MA1310: Module 5 Vectors and Conic Sections Exercise 5.2 Graphing Ellipse, Hyperbola, and Parabola Answer the following questions to complete this exercise: 1. Describe an ellipse in two sentences. 2. Find the standard form of the equation of the ellipse and give the location of its foci. The standard form of the equation of an ellipse with the center at the origin and major and minor axes of lengths 2a and 2b (where a and b are positive, and a2 > b2) is: x2 y2 1 a2 b2 or x2 y2 1 b2 a2 The location of foci are at (-c, 0) and (c, 0) where c2 = a2 - b2. 3. Graph the ellipse. a. 1 (x 2)2 (y 1)2 1 and choose the correct graph from the given graphs: 9 4 MA1310: Module 5 Vectors and Conic Sections Exercise 5.2 Graphing Ellipse, Hyperbola, and Parabola b. c. d. 2 MA1310: Module 5 Vectors and Conic Sections Exercise 5.2 Graphing Ellipse, Hyperbola, and Parabola [Hint: To graph this ellipse, find the center (h, k) by comparing the given equation with the standard form of equation centered at (h, k). Next, find a and b. Find the vertices (h - a, k) and (h + a, k). Find the foci (h + c, k) and (h - c, k).] 4. A semi-elliptic archway has a height of 20 feet and a width of 50 feet, as shown in the figure below. Can a truck 14 feet high and 10 feet wide drive under the archway without going into the other lane? 20 ft 50 ft 5. Find the standard form of the equation of the hyperbola whose graph is given below. 6. Find the vertices of the hyperbola. x2 y2 1 25 49 7. What is a parabola? Describe in two sentences. 3 MA1310: Module 5 Vectors and Conic Sections Exercise 5.2 Graphing Ellipse, Hyperbola, and Parabola 8. The giant Arecibo telescope in Puerto Rico has a parabolic dish with a diameter of 300 meters and a depth of 50 meters. (Source: National Astronomy and Ionosphere Center, Arecibo Observatory.) a. Find an equation in the form that describes a cross-section of this dish. b. If the receiver is located at the focus, how far should it be from the vertex? y (-150, 50) (150, 50) x Submission Requirements: Submit your response in a Microsoft Word document of the following specifications: Font: Arial; Point 12 Spacing: Double Evaluation Criteria: Did you show the steps to solve each problem? Did you write thorough explanations for the short-answer questions? Did you accurately choose a problem that fits the criteria for each rule? Were the answers submitted in an organized fashion that was legible and easy to follow? Were the answers correct? 4 (1) Ellipse:- It locus of a point in a plane which moves in the plane in such a way that the ratios of its distance from a fixed point( focus) in same plane to its distance from a fixed line( directrix )is always constant, which is always less than 1. ....................................................................................... (2) The equation of the ellipse is 2 2 or x y 2 1 2 4 9 x2 y 2 1 16 81 Here a=4, b=9 e 1 a2 16 65 1 2 b 81 9 The location of foci are (0, 65), (0, 65) (because foci are (0,be), (0,-be)) .............................................................................................. (3) The graph is option b. ............................................................................................. (4) On completing full ellipse we get, following figure Here, a=25 and b=20 Also center of ellipse is (0,0). Therefore, its equation is x2 y2 2 1 252 20 or x2 y2 1 625 400 (i) Put y=14 in (i), we get x 2 142 1 625 400 x2 196 204 1 x 2 625( ) 625 400 400 x 17.85 ft This means that at the height clearance of 14 feet there is more than 17 feet of road width allowed for 10 feet wide truck.(there is 25 feet width available) Therefore, archway can easily allow the truck to go through. .................................................................................... (5) Here a=4, b=2 Therefore equation is x2 y2 1 a 2 b2 Or x2 y 2 1 16 4 ................................................................................................................. (6) On comparing 2 2 x y 1 25 49 with 2 2 , we get x y 2 1 2 a b a=5, b=7 Therefore, co-ordinates of vertices are (5, 0), (-5, 0). [(a,0),(-a,0)] .................................................... (7) Parabola:- it is locus of point which moves in a plane such that its distance from fixed point in plane is always equal to its distance from a fixed line in the same plane. ........................................................... (8) Since parabola has vertex (0, 0) and it is upward parabola. Its equation is ... (i) x 2 4ay Since parabola passes through (150, 50) Therefore from (i) 1502 4a(50) 4a 1502 a 112.5 50 Therefore, the equation of parabola becomes x 2 450 y y x2 450 .......................................... (b) Now, focus is given by (0,a) that is (0,112.5) and vertex is (0,0). Therefore, distance= (0 0)2 (0 112.5)2 =112.5 meters