and is the set of all points (x, ) in a coordinate plane that satisf the equation. Some equations have graphs with precise geometric descriptions. For example, the graph of the equation = x + 3 is the line with a slope of, passing through the point (0, 3). This geometric description uses the familiar concepts of line, slope, and point. The equation = x + 3 is an algebraic description of the line. In this lesson, we will investigate how to translate between geometric descriptions and algebraic descriptions of circles. We have alread learned how to use the Pthagorean Theorem to find missing dimensions of right triangles. Now, we will see how the Pthagorean Theorem leads us to the distance formula, which leads us to the standard form of the equation of a circle. Ke Concepts The standard form of the equation of a circle is based on the distance formula. The distance formula, in turn, is based on the Pthagorean Theorem. The Pthagorean Theorem states that in an right triangle, the square of the hpotenuse is equal to the sum of the squares of the legs. c a b a + b = c If r represents the distance between the origin and an point (x, ), then x + = r. (x, ) (x, ) r r (0, 0) x x x x (0, 0) x can be positive, negative, or zero because it is a coordinate. can be positive, negative, or zero because it is a coordinate. r cannot be negative because it is a distance. U6-4 Unit 6: Modeling Geometr

11 A circle is the set of all points in a plane that are equidistant from a reference point in that plane, called the center. The set of points forms a -dimensional curve that measures 360º. The center of a circle is the point in the plane of the circle from which all points on the circle are equidistant. The center is in the interior of the circle. The radius of a circle is the distance from the center to a point on the circle. For a circle with center (0, 0) and radius r, an point (x, ) is on that circle if and onl if x + = r. (x, ) (0, 0) r x The distance formula is used to find the distance between an two points on a coordinate plane. The distance formula states that the distance d between A (x 1, 1 ) and B (x, ) ( ) + ( ) is d= x x 1 1 The distance formula is based on the Pthagorean Theorem.. Look at this diagram of a right triangle with points A, B, and C. The distance d between points A and B is unknown. A (x 1, 1 ) 1 d C (x 1, ) x x 1 B (x, ) x U6-5 Lesson 1: Deriving Equations

12 The worked example that follows shows how the distance formula is derived from the Pthagorean Theorem, using the points from the diagram to find d: AB = BC + AC Pthagorean Theorem d = x x d = x x Substitute values for sides AB, BC, and AC of the triangle. ( 1) + ( 1) Simplif. All squares are nonnegative. ( ) + ( ) d= x x 1 1 Take the square of each side of the equation to arrive at the distance formula. For a circle with center (h, k) and radius r, an point (x, ) is on that circle if and onl if ( x h) + ( k) = r. Squaring both sides of this equation ields the standard form of an equation of a circle with center (h, k) and radius r: (x h) + ( k) = r. (x, ) r k (h, k) x h (x, k) x If a circle has center (0, 0), then its equation is (x 0) + ( 0) = r, or x + = r. If the center and radius of a circle are known, then either of the following two methods can be used to write an equation for the circle: Appl the Pthagorean Theorem to derive the equation. U6-6 Unit 6: Modeling Geometr

13 Or, substitute the center coordinates and radius directl into the standard form. The general form of an equation of a circle is Ax + B + Cx + D + E = 0, where A = B, A 0, and B 0. If an one of the following three sets of facts about a circle is known, then the other two can be determined: center (h, k) and radius r standard equation: (x h) + ( k) = r general equation: Ax + B + Cx + D + E = 0 The general form of the equation of a circle comes from expanding the standard form of the equation of the circle. The standard form of the equation of a circle comes from completing the square from the general form of the equation of a circle. b Ever perfect square trinomial has the form x + bx + because it is the b b square of a binomial: x + bx + x = +. b Completing the square is the process of determining the value of and adding b it to x + bx to form the perfect square trinomial, x + bx +. U6-7 Lesson 1: Deriving Equations

14 Guided Practice Example 1 Derive the standard equation of the circle with center (0, 0) and radius Sketch the circle (x, ) x U6-8 Unit 6: Modeling Geometr

15 . Use the Pthagorean Theorem to derive the standard equation. In order to use the Pthagorean Theorem, there must be a right triangle. To create a right triangle, draw a line from point (x, ) that is perpendicular to the horizontal line through the circle. Label the resulting sides of the triangle x and (x, ) 10 5 x x Substitute the values for each side of the triangle into the formula for the Pthagorean Theorem, a + b + c. a + b + c x + = 5 x + = 5 Pthagorean Theorem Substitute values from the triangle. Simplif. The standard equation is x + = 5. U6-9 Lesson 1: Deriving Equations

16 Example Derive the standard equation of the circle with center (, 1) and radius 4. Then use a graphing calculator to graph our equation. 1. Sketch the circle (x, ) 10 4 (, 1) x Use the Pthagorean Theorem to derive the standard equation. Create a right triangle. Draw lines from point (x, ) and point (, 1) that meet at a common point and are perpendicular to each other. The length of the base of the triangle is equal to the absolute value of the difference of the x-coordinates of the endpoints. The height of the triangle is equal to the absolute value of the difference of the -coordinates of the endpoints. (continued) U6-10 Unit 6: Modeling Geometr

17 (x, ) 10 1 (x, 1) 4 (, 1) x x Substitute the resulting values for the sides of the triangle into the Pthagorean Theorem. a + b + c Pthagorean Theorem x + 1 = 4 Substitute values from the triangle. (x ) + ( 1) = 4 All squares are nonnegative, so replace the absolute value smbols with parentheses. (x ) + ( 1) = 16 Simplif. The standard equation is (x ) + ( 1) = 16. U6-11 Lesson 1: Deriving Equations

18 3. Solve the standard equation for to obtain functions that can be graphed. (x ) + ( 1) = 16 Standard equation ( 1) = 16 (x ) Subtract (x ) from both sides. ( ) 1=± 16 x If a = b, then a=± b. ( ) = 1± 16 x Add 1 to both sides to solve for. ( ) 4. Now graph the two functions, = x and ( ) = 1 16 x. On a TI-83/84: Step 1: Press [Y=]. Step : At Y 1, tpe in [1][+][ ][16][ ][(][X, T, θ, n][ ][][)][x ][)]. Step 3: At Y, tpe in [1][ ][ ][16][ ][(][X, T, θ, n][ ][][)][x ][)]. Step 4: Press [WINDOW] to change the viewing window. Step 5: At Xmin, enter [( )][9]. Step 6: At Xmax, enter [9]. Step 7: At Xscl, enter [1]. Step 8: At Ymin, enter [( )][6]. Step 9: At Ymax, enter [6]. Step 10: At Yscl, enter [1]. Step 11: Press [GRAPH]. On a TI-Nspire: Step 1: Press the [home] ke. Step : Arrow to the graphing icon and press [enter]. Step 3: At the blinking cursor at the bottom of the screen, tpe [1][+] [ ][16][ ][(][x][ ][][)][x ][enter]. (continued) U6-1 Unit 6: Modeling Geometr

19 Step 4: Move the cursor to the bottom left of the screen and click on the double right-facing arrows. Step 5: At the blinking cursor, tpe [1][ ][ ][16][ ][(][x][ ][][)][x ] [)][enter]. Step 6: Change the viewing window b pressing [menu], using the arrows to navigate down to number 4: Window/Zoom, and clicking the center button of the navigation pad. Step 7: Choose 1: Window settings b pressing the center button. Step 8: Enter in an appropriate XMin value, 9, b pressing [( )] and [9], then press [tab]. Step 9: Enter in an appropriate XMax value, [9], then press [tab]. Step 10: Leave the XScale set to Auto. Press [tab] twice to navigate to YMin and enter an appropriate YMin value, 6, b pressing [( )] and [6]. Step 11: Press [tab] to navigate to YMax. Enter [6]. Press [tab] twice to leave YScale set to auto and to navigate to OK. Step 1: Press [enter]. Step 13: Press [menu] and select : View and 5: Show Grid x U6-13 Lesson 1: Deriving Equations

20 Example 3 Write the standard equation and the general equation of the circle that has center ( 1, 3) and passes through ( 5, 5). 1. Sketch the circle ( 5, 5) r 6 4 ( 1, 3) x U6-14 Unit 6: Modeling Geometr

21 . Use the distance formula to find the radius, r. ( ) + ( ) d= x x 1 1 Distance formula Substitute ( 1, 3) and ( 5, 5) for r = [( 5) ( 1) ] + ( 5 3) (x 1, 1 ) and (x, ). r = 4 Simplif. ( ) +( ) r = r = 0 r = r = 4 5 r = Write 0 as a product with a perfect square factor. Appl the propert ab = a b. Simplif. 3. Substitute the center and radius directl into the standard form. (x h) + ( k) = r Standard form ( ) ( ) = ( ) x (x + 1) + ( 3) = 0 The standard equation is (x + 1) + ( 3) = 0. Substitute values into the equation, using the center ( 1, 3), and the radius 5. Simplif to obtain the standard equation. U6-15 Lesson 1: Deriving Equations

22 4. Square the binomials and rearrange terms to obtain the general form. (x + 1) + ( 3) = 0 Standard equation (x + 1)(x + 1) + ( 3)( 3) = 0 Expand the factors. x + x = 0 x + x = 0 x + x = 0 x + + x 6 10 = 0 The general equation is x + + x 6 10 = 0. Square the binomials to obtain trinomials. Combine the constant terms on the left side of the equation. Subtract 0 from both sides to get 0 on the right side. Rearrange terms in descending order to obtain the general equation. U6-16 Unit 6: Modeling Geometr

23 Example 4 Find the center and radius of the circle described b the equation x + 8x + + = Rewrite the equation in standard form. x + 8x + + = 0 x + 8x + = General form of the equation Subtract from both sides to get the constant term on one side. x 8x+ + = Group same-variable terms. Next, complete the square for both variables. Add the same values to both sides of the equation as shown: 8 8 x 8x = + + x 8x = Simplif the equation shown above. (x 4) + ( + 1) = 15 Write the perfect square trinomials as squares of binomials. The standard equation is (x 4) + ( + 1) = 15.. Determine the center and radius. Write the standard equation (x 4) + ( + 1) = 15 from step 1. ( x 4) + ( 1) = Rewrite to match the form ( 15) (x h) + ( k) = r. For the equation (x h) + ( k) = r, the center is (h, k) and the radius ( ) + ( ) is r, so for the equation x , the center is (4, 1) and the radius is 15. = ( ) U6-17 Lesson 1: Deriving Equations

24 Example 5 Find the center and radius of the circle described b the equation 4x x = Rewrite the equation in standard form. 4x x = 0 x + + 5x = 0 x + + 5x 10 = 9 x + 5x+ 10= 9 General form of the equation Divide each term on both sides b 4 to make the leading coefficient 1. Subtract 9 from both sides to get the constant term on one side. Combine like terms. Next, complete the square for both variables. Add the same values to both sides of the equation, as shown below: x + 5x = + 5 x + 5x = Simplif the equation from above. 5 x+ + ( 5 ) = 9 Write the perfect square trinomials as squares of binomials. 4 5 The standard equation is x+ + 5 ( ) = 9. 4 U6-18 Unit 6: Modeling Geometr

25 . Determine the center and radius. 5 x+ + ( ) 5 = 9 4 x 5 + ( ) = 3 5 Write the standard equation from step 1. Rewrite to match the form (x h) + ( k) = r. For the equation (x h) + ( k) = r, the center is (h, k) and the radius is r, so for the equation x , 5 and the radius is 3. 3 ( ) =, the center is U6-19 Lesson 1: Deriving Equations

26 PRACTICE UNIT 6 MODELING GEOMETRY Lesson 1: Deriving Equations Practice 6.1.1: Deriving the Equation of a Circle For problems 1 4, write the standard equation of the circle described. 1. The center is (0, 0) and the radius is 5.. The center is (3, 3) and the radius is The center is (1.1, ) and the radius is The center is (5, ) and the circle passes through (0, 6). Use the provided information in each problem that follows to solve. 5. Write the general equation of the circle with center ( 3.5, 1) and radius Find the center and radius of the circle described b the equation x + 8x = 0. U6-0 Unit 6: Modeling Geometr continued

27 PRACTICE UNIT 6 MODELING GEOMETRY Lesson 1: Deriving Equations 7. Find the center and radius of the circle described b the equation 133 4x x+ 4 + = A particular radio station emits a strong signal within a 3-mile radius. The station is located at (10, 1) on a coordinate plane whose units represent miles. What is the standard equation of the outer boundar of the region that receives a strong signal? If Marc lives at (15, 18), does she receive a strong signal? Explain. 9. A furniture store offers free deliver to homes that are 40 miles or fewer from the store. The store is located at (0, 0) on a coordinate plane whose units represent miles. What is the standard equation of the outer boundar of the free deliver region? Will a customer at , get free deliver? Explain. ( ) 10. Mr. Beck is a high school math teacher who coaches the track-and-field team. He makes a drawing of the team s practice field on a coordinate plane, using feet as the distance unit. He draws a quarter-circle on the coordinate plane to represent a boundar of the region for discus-throw practice. He writes the equation x ,600 = 0 to represent the full circle. What are the center and radius of the full circle? U6-1 Lesson 1: Deriving Equations

28 Lesson 6.1.: Deriving the Equation of a Parabola Introduction Earlier we studied the circle, which is the set of all points in a plane that are equidistant from a given point in that plane. We have investigated how to translate between geometric descriptions and algebraic descriptions of circles. Now we will investigate how to translate between geometric descriptions and algebraic descriptions of parabolas. Ke Concepts A quadratic function is a function that can be written in the form f (x) = ax + bx + c, where a 0. The graph of an quadratic function is a parabola that opens up or down. A parabola is the set of all points that are equidistant from a given fixed point and a given fixed line that are both in the same plane as the parabola. That given fixed line is called the directrix of the parabola. The fixed point is called the focus. The parabola, directrix, and focus are all in the same plane. The vertex of the parabola is the point on the parabola that is closest to the directrix. Focus (F) p Vertex (V) p U6- Unit 6: Modeling Geometr Directrix Parabolas can open in an direction. In this lesson, we will work with parabolas that open up, down, right, and left. As with circles, there is a standard form for the equation of a parabola; however, that equation differs depending on which direction the parabola opens (right/left or up/down).

29 Parabolas That Open Up or Down The standard form of an equation of a parabola that opens up or down and has vertex (h, k) is (x h) = 4p( k), where p 0. F (h, k + p) Directrix = k p V (h, k) V (h, k) Directrix = k p F (h, k + p) Parabolas That Open Right or Left p > 0 p < 0 The standard form of an equation of a parabola that opens right or left and has vertex (h, k) is ( k) = 4p(x h), where p 0. Directrix Directrix V (h, k) F (h + p, k) F (h + p, k) V (h, k) x = h p p > 0 p < 0 x = h p U6-3 Lesson 1: Deriving Equations

30 All Parabolas For an parabola, the focus and directrix are each p units from the vertex. Also, the focus and directrix are p units from each other. If the vertex is (0, 0), then the standard equation of the parabola has a simple form. (x 0) = 4p( 0) is equivalent to the simpler form x = 4p. ( 0) = 4p(x 0) is equivalent to the simpler form = 4px. Either of the following two methods can be used to write an equation of a parabola: Appl the geometric definition to derive the equation. Or, substitute the vertex coordinates and the value of p directl into the standard form. U6-4 Unit 6: Modeling Geometr

31 Guided Practice 6.1. Example 1 Derive the standard equation of the parabola with focus (0, ) and directrix = from the definition of a parabola. Then write the equation b substituting the vertex coordinates and the value of p directl into the standard form. 1. To derive the equation, begin b plotting the focus. Label it F (0, ). Graph the directrix and label it =. Sketch the parabola. Label the vertex V. F (0, ) V =. Let A (x, ) be an point on the parabola. A (x, ) F (0, ) V B (x, ) = Point A is equidistant from the focus and the directrix. The distance from A to the directrix is the vertical distance AB, where B is on the directrix directl below A. Since the directrix is at =, the -coordinate of B is. Because B is directl below A, it has the same x-coordinate as A. So B has coordinates (x, ). U6-5 Lesson 1: Deriving Equations

32 3. Appl the definition of a parabola to derive the standard equation using the distance formula. Since the definition of a parabola tells us that AF = AB, use the graphed points for AF and AB to appl the distance formula to this equation. [ ] ( ) + ( ) = ( ) + x 0 x x ( ) ( ) = ( + ) Simplif. x + x + ( ) = ( + ) x = ( + ) ( ) x = ( ) ( 4 + 4) x = x = 8 Square both sides to ield an equivalent equation. (See note.) Subtract ( ) from both sides to get all x terms on one side and all terms on the other side. Square the binomials. Distribute the negative sign. Simplif. The standard equation is x = 8, or (x 0) = 8( 0). Note: Squaring both sides of an equation sometimes does not ield an equivalent equation. For example, the onl solution to x = 6 is 3, but squaring both sides of that equation ields 4x = 36, which has two solutions: 3 and 3. Note that if 3 is substituted for x in the equation x = 6, the value of the left side is negative. However, ever ordered pair (x, ) that satisfies x + ( ) = ( + ) also satisfies ( ) = ( + ) and vice versa because all terms in both x + equations are squared terms and therefore nonnegative. Note that if an negative value is substituted for x or in these equations, both sides of the equation remain nonnegative. U6-6 Unit 6: Modeling Geometr

33 4. To write the equation using the standard form, first determine the coordinates of the vertex and the value of p. F (0, ) A (x, ) V (0, 0) B (x, ) = For an parabola, the distance between the focus and the directrix is p. In this case, p = ( )= 4, thus p =. The parabola opens up, so p is positive, and therefore p =. For an parabola, the distance between the focus and the vertex is p. Since p =, the vertex is units below the focus. Therefore, the vertex coordinates are (0, 0). 5. Use the results found in step 4 to write the equation. (x h) = 4p( k) Standard form for a parabola that opens up or down (x 0) = 4()( 0) Substitute the values for h, p, and k. x = 8 Simplif. The standard equation is x = 8, or (x 0) = 8( 0). The results found in steps 3 and 5 match; so, either method of finding the equation (deriving it using the definition or writing the equation using the standard form) will ield the same equation. U6-7 Lesson 1: Deriving Equations

34 Example Derive the standard equation of the parabola with focus ( 1, ) and directrix x = 7 from the definition of a parabola. Then write the equation b substituting the vertex coordinates and the value of p directl into the standard form. 1. To derive the equation, begin b plotting the focus. Label it F ( 1, ). Graph the directrix and label it x = 7. Sketch the parabola. Label the vertex V. x = 7 F ( 1, ) V U6-8 Unit 6: Modeling Geometr

35 . Let A (x, ) be an point on the parabola. A (x, ) B (7, ) F ( 1, ) V x = 7 Point A is equidistant from the focus and the directrix. The distance from A to the directrix is the horizontal distance AB, where B is on the directrix directl to the right of A. The x-value of B is 7 because the directrix is at x = 7. Because B is directl to the right of A, it has the same -coordinate as A. So, B has coordinates (7, ). U6-9 Lesson 1: Deriving Equations

36 3. Appl the definition of a parabola to derive the standard equation using the distance formula. Since the definition of a parabola tells us that AF = AB, use the graphed points for AF and AB to appl the distance formula to this equation. [ x ( 1) ] + ( ) = ( x 7) + ( ) ( x+ 1) + ( ) = ( x 7) Simplif. (x + 1) + ( ) = (x 7) Square both sides. ( ) = (x 7) (x + 1) Subtract (x + 1) from both sides to get all x terms on one side and all terms on the other side. ( ) = (x 14x + 49) (x + x + 1) Square the binomials on the right side. ( ) = x 14x + 49 x x 1 Distribute the negative sign. ( ) = 16x + 48 Simplif. ( ) = 16(x 3) Factor on the right side to obtain the standard form ( k) = 4p(x h). The standard equation is ( ) = 16(x 3). U6-30 Unit 6: Modeling Geometr

37 4. Write the equation using standard form. To write the equation using the standard form, first determine the coordinates of the vertex and the value of p. A (x, ) B (7, ) F ( 1, ) V (3, ) x = 7 For an parabola, the distance between the focus and the directrix is p. In this case, p = 7 ( 1)= 8, thus p = 4. The parabola opens left, so p is negative, and therefore p = 4. For an parabola, the distance between the focus and the vertex is p. Since p = 4, the vertex is 4 units right of the focus. Therefore, the vertex coordinates are (3, ). 5. Use the results found in step 4 to write the equation. ( k) = 4p(x h) Standard form for a parabola that opens right or left ( ) = 4( 4)(x 3) Substitute the values for h, p, and k. ( ) = 16(x 3) Simplif. The standard equation is ( ) = 16(x 3). The results shown in steps 3 and 5 match; so, either method of finding the equation (deriving it using the definition or writing the equation using the standard form) will ield the same equation. U6-31 Lesson 1: Deriving Equations

38 Example 3 Derive the standard equation of the parabola with focus (0, p) and directrix = p, where p is an real number other than Sketch diagrams showing the two possible orientations of the parabola. Include the focus, directrix, and vertex. F (0, p) = p V x V x = p F (0, p) Because the focus is (0, p) and the directrix is = p, the parabola can open either up or down. The vertex is the origin because it is equidistant from the focus and directrix. U6-3 Unit 6: Modeling Geometr

39 . Let A (x, ) be an point on the parabola. F (0, p) A (x, ) = p B (x, p) V x V x B (x, p) = p F (0, p) A (x, ) Point A is equidistant from the focus and the directrix. The distance from A to the directrix is the vertical distance AB, where B is on the directrix directl above or below A. Because B is directl above or below A, it has the same x-coordinate as A. So B has coordinates (x, p). 3. Appl the definition of a parabola to derive the standard equation using the distance formula. Since the definition of a parabola tells us that AF = AB, use the graphed points for AF and AB to appl the distance formula to this equation. [ ] ( ) + ( ) = ( ) + x p x x ( p) 0 ( ) = ( + ) Simplif. x + p p x + ( p) = ( + p) x = ( + p) ( p) x = ( + p + p ) ( p + p ) x = + p + p + p p x = 4p Square both sides. Subtract ( p) from both sides to get all x terms on one side and all terms on the other side. Square the binomials on the right side. Distribute the negative sign. Simplif. The standard equation is x = 4p, or (x 0) = 4p( 0). U6-33 Lesson 1: Deriving Equations

40 Example 4 Write the standard equation of the parabola with focus ( 5, 6) and directrix = 3.4. Then use a graphing calculator to graph our equation. 1. Plot the focus and graph the directrix. Sketch the parabola. Label the vertex V. = 3.4 V x F ( 5, 6). To write the equation, first determine the coordinates of the vertex and the value of p. = 3.4 V ( 5, 1.3) x F ( 5, 6) The distance between the focus and the directrix is p. So p = 34. ( 6)= 94., and p = 47.. The parabola opens down, so p is negative, and therefore p = 4.7. The distance between the focus and the vertex is p, so the vertex is 4.7 units above the focus. Add to find the -coordinate of the vertex: = 1.3. The vertex coordinates are ( 5, 1.3). U6-34 Unit 6: Modeling Geometr

41 3. Use the results found in step to write the equation. (x h) = 4p( k) Standard form for a parabola that opens down [x ( 5)] = 4( 4.7)[ ( 1.3)] Substitute the values for h, p, and k. (x + 5) = 18.8( + 1.3) Simplif. The standard equation is (x + 5) = 18.8( + 1.3). 4. Solve the standard equation for to obtain a function that can be graphed. (x + 5) = 18.8( + 1.3) Standard equation 1 ( x+ 5) = Multipl both sides b ( x+ 5) 13. = Add 1.3 to both sides Graph the function using a graphing calculator. On a TI-83/84: Step 1: Press [Y=]. Step : At Y 1, tpe in [(][( )][1][ ][18.8][)][(][X, T, θ, n][+][5][)][x ] [ ][1.3]. Step 3: Press [WINDOW] to change the viewing window. Step 4: At Xmin, enter [( )][1]. Step 5: At Xmax, enter [1]. Step 6: At Xscl, enter [1]. Step 7: At Ymin, enter [( )][8]. Step 8: At Ymax, enter [8]. Step 9: At Yscl, enter [1]. Step 10: Press [GRAPH]. (continued) U6-35 Lesson 1: Deriving Equations

42 On a TI-Nspire: Step 1: Press the [home] ke. Step : Arrow to the graphing icon and press [enter]. Step 3: At the blinking cursor at the bottom of the screen, enter [(] [( )][1][ ][18.8][)][(][x][+][5][)][x ][ ][1.3]. Step 4: Change the viewing window b pressing [menu], arrowing down to number 4: Window/Zoom, and clicking the center button of the navigation pad. Step 5: Choose 1: Window settings b pressing the center button. Step 6: Enter in an appropriate XMin value, 1, b pressing [( )] and [1], then press [tab]. Step 7: Enter in an appropriate XMax value, [1], then press [tab]. Step 8: Leave the XScale set to Auto. Press [tab] twice to navigate to YMin and enter an appropriate YMin value, 8, b pressing [( )] and [8]. Step 9: Press [tab] to navigate to YMax. Enter [8]. Press [tab] twice to leave YScale set to auto and to navigate to OK. Step 10: Press [enter]. Step 11: Press [menu] and select : View and 5: Show Grid x U6-36 Unit 6: Modeling Geometr

43 Example 5 The following diagram shows a plan for a top view of a stage. The back wall is to be on a parabolic curve from A to B so that all sound waves coming from point F that hit the wall are redirected in parallel paths toward the audience. F is the focus of the parabola and V is the vertex. A V F B An engineer draws the parabola on a coordinate plane, using feet as the unit of distance. The focus is ( 7, 0), the directrix is x = 5, and points A and B are on the -axis. What is the equation of the parabola? What is the width of the stage, AB? A (0, 1 ) V F ( 7, 0) x B (0, ) x = 5 U6-37 Lesson 1: Deriving Equations

44 1. To write the equation, first determine the coordinates of the vertex and the value of p. A (0, 1 ) V ( 16, 0) F ( 7, 0) x B (0, ) x = 5 The distance between the focus and the directrix is p, so p = 7 ( 5)= 18 and p = 9. The parabola opens right, so p is positive, and therefore p = 9. For an parabola, the distance between the focus and the vertex is p. In this case p = 9, so the vertex is 9 units left of the focus and therefore the vertex coordinates are ( 16, 0). U6-38 Unit 6: Modeling Geometr

45 . Use the results found in step 1 to write the equation. ( k) = 4p(x h) Standard form for a parabola that opens right or left ( 0) = 4(9)[x ( 16)] Substitute the values for h, p, and k. = 36(x + 16) Simplif. The standard equation is = 36(x + 16). 3. To find the width of the stage, AB, first find the -intercepts 1 and. = 36(x + 16) Equation of the parabola from step = 36(0 + 16) Substitute 0 for x. = 576 =± 576 =± 4 So, 1 = 4 and = 4. AB = 1 = 4 ( 4) = 48 The width of the stage is 48 feet. Simplif. U6-39 Lesson 1: Deriving Equations

46 PRACTICE UNIT 6 MODELING GEOMETRY Lesson 1: Deriving Equations Practice 6.1.: Deriving the Equation of a Parabola For problems 1 4, derive the standard equation of the parabola with the given focus and directrix. Also, write the equation that shows how ou applied the distance formula. 1. focus: (0, 3); directrix: = 3. focus: (4, 0); directrix: x = 4 3. focus: (5, 0); directrix: x = 3 4. focus: (4, 8); directrix: = 4 For problems 5 and 6, write the standard equation of the parabola with the given focus and directrix. 5. focus: ( 3, ); directrix: x = 1 6. focus: (6., 1.8); directrix: = 0. Use what ou know about parabolas to solve problems Identif the vertex, focus, and directrix of the parabola whose equation is 1 ( + ) = ( x+ 4). U6-40 Unit 6: Modeling Geometr continued

47 PRACTICE UNIT 6 MODELING GEOMETRY Lesson 1: Deriving Equations 8. The diagram below shows a parabolic flashlight reflector. Light ras from the center of the bulb at point F are reflected in parallel paths to form a beam of light. A cross section of the reflector is a section of a parabola. F Cross-section view The parabola is placed on a coordinate plane whose unit of distance is inches. The focus F is (0.5, 0) and the directrix is x = 0.5. What is the standard equation of the parabola? 9. The diagram below shows a railroad tunnel opening that is a parabolic curve. Height P Q x-axis The diagram is placed on a coordinate plane so that points P and Q are on the x-axis, the focus is (15, 10.5), and the directrix is = The unit of distance on the grid is feet. What is the standard equation of the parabola? What is the height of the opening? What is PQ, the width of the opening at ground level? Sketch the parabola, showing the coordinates of P, Q, and the vertex. continued U6-41 Lesson 1: Deriving Equations

48 PRACTICE UNIT 6 MODELING GEOMETRY Lesson 1: Deriving Equations 10. The diagram below shows a radio telescope dish. Incoming light ras reflect off of the dish and toward the feed, located at point F. A cross section of the dish is a section of a parabola. The feed is 48 inches above the vertex. The diameter of the dish at the top is 10 feet. F Cross-section view Depth An astronom student draws the parabola on a coordinate plane so that the vertex is at the origin. What is the equation of the parabola on the plane? What is the depth of the dish? U6-4 Unit 6: Modeling Geometr

49 UNIT 6 MODELING GEOMETRY Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas Common Core Georgia Performance Standard MCC9 1.G.GPE.4 Essential Questions 1. How can ou determine whether a point is on a graph?. What are some formulas that help describe how points and figures in a coordinate plane are situated with respect to each other? 3. How are equations of quadratic functions related to certain tpes of parabolas and their equations? WORDS TO KNOW axis of smmetr of a parabola circle directrix of a parabola distance formula the line through the vertex of a parabola about which the parabola is smmetric. The equation of the axis of smmetr is x = b a. the set of all points in a plane that are equidistant from a reference point in that plane, the center. The set of points forms a -dimensional curve that measures 360º. a line that is perpendicular to the axis of smmetr of a parabola and that is in the same plane as both the parabola and the focus of the parabola; the fixed line referenced in the definition of a parabola a formula that states the distance between points (x 1, 1 ) and (x, ) is equal to x x ( ) + ( ) 1 1 U6-43 Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas

50 focus of a parabola midpoint formula parabola a fixed point on the interior of a parabola that is not on the directrix of the parabola but is on the same plane as both the parabola and the directrix; the fixed point referenced in the definition of a parabola a formula that states the midpoint of the line segment created b connecting (x 1, 1 ) and (x, ) is x1+ x 1+, the set of all points that are equidistant from a fixed line, called the directrix, and a fixed point not on that line, called the focus. The parabola, directrix, and focus are all in the same plane. The vertex of the parabola is the point on the parabola that is closest to the directrix. b perfect square trinomial a trinomial of the form x + bx + quadratic function slope formula standard form of an equation of a circle standard form of an equation of a parabola theorem vertex of a parabola a function that can be written in the form f (x) = ax + bx + c, where a 0. The graph of an quadratic function is a parabola. a formula that states the slope of the line through (or the line segment connecting) A (x 1, 1 ) and 1 B (x, ) is x x 1 (x h) + ( k) = r, where (h, k) is the center and r is the radius (x h) = 4p( k) for parabolas that open up or down; ( k) = 4p(x h) for parabolas that open right or left. For all parabolas, p 0 and the vertex is (h, k). a statement that is shown to be true the point on a parabola that is closest to the directrix and lies on the axis of smmetr U6-44 Unit 6: Modeling Geometr

51 Recommended Resources IXL Learning. Lines in the coordinate plane: Slopes of lines. This interactive website gives a series of problems involving slopes of lines and scores them immediatel. If the user submits a wrong answer, the correct answer is given and then a thorough explanation is provided, including the slope formula and how it is applied to the problem. IXL Learning. Lines in the coordinate plane: Slopes of parallel and perpendicular lines. This interactive website gives a series of problems involving slopes of lines and whether the lines are parallel, perpendicular, or neither, and scores the problems immediatel. If the user submits a wrong answer, the correct answer is given and then a thorough explanation is provided, including the slope conditions required for parallel and perpendicular lines, and how those conditions are applied to the problem. IXL Learning. Points, lines, and segments: Midpoint formula. This interactive website gives a series of problems asking for midpoints of line segments and scores them immediatel. If the user submits a wrong answer, the correct answer is given and then a thorough explanation is provided, including the midpoint formula and how it is applied to the problem. U6-45 Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas

52 Lesson 6..1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas Introduction A theorem is statement that is shown to be true. Some important theorems have names, such as the Pthagorean Theorem, but man theorems do not have names. In this lesson, we will appl various geometric and algebraic concepts to prove and disprove statements involving circles and parabolas in a coordinate plane. If a statement is proven, it is a theorem. If a statement is disproved, it is not a theorem. The directions for most problems will have the form Prove or disprove, meaning we will work through those problems to discover whether each statement is true or false. Then, at the end of the work, we will state whether we have proved or disproved the statement. Ke Concepts A theorem is an statement that is proven or can be proved to be true. The standard form of an equation of a circle with center (h, k) and radius r is ( ) + ( ) = x h k r. This is based on the fact that an point (x, ) is on the ( ) + ( ) = circle if and onl if x h k r. b Completing the square is the process of determining the value of b and adding it to x + bx to form the perfect square trinomial x + bx +. A quadratic function can be represented b an equation of the form f (x) = ax + bx + c, where a 0. The graph of an quadratic function is a parabola that opens up or down. A parabola is the set of all points that are equidistant from a fixed line, called the directrix, and a fixed point not on that line, called the focus. The parabola, directrix, and focus are all in the same plane. The distance between the focus and a point on the parabola is the same as the distance from that point to the directrix. The vertex of the parabola is the point on the parabola that is closest to the U6-46 Unit 6: Modeling Geometr

53 directrix. Ever parabola is smmetric about a line called the axis of smmetr. The axis of smmetr intersects the parabola at the vertex. The x-coordinate of the vertex is b a. b The -coordinate of the vertex is f a. The standard form of an equation of a parabola that opens up or down and has vertex (h, k) is (x h) = 4p( k), where p 0 and p is the distance between the vertex and the focus and between the vertex and the directrix. Parabolas that open up or down represent functions, and their equations can be written in either of the following forms: = ax + bx + c or (x h) = 4p( k). If one form is known, the other can be found. The standard form of an equation of a parabola that opens right or left and has vertex (h, k) is ( k) = 4p(x h), where p 0 and p is the distance between the vertex and the focus and between the vertex and the directrix. In an parabola: The focus and directrix are each p units from the vertex. The focus and directrix are p units from each other. U6-47 Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas

54 Guided Practice 6..1 Example 1 Given the point A ( 6, 0), prove or disprove that point A is on the circle centered at ( ) the origin and passing through, Draw a circle on a coordinate plane using the given information. You do not et know if point A lies on the circle, so don t include it in our diagram. In the diagram that follows, the name P is assigned to the origin and G is assigned to the known point on the circle. To help in plotting points, ou can use a calculator to find decimal approximations. P (0, 0) x G (, 4 ) U6-48 Unit 6: Modeling Geometr

55 . Find the radius of the circle using the distance formula. Use the known points, P and G, to determine the radius of the circle. ( ) + ( ) r= x x 1 1 Distance formula [ ] + r = ( ) ( 0) 4 0 r = ( ) ( ) + 4 ( ) ( ) Substitute (0, 0) and (, 4 ) for (x 1, 1 ) and (x, ). Simplif, then solve. r = 4+ 3 r = 36 r = 6 The radius of the circle is 6 units. For point A to be on the circle, it must be precisel 6 units awa from the center of the circle. U6-49 Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas

56 3. Find the distance of point A from the center P to determine whether it is on the circle. The coordinates of point P are (0, 0). The coordinates of point A are ( 6, 0). ( ) + ( ) AP = x x 1 1 Distance formula AP = [( 6) (0)] + [(0) (0) ] Substitute (0, 0) and ( 6, 0) for (x 1, 1 ) and (x, ). AP = AP = ( 6) ( 6) +( 0) Simplif, then solve. AP = 36 AP = 6 Point A is 6 units from the center, and since the radius of the circle is 6 units, point A is on the circle. The original statement has been proved, so it is a theorem. A ( 6, 0) P (0, 0) x G (, 4 ) U6-50 Unit 6: Modeling Geometr

57 Example Prove or disprove that the quadratic function graph with vertex ( 4, 0) and passing through (0, 8) has its focus at ( 4, 1). 1. Sketch the graph using the given information. (0, 8) ( 4, 0) x U6-51 Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas

58 . Derive an equation of the parabola from its graph. The parabola opens up, so it represents a function. Therefore, its equation can be written in either of these forms: = ax + bx + c or (x h) = 4p( k). We were given a vertex and a point on the parabola; therefore, we ll use the form (x h) = 4p( k) and the vertex to begin deriving the equation. The vertex is ( 4, 0), so h = 4 and k = 0. (x h) = 4p( k) Standard form of an equation for a parabola that opens up or down [x ( 4)] = 4p( 0) Substitute the vertex ( 4, 0) into the equation. (x + 4) = 4p Simplif, but do not expand the binomial. The equation of the parabola is (x + 4) = 4p. 3. Substitute the given point on the parabola into the standard form of the equation to solve for p. The point given is (0, 8). (x + 4) = 4p Simplified equation from step (0 + 4) = 4p(8) Substitute the point (0, 8) into the equation. 16 = 3p Simplif and solve for p. p = 1 U6-5 Unit 6: Modeling Geometr

59 4. Use the value of p to determine the focus. p is positive, so the focus is directl above the vertex. p = 1, so the focus is 1 unit above the vertex. The vertex is ( 4, 0), so the focus is 4, 1. This result disproves the statement that the quadratic function graph with vertex ( 4, 0) and passing through (0, 8) has its focus at ( 4, 1). The statement has been disproved, so it is not a theorem. Instead, the following statement has been proved: The quadratic function graph with vertex ( 4, 0) and passing through (0, 8) has its focus at 4, 1. (0, 8) F ( 4, ) 1 V( 4, 0) x U6-53 Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas

60 Example 3 The following information is given about a parabola: The vertex V is at (0, 0). The focus F is at (p, 0), with p > 0. The line segment through F is perpendicular to the axis of smmetr and connects two points of the parabola. Prove that the line segment through F has length 4p. 1. Make a sketch using the given information. A (p, 1 ) V (0, 0) F (p, 0) x B (p, ) The vertex is (0, 0) and the focus is F (p, 0) with p > 0, so F is on the positive x-axis. The axis of smmetr is the x-axis. The line segment through F, perpendicular to the axis of smmetr, and connecting two points of the parabola is the vertical line segment, AB. Because the segment is vertical, both A and B have p as their x-coordinate, matching the x-coordinate of F. The -coordinates of A and B are unknown; the are named 1 and in the diagram. U6-54 Unit 6: Modeling Geometr

61 . Because the parabola opens to the right, its equation has the form ( k) = 4p(x h). Use this equation to solve for 1 and in terms of p. ( k) = 4p(x h) Standard form of the equation for a parabola that opens right or left ( 0) = 4p(x 0) Substitute (0, 0) for (h, k). = 4px Simplif. ( 1 ) = 4p p Substitute the coordinates of point A. ( 1 ) = 4p Simplif. 1 = 4 p 1 = p Solve for 1. 1 is positive, so take the positive square root of 4p. ( ) = 4p p Substitute the coordinates of point B into the equation = 4px. ( ) = 4p Simplif. = 4 p = p Solve for. is negative, so take the negative square root of 4p. 3. Use the results from step to find the length of AB. AB = 1 = p ( p) = p + p = 4p For the parabola with vertex V (0, 0) and focus F (p, 0) with p > 0, the line segment through F has length 4p. The original statement has been proved; therefore, it is a theorem. U6-55 Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas

62 Example 4 Prove or disprove that the points A (4, ), B (, 5), C (6, 5), and D ( 4, 10) are all on the quadratic function graph with vertex V (, 1) that passes through E (0, ). 1. Make a sketch with the given information. A quadratic function graph is a parabola. You do not et know if an of the points A, B, C, or D lie on the parabola, so do not show them in our sketch. E (0, ) V (, 1) x U6-56 Unit 6: Modeling Geometr

63 . Derive an equation of the parabola from its graph. The parabola opens up, so it represents a function. Therefore, its equation can be written in either of these forms: = ax + bx + c or (x h) = 4p( k). We were given a vertex and a point on the parabola, so we ll use the form (x h) = 4p( k) and the vertex to begin deriving the equation. The vertex is (, 1), so h = and k = 1. (x h) = 4p( k) Standard form of the equation for a parabola that opens up or down (x ) = 4p( 1) Substitute (, 1) for (h, k). 3. Continue to derive the equation of the parabola b finding p. Use the given point (0, ) and the equation derived from step. (x ) = 4p( 1) Derived equation [(0) ] = 4p[() 1] Substitute point E (0, ) for (x, ). ( ) = 4p(1) Simplif, then solve for p. 4 = 4p p = 1 U6-57 Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas

64 4. Convert the standard form of the equation into the general form of the equation. Use the value of p = 1 and the standard form of the equation derived in step. (x ) = 4p( 1) Standard form of the equation with the vertex substituted (x ) = 4(1)( 1) Standard form of the equation with the value of p substituted x 4x + 4 = 4( 1) Expand the binomial. 1 x x+ 1= 1 4 Divide both sides of the equation b 4. 1 x x+ = Add 1 to both sides. 4 1 The equation of the parabola is = x x Determine whether the points A, B, C, and D are on the parabola b substituting their coordinates into the equation. A (4, ): ( ) ( ) = B (, 5): ( ) ( ) 5= C (6, 5): ( ) ( ) 5= D ( 4, 10): ( ) ( ) 10= A, B, C, and D are all on the parabola. The statement has been proved, so it is a theorem. Yes, the equation is true, so A is on the parabola. Yes, the equation is true, so B is on the parabola. Yes, the equation is true, so C is on the parabola. Yes, the equation is true, so D is on the parabola. U6-58 Unit 6: Modeling Geometr

65 Example 5 Prove or disprove that P (, 1), Q (6, 5), R (8, 1), and S (0, 3) are vertices of a rectangle that is inscribed in the circle centered at C (3, 1) and passing through A( 11, + 1). 1. Make sketches using the given information. You do not et know if an of the points P, Q, R, or S lie on the circle, so show the polgon on a separate coordinate sstem. To help in plotting point A, ou can use a calculator to find a decimal approximation. A (1, 1 + 1) Q (6, 5) C (3, 1) x P (, 1) R (8, 1) x S (0, 3). Find the radius of the circle. ( ) = ( ) +( ) ( ) + + radius = AC = radius = AC = 4+ 1= 5= 5 The radius of the circle is 5 units. U6-59 Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas

66 3. Find the distance of points P, Q, R, and S from the center C to determine whether the are on the circle. PC = QC = RC = SC = ( ) ( ) = ( ) +( ) = = ( 3 6) + ( 1 5) = ( 3) + ( 4) = 9+ 16= 5= 5 ( 3 8) + ( 1 1) = ( 5) +( 0) = 5 = 5 ( 3 0) + 1 ( 3) = ( 3 ) +( 4 ) = = 5 = 5 P, Q, R, and S are each 5 units from the center, and since the radius is also 5 units, the points are all on the circle. 4. Determine whether PQRS is a rectangle. Use slopes to identif parallel and perpendicular segments. 5 1 slope of PQ = 6 4 ( ) = 8 = slope of RS = 3 1 = = slope of QR = 5 = 4 = 8 6 ( ) 1 slope of SP = 3 = 4 0 = PQ is parallel to RS and QR is parallel to SP because the have equal slopes. Therefore, PQRS is a parallelogram because it has both pairs of opposite sides parallel. PQ is perpendicular to QR because the product of their slopes is 1: 1 1 ( )=. Thus, PQRS is a rectangle because it is a parallelogram with a right angle. U6-60 Unit 6: Modeling Geometr

67 5. Steps and 3 allowed us to determine that P, Q, R, and S are all on the circle. Step 4 shows that PQRS is a rectangle. Therefore, P (, 1), Q (6, 5), R (8, 1), and S (0, 3) are vertices of a rectangle that is inscribed in the circle centered at C (3, 1) and passing through A( 11, + 1). The statement has been proved, so it is a theorem. A (1, 1 + 1) Q (6, 5) P (, 1) C (3, 1) R (8, 1) x S (0, 3) U6-61 Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas

68 PRACTICE UNIT 6 MODELING GEOMETRY Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas Practice 6..1: Using Coordinates to Prove Geometric Theorems About Circles and Parabolas Use the given information to prove or disprove each statement. Justif our reasoning. ( ) 1. Prove or disprove that point Q 1, lies on the circle centered at the origin R and passing through the point A (0, 3).. Prove or disprove that point A (0, 7) is on the circle centered at the origin R and passing through the point P ( 5, 6). 3. Given the points P (, ), Q (4, 8), and R (0, 0), prove or disprove that the points are on the parabola with focus F 0, 1 and directrix = Given the points A (, 8), B (1, ), and C (.5, 10.15), prove or disprove 1 that the points are on the quadratic function graph with focus F 0, 8 and directrix = Prove or disprove that the points A (5, 1), B (, ), C (6, ), and D (1, 7) are all on the quadratic function graph with vertex V (4, ) that passes through E (0, 14). U6-6 Unit 6: Modeling Geometr continued

69 PRACTICE UNIT 6 MODELING GEOMETRY Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas 6. Prove or disprove that the point A (5, 6) lies on the parabola with focus F (5, 1) and directrix x = Prove or disprove that the points A ( 5, 1), B (5, 1), and C (0, 13) are the vertices of an isosceles triangle inscribed in the circle centered at the origin Q and passing through the point P ( 10, 69). 8. The diagram below shows a target at a carnival dart game. The diagram is on a coordinate sstem. A plaer wins a prize b hitting the shaded ring. The ring is formed b two circles. Both circles have center C (1, 1). One circle passes through P 1 (6, 0) and the other circle passes through P (1, 3). Natasha throws a dart and hits the point Q (19, 4). Does she get a prize? Justif our answer. C (1, 1) x continued U6-63 Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas

70 PRACTICE UNIT 6 MODELING GEOMETRY Lesson : Using Coordinates to Prove Geometric Theorems About Circles and Parabolas 9. An art student created the following graph to represent the letter M. To create the image, she graphed two parabolas intersecting at the point A. The parabolas are described as follows: a parabola with vertex V 1 (6, 8) and focus F 1 (6, 5) a parabola with vertex V (18, 8) and focus F (18, 5) Prove or disprove that point A has coordinates (1, 5). A x 10. The diagram below represents a suspension bridge. The curve is a portion of a parabola. The parabola has vertex V (0, 10) and passes through the point (0, 10.4). Prove or disprove that all points on the parabola are equidistant from the point (0, 60) and the line = 40. U6-64 Unit 6: Modeling Geometr

71 UNIT 6 MODELING GEOMETRY Lesson 3: Solving Sstems of Linear Equations and Circles Common Core Georgia Performance Standard MCC9 1.A.REI.7 Essential Questions 1. Given a line and a circle in the same plane, how man points of intersection are possible?. Given a sstem of equations, one of a line and one of a circle, how are the solutions of the sstem related to points of intersection of the graphs? 3. Given a sstem of equations, one of a line and one of a circle, how are the solutions of the sstem related to solutions of a quadratic equation in one variable? WORDS TO KNOW circle the set of all points in a plane that are equidistant from a reference point in that plane, the center. The set of points forms a -dimensional curve that measures 360º. distance formula a formula that states the distance between points (x 1, 1 ) ( ) ( 1) + 1 and (x, ) is equal to x x linear equation an equation that can be written in the form ax + b = c, where a, b, and c are rational numbers; can also be written as = mx + b, in which m is the slope, b is the -intercept, and the graph is a straight line. The solutions to the linear equation are the infinite set of points on the line. midpoint formula a formula that states the midpoint of the line segment created b connecting (x 1, 1 ) and (x, ) is x1+ x 1+, U6-65 Lesson 3: Solving Sstems of Linear Equations and Circles

72 quadratic equation an equation that can be written in the form ax + bx + c = 0, where x is the variable, a, b, and c are constants, and a 0 quadratic formula standard form of an equation of a circle sstem of equations Zero Product Propert a formula that states the solutions of a quadratic equation of the form ax + bx + c = 0 are given b b b ac x = ± 4. A quadratic equation in this form a can have no real solutions, one real solution, or two real solutions. (x h) + ( k) = r, where (h, k) is the center and r is the radius a set of equations with the same unknowns if the product of two factors is 0, then at least one of the factors is 0 U6-66 Unit 6: Modeling Geometr

73 Recommended Resources IXL Learning. Factoring: Factor polnomials. This interactive website gives a series of problems asking the user to factor polnomials and scores the answers immediatel. Incorrect answers are corrected and explained step-b-step. The user will be able to appl this skill when solving sstems involving a linear equation and an equation of a circle. IXL Learning. Quadratic equations: Solve a quadratic equation b factoring. This interactive website provides users with quadratic equations to solve b factoring and scores the answers immediatel. Incorrect answers are corrected and explained step-b-step. The user will be able to appl this skill when solving sstems involving a linear equation and an equation of a circle. IXL Learning. Quadratic equations: Solve a quadratic equation using the quadratic formula. This interactive website provides users with quadratic equations to solve b using the quadratic formula and scores the answers immediatel. Incorrect answers are corrected and explained step-bstep. The user will need to be able solve quadratic equations with the quadratic formula when solving sstems involving a linear equation and an equation of a circle. U6-67 Lesson 3: Solving Sstems of Linear Equations and Circles

74 Lesson 6.3.1: Solving Sstems of Linear Equations and Circles Introduction We have worked with linear equations and equations of circles. We have solved sstems of equations, including sstems involving linear equations and equations of parabolas. Now we will bring together skills and concepts from these topics and appl them in this lesson to solve sstems of linear equations and equations of circles. Ke Concepts A sstem of equations is a set of equations with the same unknowns. If a twoequation sstem has a linear equation and a circle equation, then the sstem can have no real solutions, one real solution, or two real solutions. A solution is an ordered pair; its graphical representation is a point at which the line and circle intersect. No real solution One real solution Two real solutions Sstems of equations can be solved b using graphical and/or algebraic methods. The graphical method is sometimes useful onl for estimating solutions. No matter which method is used to solve a sstem, the solution(s) should be checked algebraicall b substitution. A sstem with two real solutions will have two points of intersection between the line and the circle. A sstem with one real solution will have exactl one point of intersection. A sstem with no real solutions will have two complex solutions that cannot be graphed on the Cartesian coordinate plane. Remember that the standard form of an equation of a circle is (x h) + ( k) = r, where (h, k) is the center and r is the radius. A quadratic equation of the form ax + bx + c = 0 can have no real solutions, one real solution, or two real solutions. U6-68 Unit 6: Modeling Geometr

75 Using substitution, a sstem of a linear equation and a circle equation can be reduced to a single quadratic equation whose solutions lead to the solutions of the sstem. The quadratic formula states that if a quadratic equation of the form ax + bx + c = 0 has one or two real solutions, then those solutions are given b b b ac x = ± 4. a To factor a polnomial means to write it as a product of two or more polnomials. Examples: x + 6x + 9 = (x + 3)(x + 3) x + 5x 6 = (x + 6)(x 1) 3x + 10x = x(3x + 10) Note: x is a polnomial with one term, so it is a monomial. The Zero Product Propert states that if a product equals 0, then at least one of its factors is 0. Examples: If ab = 0, then a = 0 or b = 0. If (x + 6)(x 1) = 0, then x + 6 = 0 or x 1 = 0. Ever quadratic equation of the form ax + bx + c = 0 that has one or two real solutions can be solved b the quadratic formula; some can be solved more easil b factoring and using the Zero Product Propert. The distance formula states that the distance between points (x 1, 1 ) and (x, ) ( 1) + ( 1). is equal to x x The midpoint formula states that the midpoint of the line segment connecting x1+ x 1+ points (x 1, 1 ) and (x, ) is,. U6-69 Lesson 3: Solving Sstems of Linear Equations and Circles

76 Guided Practice Example 1 Write the sstem that appears to be represented b the line with equation = x and the circle shown below. Solve the sstem graphicall and then algebraicall. Check the solution(s) b substitution x 1. Write the equation of the circle. Based on the graph, the center is (6, 6) and the radius is 10. So the equation of the circle is (x 6) + ( 6) = 10.. Write the sstem. = x x = 100 ( ) ( ) Equation of the line Equation of the circle U6-70 Unit 6: Modeling Geometr

77 3. Graph the linear equation and identif an points of intersection. For each point of intersection, identif or estimate its apparent coordinates (9, 16) 8 6 (6, 6) (0, ) 4 x The linear equation = x has a slope of and -intercept of, so draw the line with slope through the point (0, ). The line intersects the circle in two points. The coordinates seem to be (0, ) and approximatel (9, 16). U6-71 Lesson 3: Solving Sstems of Linear Equations and Circles

78 4. Begin solving the sstem algebraicall. Substitute the value of in terms of x from the linear equation into the equation of the circle and then solve the resulting equation. First, write the sstem: = x ( x 6 ) + ( 6) = 100 Equation of the line Equation of the circle (x 6) + [(x ) 6] = 100 (x 6) + (x 8) = 100 Simplif. x 1x x 3x + 64 = 100 5x 44x = 100 5x 44x = 0 x(5x 44) = 0 x = 0 or 5x 44 = 0 Substitute x for into the equation of the circle. Square the binomials. Combine like terms. Subtract 100 from both sides to get 0 on the right side. Factor. Appl the Zero Product Propert. 5x = 44 Divide both sides b 5. x = 44 5 Solve. U6-7 Unit 6: Modeling Geometr

79 5. Finish solving the sstem algebraicall. Substitute the x-values from step 4 into the linear equation and find the corresponding -values. When x = 0: When x = 44 5 : = x = x = (0) = = 0 = = = = The solutions are (0, ) and 44, , or (8.8, 15.6). 6. Compare the solutions obtained algebraicall with those obtained graphicall. One solution matches: (0, ). The other solution obtained algebraicall, (8.8, 15.6), supports the estimate obtained graphicall: (9, 16). U6-73 Lesson 3: Solving Sstems of Linear Equations and Circles

80 7. Check the solutions algebraicall b substitution. First, check (0, ) b substituting it into both of the original equations in the sstem: Equation of the line: Equation of the circle: = x (x 6) + ( 6) = 100 ( ) ( ) ( ) + ( ) = = (0) = 100 = = = = 100 Then check (8.8, 15.6): Equation of the line: Equation of the circle: = x (x 6) + ( 6) = = (8.8) (8.8 6) + (15.6 6) = = 17.6 (.8) + (9.6) = = = = 100 Both solutions check. The solutions of the sstem are (0, ) and (8.8, 15.6). U6-74 Unit 6: Modeling Geometr

81 Example Solve the sstem below. Check the solution(s), then graph the sstem on a graphing calculator. 4x+ 3= 5 x + = 5 1. Solve the linear equation for one of its variables in terms of the other. 4x + 3 = 5 Equation of the line 3 = 4x = x+ 3 3 Add 4x to both sides to isolate the -term on the left side. Divide both sides b 3 to solve for in terms of x.. Use the result from step 1 to substitute for in the circle equation. x + = 5 Equation of the circle x x = Substitute 4 5 x + for x + x + x + = x + 16x + 00x + 65 = 5 5x + 00x + 65 = 5 5x + 00x = 0 x + 8x + 16 = 0 (x + 4)(x + 4) = 0 Factor. Square the binomial. Multipl both sides b 9 to eliminate fractions. Combine like terms. Subtract 5 from both sides to get 0 on one side. Divide both sides b 5 to simplif. x + 4 = 0 and x + 4 = 0 Set the factors equal to 0. x = 4 and x = 4 Solve. U6-75 Lesson 3: Solving Sstems of Linear Equations and Circles

82 3. Substitute the x-value from step into the linear equation and find the corresponding -value. 4 5 = x = ( 4)+ 3 3 = Linear equation solved for Substitute 4 for x. Simplif, then solve. = 9 3 = 3 The solution to the sstem of equations is ( 4, 3). 4. Check the solution b substituting it into both original equations. Equation of the line: Equation of the circle: 4x + 3 = 5 x + = 5 4( 4) + 3(3) = 5 ( 4) + 3 = = = 5 5 = 5 5 = 5 The solution checks. The solution of the sstem is ( 4, 3). U6-76 Unit 6: Modeling Geometr

83 5. Graph the sstem on a graphing calculator. First, solve the circle equation for to obtain functions that can be graphed. x + = 5 = 5 x Equation of the circle Subtract x from both sides to isolate on one side. =± 5 x If a = b, then a=± b because a can be negative. 4 5 = x The linear equation solved for is Graph the linear function and the two functions represented b the circle. On a TI-83/84: Step 1: Press [Y=]. Step : At Y 1, tpe in [(][4][ ][3][)][X, T, θ, n][+][5][ ][3]. Step 3: At Y, tpe in [ ][5][ ][X, T, θ, n][x ][)]. Step 4: At Y 3, tpe in [( )][ ][5][ ][X, T, θ, n][x ][)]. Step 5: Press [WINDOW] to change the viewing window. Step 6: At Xmin, enter [( )][1]. Step 7: At Xmax, enter [1]. Step 8: At Xscl, enter [1]. Step 9: At Ymin, enter [( )][8]. Step 10: At Ymax, enter [8]. Step 11: At Yscl, enter [1]. Step 1: Press [GRAPH]. On a TI-Nspire: Step 1: Press the [home] ke. Step : Navigate to the graphs icon and press [enter]. Step 3: At the blinking cursor at the bottom of the screen, tpe in [(][4][ ][3][)][x][+][5][ ][3] and press [enter]. (continued) U6-77 Lesson 3: Solving Sstems of Linear Equations and Circles

84 Step 4: Move the cursor over to the double arrow at the bottom left of the screen and press the center ke of the navigation pad to open another equation prompt. Step 5: At the blinking cursor, tpe in [ ][5][ ][x][x ] and press [enter]. Step 6: Move the cursor over to the double arrow at the bottom left of the screen and press the center ke of the navigation pad to open another equation prompt. Step 7: At the blinking cursor, tpe in [( )][ ][5][ ][x][x ] and press [enter] x U6-78 Unit 6: Modeling Geometr

85 Example 3 Determine whether the line with equation x = 5 intersects the circle centered at ( 3, 1) with radius 4. If it does, then find the coordinates of the point(s) of intersection. 1. Write the sstem of equations represented b the line and the circle. The equation of the circle with center (h, k) and radius r is (x h) + ( k) = r, so the equation of the circle is [x ( 3)] + ( 1) = 4, or (x + 3) + ( 1) = 16. The sstem is: x = 5 x = 16 ( ) ( ) Equation of the line Equation of the circle. Substitute the x-value from the linear equation into the equation of the circle and then solve the resulting equation. (x + 3) + ( 1) = 16 Equation of the circle ( 5 + 3) + ( 1) = 16 Substitute 5 for x. 4 + ( 1) = 16 Simplif = 16 Square the binomial. + 5 = = 0 Combine like terms. Subtract 16 from both sides to get 0 on the right side. Appl the quadratic formula. The solutions of ax + bx + c = 0 are b b ac x = ±, so the solutions of an equation in the form a 4 b b ac a + b + c = 0 are = ±. a 4 (continued) U6-79 Lesson 3: Solving Sstems of Linear Equations and Circles

86 = ( )± ( ) 41 ( ) 11 1 = ± ( ) ( ) Appl the quadratic formula. Simplif. = ± 48 The solutions to the quadratic equation are = + 48 = and 3. The linear equation in the sstem is x = 5, so an solution must have 5 as its x-coordinate. Use this fact along with the results from step to write the solutions of the sstem. + The solutions of the sstem are 5, 48 ( 5446,. ) and 5, 48 ( 5, 46. ). These are the coordinates of the points at which the line intersects the circle, as shown on the following graph. 6 ( 5, 4.46) ( 3, 1) 4 0 x ( 5,.46) 4 U6-80 Unit 6: Modeling Geometr

87 Example 4 Determine whether the line with equation = x 5 intersects the circle centered at the origin with radius 3. If it does, then find the coordinates of the point(s) of intersection. 1. Write the sstem of equations. = x 5 x + = 9 Equation of the line Equation of the circle. Substitute the value of in terms of x from the linear equation into the equation of the circle and then solve the resulting equation. x + = 9 Equation of the circle x + (x 5) = 9 Substitute x 5 for. x + x 10x + 5 = 9 x 10x + 5 = 9 x 10x + 16 = 0 x 5x + 8 = 0 Square the binomial. Combine like terms. Subtract 9 from both sides to get 0 on the right side. Divide both sides b to simplif. x = ( 5 )± x = ± 5 3 ( ) ( )( ) ( ) Appl the quadratic formula. Simplif. x = 5± 7 5 = ± i 7 (continued) U6-81 Lesson 3: Solving Sstems of Linear Equations and Circles

88 The quadratic equation has no real solutions because 7 is not a real number. The solutions are complex and cannot be graphed on the Cartesian coordinate plane. The complex solutions are x and x = 5 i 7. = 5+ i 7 The sstem has no real solutions, so the line does not intersect the circle, as shown in the following graph x 4 6 U6-8 Unit 6: Modeling Geometr

89 Example 5 Solve the sstem. 1 = x+ 1 4 ( x+ 3) + ( + ) = 3 1. Substitute the value of in terms of x from the linear equation into the equation of the circle and then solve the resulting equation. (x + 3) + ( + ) = 3 Equation of the circle 1 ( x+ ) + x = 3 4 ( x+ 3 1 ) + x+ 3 4 = 3 Substitute the value of in terms of x for. Simplif. 1 3 x + 6x+ 9+ x x+ 9= 3 Square the binomials x + 96x x 4x = x + 7x + 88 = x + 7x 80 = 0 Multipl both sides b 16 to eliminate fractions. Combine like terms. Subtract 368 from both sides to get 0 on the right side. x = ± ( )( 80 ) 17 ( ) x = 7 ± Appl the quadratic formula. Simplif. x = 7 ± 10, (continued) U6-83 Lesson 3: Solving Sstems of Linear Equations and Circles

90 The solutions to the quadratic equation are x = , and x = 7 10, Substitute the approximate x-values into the linear equation. 1 = x+ 1 Equation of the line 4 Substitute 0.91 for x: Substitute 5.15 for x: 1 ( ) ( ) The approximate solutions of the sstem are (0.91, 0.77) and ( 5.15,.9), as shown in the following graph. ( 5.15,.9) 4 (0.91, 0.77) x ( 3, ) 4 6 U6-84 Unit 6: Modeling Geometr