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Lesson 34: Are All Parabolas Congruent?

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Lesson 34: Are All Parabolas Congruent? NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 34 M1 ALGEBRA II Lesson 34: Are All Parabolas Congruent? Student Outcomes . Students learn the vertex form of the equation of a parabola and how it arises from the definition of a parabola. Students perform geometric operations, such as rotations, reflections, and translations, on arbitrary parabolas to discover standard representations for their congruence classes. In doing so, they learn that all parabolas 1 with the same distance 푝 between the focus and the directrix are congruent to the graph of 푦 = 푥2. 2푝 Lesson Notes This lesson builds upon the previous lesson and applies transformations to show that all parabolas with the same distance between their focus and directrix are congruent. Recall that two figures in the plane are congruent if there exists a finite sequence of rigid motions that maps one onto the other, so it makes sense to discuss congruency of parabolas. The lesson closes with a theorem and proof detailing the answer to the question posed in the lesson title. By using transformations in this lesson to determine the conditions under which two parabolas are congruent, this lesson builds coherence with the work students did in Geometry. This lesson specifically asks students to consider how we can use transformations to prove two figures are congruent. Additionally, the lesson reinforces the connections between geometric transformations and transformations of the graphs of functions. There are many opportunities to provide scaffolding in this lesson for students who are not ready to move quickly to abstract representation. Use technology, patty paper or transparencies, and simple hand-drawn graphs as appropriate to support student learning throughout this lesson. Use the anchor poster created in Lesson 33, and keep key vocabulary words and formulas (e.g., the distance formula) displayed for student reference. Consider breaking this lesson up into two days; on the first day, explore the definition of congruent parabolas, sketch parabolas given their focus and directrix, and explore the consequences of changing the distance between the focus and directrix, 푝. On the second day, derive the analytic equation for a parabola with a given focus and directrix and vertex at the origin, and prove the theorem on parabola congruence. Classwork Opening Exercise (7 minutes) Allow students to discuss their approaches to this exercise with a partner or in small groups. Keep encouraging students to consider the definition of a parabola as they try to sketch the parabolas. Encourage students who draw a haphazard curve to consider how they could make sure their graph is the set of points equidistant from the focus and directrix. Throughout this lesson, provide students with access to graphing calculators or other computer graphing software that they can use to test and confirm conjectures. Model this further, and provide additional scaffolding by using an online applet located at the website http://www.intmath.com/plane-analytic-geometry/parabola-interactive.php. Within the applet, students can move either the focus or the directrix to change the value of 푝. They can also slide a point along the parabola noting the equal distances between that point and the focus and that point and the directrix. Depending on the level of students, the teacher can begin these exercises by moving directly to the applet or by having them start the exercises and use the applet later in this section. Lesson 34: Are All Parabolas Congruent? 382 This work is licensed under a 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 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 34 M1 ALGEBRA II Opening Exercise Are all parabolas congruent? Use the following questions to support your answer. Scaffolding: . a. Draw the parabola for each focus and directrix given below. If students struggle with vocabulary, refer them The solution is shown below. back to the anchor poster. As an extension, ask students to derive the equation for the parabola as was shown in Example 1 of the previous lesson. b. What do we mean by congruent parabolas? Two parabolas would be congruent if we could find a sequence of rigid motions that takes one parabola onto the other. We could translate the vertex of the first one onto the vertex of the second one, then rotate the image of the first one so that the directrices are parallel and both parabolas open in the same direction. If the transformed first parabola coincides with the original second parabola, then the two original parabolas are congruent. c. Are the two parabolas from part (a) congruent? Explain how you know. These two parabolas are not congruent. They have the same vertex but different 풚-values for each 풙 in the domain, except for the point (ퟎ, ퟎ). If we translate the first one somewhere else, then the vertices will not align. If we reflect or rotate, then both parabolas will not open upward. There is no rigid transformation or set of transformations that takes the graph of one parabola onto the other. MP.3 d. Are all parabolas congruent? No, we just found two that are not congruent. e. Under what conditions might two parabolas be congruent? Explain your reasoning. Once we align the vertices and get the directrices parallel and parabolas opening in the same direction through rotation or reflection, the parabolas will have the same shape if the focus and directrix are aligned. Thus, two parabolas will be congruent if they have the same distance between the focus and directrix. Debrief this exercise using the following questions, which can also be used as scaffolds if students are struggling to begin this problem. During the debrief, record student thinking on chart paper to be used for reference at the end of this lesson as a means to confirm or refute their conjectures. At some point in this discussion, students should recognize that it would be nice if there was a name for the distance between the focus and directrix of the parabola. When appropriate, let them know that this distance is denoted by 푝 in this lesson and lessons that follow. If they do not mention it here, it is brought into the discussion in Exercise 5. Lesson 34: Are All Parabolas Congruent? 383 This work is licensed under a 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 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 34 M1 ALGEBRA II . 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 푦 = −1? Along the 푦-axis, the distance between the focus and the directrix is 2, so one point, the vertex, will be halfway between them at (0,0). Since the distance between the focus and the directrix is 푝 = 2 units, if we go 2 units to the left and right of the focus, we will be able to locate two more points on the parabola at (2,1) and (−2,1). 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 푦 = −2? The distance between the focus and the directrix is 푝 = 4 units, so the vertex is halfway between the focus and directrix at (0,0). Moving right and left 4 units from the focus gives points (4,2) and (−4,2). Generalize this process of finding three points on a parabola with a given focus and directrix. Find the vertex halfway between the focus and directrix, and let 푝 be the distance between the focus and directrix. Then, sketch a line through the focus parallel to the directrix. Locate the two points 푝 units along that line in either direction from the focus. The vertex and these two points are all on the parabola. How does this process help us determine if two parabolas are congruent? Two parabolas will be congruent if the two points found through this process are the same distance away from the focus. That is, two parabolas will be congruent if they have the same value of 푝 or the same distance between the focus and directrix. Exercises 1–5 (5 minutes) Students practice drawing parabolas given a focus and a directrix. These parabolas are NOT all oriented with a horizontal directrix. Let students struggle with how to construct a sketch of the parabola. Remind them of their work in the Opening Exercise. The teacher may choose to do one problem together to model the process. Draw a line perpendicular to the directrix through the focus. Locate the midpoint of the segment connecting the focus and directrix. Then, create a square on either side of this line with a side length equal to the distance between the focus and the directrix. One vertex of each square that is not on the directrix or axis of symmetry is another point on the parabola. If dynamic geometry software is available, students could also model this construction using technology. Exercises 1–5 1. Draw the parabola with the given focus and directrix. Lesson 34: Are All Parabolas Congruent? 384 This work is licensed under a 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 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
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