Parabola PDF - Engineering Mathematics 2

Engineering Mathematics 2 Parabola MPS Department|FEU Institute of Technology  OBJECTIVES At the end of the discussion, the learner should be able to: Define analytically and graph a parabola Determine the opening of a parabola Determine the standard...

Engineering Mathematics 2 Parabola MPS Department|FEU Institute of Technology  OBJECTIVES At the end of the discussion, the learner should be able to: Define analytically and graph a parabola Determine the opening of a parabola Determine the standard form of the equation of a parabola Identify and define parts of the graph of a parabola Find the standard equation of a parabola given condition Determine the essential parts of a parabola Find the general equation of a parabola given conditions Recall that a parabola is a type of conic section whose eccentricity is 𝑒 = 1. Thus, we can define a parabola as the locus of a point that moves in a plane so that its distance from a fixed point (Focus) is equal to its distance from a fixed line (Directrix). Let 𝑃 𝑥, 𝑦 be a point on a parabola with focus at 𝑎, 0 and directrix 𝑀. Then 𝑃𝐹 = 𝑃𝐷. Now, 𝑃𝐷 = 𝑥 + 𝑎 and 𝑃𝐹 = 𝑥−𝑎 2 + 𝑦−0 2 = 𝑥−𝑎 2 + 𝑦2. Thus, 𝑥 + 𝑎 = 𝑥 − 𝑎 2 + 𝑦2. 𝑥 + 𝑎 2 = 𝑥 − 𝑎 2 + 𝑦2. 𝑥 2 + 2𝑎𝑥 + 𝑎2 = 𝑥 2 − 2𝑎𝑥 + 𝑎2 + 𝑦 2. 𝑦 2 = 4𝑎𝑥.  The perpendicular line to the directrix passing through the focus of a conic is called its principal axis. In a parabola, axis of symmetry is equivalent to principal axis. The points that cut through the principal axis are called the vertices of a conic. The chord through the focus perpendicular to the principal axis is called the Latus Rectum. By translation of axes (reading assignment), we obtain the following standard equations with respect to their principal axis: Let 𝑎 be the directed distance from the vertex to the focus of a parabola. Then The equation of the parabola with principal axis parallel to the x-axis and vertex at 𝑉 ℎ, 𝑘 is 𝒚 − 𝒌 𝟐 = 𝟒𝒂 𝒙 − 𝒉 or 𝒚 − 𝒌 𝟐 = −𝟒𝒂 𝒙 − 𝒉. The equation of the parabola with principal axis parallel to the y-axis and vertex at 𝑉 ℎ, 𝑘 is 𝒙 − 𝒉 𝟐 = 𝟒𝒂 𝒚 − 𝒌 or 𝒙 − 𝒉 𝟐 = −𝟒𝒂 𝒚 − 𝒌. The length of the latus rectum is 4𝑎. Parabola can also be classified in terms of its opening. In this course, we only consider four types of opening of parabola. Other openings can be obtained by rotation of axes. Opening: 1.) Opens to the right 2.) Opens to the left 3.) Opens Upward 4.) Opens downward  Let 𝑎 be positive real number. Then we have the following observations: If 𝑎 = distance from vertex to focus or from vertex to directrix then 2𝑎 = distance from focus to an end of latus rectum or a directrix and 4𝑎 = length of latus rectum. Also, the midpoint of the endpoint of latus rectum is the focus. The direction of the focus in reference to the vertex is the same as the opening of the parabola. e.g., if the parabola open downward, then the its focus must be below the vertex. Moreover, the vertex can be found on the axis of symmetry and is between the focus and the directrix. Let 𝑎 be the distance from the vertex to the focus of a parabola. Then The equation of the parabola with principal axis parallel to the x-axis and vertex at 𝑉 ℎ, 𝑘 and opens to the right is 𝒚 − 𝒌 𝟐 = 𝟒𝒂 𝒙 − 𝒉. The equation of the parabola with principal axis parallel to the x-axis and vertex at 𝑉 ℎ, 𝑘 and opens to the left is 𝒚 − 𝒌 𝟐 = −𝟒𝒂 𝒙 − 𝒉. The equation of the parabola with principal axis parallel to the y-axis and vertex at 𝑉 ℎ, 𝑘 and opens upward is 𝒙 − 𝒉 𝟐 = 𝟒𝒂 𝒚 − 𝒌. The equation of the parabola with principal axis parallel to the y-axis and vertex at 𝑉 ℎ, 𝑘 and opens downward is 𝒙 − 𝒉 𝟐 = −𝟒𝒂 𝒚 − 𝒌. In summary, we have the following essential details of a parabola: Principal axis/axis of symmetry/axis of parabola Directrix Focus Vertex End points of Latus Rectum Consider the parabola 𝑥 2 − 4𝑦 = 0. Find the following: Principal axis/axis of symmetry/axis of parabola Opening Directrix Focus Vertex Length of the Latus Rectum End points of Latus Rectum Transforming 𝑥 2 − 20𝑦 = 0 into its standard form, 𝑥 2 = 20𝑦. 𝑥−0 2 =4 5 𝑦−0. From this result, we say that the parabola 𝑥 2 − 20𝑦 = 0 has vertex at the origin, it opens upward, and the focus is 5 units away from the vertex. Thus, if we move 5 units above the origin, we get 0,5 as the focus. Moreover, moving 5 units downward, we’ll have the directrix. Also, the length of the latus rectum is 20 units. Hence, we have its endpoints at 10,5 and −10,5. desmos.com Summarizing the results: Principal axis/axis of symmetry/axis of parabola: 𝑦 − 𝑎𝑥𝑖𝑠 Opening: Upward Directrix: 𝑥 = −5 Focus: 0,5 Vertex: 0,0 Length of the Latus Rectum: 20 units End points of Latus Rectum: 10,5 and 10,5 Consider the parabola𝑦 2 − 8𝑥 − 4𝑦 + 20 = 0. Find the following: Principal axis/axis of symmetry/axis of parabola Opening Directrix Focus Vertex Length of the Latus Rectum End points of Latus Rectum First, let us transform the equation into standard form 𝑦 2 − 8𝑥 − 4𝑦 + 20 = 0. 𝑦 2 − 4𝑦 = 8𝑥 − 20. 𝑦 2 − 4𝑦 + 4 = 8𝑥 − 20 + 4. 𝑦−2 2 =8 𝑥−2. 𝑦−2 2 =4 2 𝑥−2. Thus, the parabola opens to the right with vertex at 2,2 , its focus is at 4,2 , and its directrix is the line 𝑥 = 0. desmos.com From the previous result, we can now determine the end points of the latus rectum by moving 4 units from the focus both upward and downward directions (refer to figure 1). Thus, the graph of the parabola is (see figure 2) desmos.com figure 1 figure 2 We have the following results: Principal axis/axis of symmetry/axis of parabola: 𝑦 = 2 Opening: Right Directrix: 𝑥 = 0 Focus: 4,2 Vertex: 2,2 Length of the Latus Rectum: 8 units End points of Latus Rectum: 4, −2 and 4,6 In this section, we are to determine equations of parabola given some conditions. Let 𝑃 𝑥, 𝑦 be any point on a parabola. Then the general equation of a parabola is given by 𝑨𝒙𝟐 + 𝑫𝒙 + 𝑬𝒚 + 𝑭 = 𝟎 (1) whenever the parabola has principal axis parallel to y-axis. 𝐁𝒚𝟐 + 𝑫𝒙 + 𝑬𝒚 + 𝑭 = 𝟎 (2) whenever the parabola has principal axis parallel to x-axis. Find the equation of the parabola with vertex at the origin, axis at x-axis, and passing through (1, −2). Solution: Since the axis of symmetry is the x-axis and the vertex is place at the left of the point, then we conclude that this parabola opens to the right. Thus, it has standard equation of the form 𝑦 − 𝑘 2 = 4𝑎 𝑥 − ℎ.  𝑦 − 𝑘 2 = 4𝑎 𝑥 − ℎ. Substituting ℎ, 𝑘 = 0,0 , we get 𝑦 − 0 2 = 4𝑎 𝑥 − 0. 𝑦 2 = 4𝑎𝑥. Now, the parabola passes through 1, −2. This implies 1, −2 is a solution to the above equation. Hence, −2 2 = 4𝑎 1. Thus, 𝑎 = 1. Therefore, the standard equation is 𝑦 2 = 4𝑥. Determine the general equation of the Parabola defined by vertex at 3, 1 and focus at 3, −5. Solution: The focus is located below the vertex. Hence, this parabola opens downward. From this, we say that its standard equation is of the form 𝑥 − ℎ 2 = −4𝑎 𝑦 − 𝑘.  𝑥 − ℎ 2 = −4𝑎 𝑦 − 𝑘. Now, we only need to solve for 𝑎. By the definition of 𝑎, it can be obtained by computing the distance between the vertex and the focus. We have, 𝑎 = 3 − 3 2 + 1 + 5 2. (or you can simply count the distance from the graph shown previously) = 6. Substituting 𝑎 and the vertex to the standard equation 𝑥 − 3 2 = −4 6 𝑦 − 1. 𝑥 2 − 6𝑥 + 9 = −24𝑦 + 24. 𝒙𝟐 − 𝟔𝒙 + 𝟐𝟒𝒚 − 𝟏𝟓 = 𝟎. Determine the general equation of the Parabola defined by focus at −2, −3 and directrix 𝑥 = 3. Solution: Since the directrix is at the right of the vertex, then the focus should be at the left of the vertex. Hence, the parabola opens to the left. It follows that this parabola has standard equation of the form 𝑦 − 𝑘 2 = −4𝑎 𝑥 − ℎ. Find the equation of a parabola opens upward whose endpoints of the latus rectum are −4,2 and 16, 2. Solution: Since the parabola opens upward, the standard equation should be of the form 𝑥 − ℎ 2 = 4𝑎 𝑦 − 𝑘. The latus rectum has length 20. This implies that the focus, which is the midpoint of the latus rectum, is 5 units away from the vertex. It follows that the focus is at 6,2 and hence, the vertex is at 6, −3. Therefore, we have the equation 𝑥 − 6 2 = 20 𝑦 + 3. By expansion, we get 𝑥 2 − 12𝑥 + 36 = 20𝑦 + 60. Simplifying, we have 𝒙𝟐 − 𝟏𝟐𝒙 − 𝟐𝟎𝒚 − 𝟐𝟒 = 𝟎. Suppose the focus of a parabola that opens to the right is at −3,4. What is the equation of its principal axis? Answer: The parabola opens to the right implies the principal axis is parallel to the x-axis. Hence, it is a horizontal line. Moreover, this line passes through the focus −3,4. Therefore, the equation of the principal axis is 𝒚 = −𝟑. References Earl Swokowski, Jeffery A. Cole, Algebra and Trigonometry with Analytic Geometry, Classic 13th Edition, Brooks Cole, 2012. Stewart, James, Calculus: Early Transcendentals, 6th edition, Brooks Cole, 2007 Comandante, Felipe L. (2007). Analytic and Solid Geometry. Manila: National Bookstore. Ymas, Sergio E. et al. (2007). Solid Mensuration. Manila: Ymas Publishing House. A.C. Burdette, Analytic Geometry Atherton H. Sprague, Essentials of Plane Trigonometry and and Analytic Geometry http://tutorial.math.lamar.edu/ https://www.wolframalpha.com/ desmos.com