Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 113 pptx

E.2 Defining Conics Algebraically 1101 F 1 and F 2 are called the foci of the ellipse. To sketch an ellipse, you can tack one end of a string to each focus and, with the point of a pen, hold the string taut. by moving around the foci you'll trace out an ellipse. d 1 + d 2 is constant d 1 F 1 F 2 d 2 Pen Figure E.5 Ellipse The farther apart the two foci, the more elongated the ellipse. The closer they are to one another, the more circular the ellipse looks. If the two foci merge to one point, then the figure is a circle. A hyperbola A hyperbola is the set of points (x, y) the difference of whose distances from two distinct fixed points F 1 and F 2 is constant. F 1 and F 2 are called the foci of the hyperbola. |d 2 – d 1 | is constant d 1 d 1 d 2 d 2 F 1 F 2 Figure E.6 E.2 DEFINING CONICS ALGEBRAICALLY Next we relate geometric and algebraic representations of conics. First the conclusions are presented. Then a road map is given. The actual trips from geometric to algebraic respresentations areleft asproblems atthe endof thesection. Signposts are provided. To find these problems, look under the appropriate conic section; they lead the section of problems for the particular conic. 1102 APPENDIX E Conic Sections Relating the Geometric and Algebraic Representations of Conics If a circle has center (0, 0) and the distance from the center is denoted by r, then the circle is given by the equation x 2 + y 2 = r 2 . y y x x r Circle: center (0, 0) radius r Figure E.7 If a parabola has focus (0, c) and directrix y =−c,then it will have a vertex at (0, 0) and be given by the equation y = 1 4c x 2 . y x c Focus directrix Parabola: focus (0, c) directrix y = – c y = – c Figure E.8 If an ellipse has foci at (c,0)and (−c,0)and the sum of the distances from any point on the ellipse to the foci is denoted by 2a, then the ellipse will have x-intercepts of (a,0)and (−a,0)and be given by the equation x 2 a 2 + y 2 b 2 = 1, where b 2 = a 2 − c 2 . (See Figure E.9(a).) If an ellipse has foci at (0, c) and (0, −c) and the sum of the distances from any point on the ellipse to the foci is denoted by 2a, then the ellipse will have y-intercepts of (0, a) and (0, −a) and be given by the equation y 2 a 2 + x 2 b 2 = 1, where b 2 = a 2 − c 2 . (See Figure E.9(b).) E.2 Defining Conics Algebraically 1103 y y xx a a c c –c –c –a –a –b –b b b Ellipse with foci (c, 0), (–c, 0) (a) Ellipse with foci (0, c), (0, –c) (b) Figure E.9 If a hyperbola has foci at (c,0)and (−c,0)and the magnitude of the difference of the distances from any point on the hyperbola to the foci is denoted by 2a, then the hyperbola will have x-intercepts of (a,0)and (−a,0)and be given by the equation x 2 a 2 − y 2 b 2 = 1, where b 2 = c 2 − a 2 . (See Figure E.10(a).) If a hyperbola has foci at (0, c) and (0, −c) and the magnitude of the difference of the distances from any point on the hyperbola to the foci is denoted by 2a, then the hyperbola will have y-intercepts of (0, a) and (0, −a) and be given by the equation y 2 a 2 − x 2 b 2 = 1, where b 2 = c 2 − a 2 . (See Figure E.10(b).) y y xx a a b b c c –a –a –b –b –c –c y = x –b a y = x – a b y = x b a y = x a b Hyperbola with foci (c, 0) (– c, 0) (a) Hyperbola with foci (0, c) (0, –c) (b) Figure E.10 1104 APPENDIX E Conic Sections Roadmaps Circle: Let (x, y) be a point on the circle with center (0, 0). Use the distance formula (or the Pythagorean Theorem) to conclude that if the distance from (x, y) to (0, 0) is r, then x 2 + y 2 = r 2 . Parabola: Let (x, y) be a point on the parabola with focus F at (0, c) and directrix L at y =−c.Using the distance formula, we know that the distance from (x, y) to F is  x 2 + (y − c) 2 and the distance from (x, y) to L is  (y + c) 2 . Equate the distances, square both sides, and simplify. Ellipse: Let (x, y) be a point on the ellipse with foci at (c,0)and (−c,0).Denote the sum of the distances from any point on the ellipse to the foci by 2a. (This distance is some positive number, and denoting it by 2a is convenient.) Using the distance formula we have that  (x + c) 2 + y 2 +  (x − c) 2 + y 2 = 2a. Isolate one radical and square both sides to eliminate it. Then isolate the other radical and square both sides to eliminate the second radical. Simplifying and denoting the quantity a 2 − c 2 by b 2 gives the desired result. Hyperbola: Let (x, y)be a point on the hyperbola with foci at (0, c) and (0, −c). Denote the magnitude (absolute value) of the differences of the distances from any point on the hyperbola to the foci by 2a. (Denoting this difference by 2a is convenient.) Using the distance formula we obtain  (x + c) 2 + y 2 −  (x − c) 2 + y 2 =±2a. Isolate one radical and square both sides to eliminate it. Then isolate the other radical and square both sides to eliminate the second radical. Simplifying and denoting the quantity c 2 − a 2 by b 2 gives the desired result. Conics Given as Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 Consider the equation Ax 2 + Cy 2 = P , where A, C, and P are constants. This is i. the equation of a circle if A = C, and the signs of A, C, and P agree; ii. the equation of an ellipse if A, C, and P have the same signs; iii. the equation of a hyperbola if A and C have opposite signs (to find out which way the hyperbola opens, look for the x- and y-intercepts—you’ll find only one pair); iv. the equation of a parabola if either A = 0orC=0. Consider the graph of 4(x − 2) 2 + 9(y − 3) 2 = 36. This is the ellipse 4x 2 + 9y 2 = 36 shifted 2 units right and 3 units up. This example can be generalized. The graph of A(x − h) 2 + C(y − k) 2 = P is the graph of the conic section Ax 2 + Cy 2 = P shifted right h units and up k units. Multiplying out gives an equation of the form Ax 2 + Cy 2 + Dx + Ey + F = 0, E.2 Defining Conics Algebraically 1105 where A, C, D, E, and F are constants. Conversely, an equation of the form Ax 2 + Cy 2 + Dx + Ey + F = 0 can, by com- pleting the square twice, be put in the form A(x − h) 2 + C(y −k) 2 =P . By looking at the signs of A and C as described above, one can determine the nature of the conic section. ◆ EXAMPLE E.1 Put the conic section 2x 2 − y 2 + 4x + 6y − 8 =0 into the form A(x − h) 2 + C(y −k) 2 =P and determine what it looks like. SOLUTION 2x 2 − y 2 + 4x + 6y − 8 =0 2x 2 + 4x − y 2 + 6y = 8 2(x 2 + 2x) −(y 2 − 6y) =8 2  (x + 1) 2 − 1  −  (y − 3) 2 − 9  = 8 2(x + 1) 2 − 2 − (y − 3) 2 + 9 = 8 2(x + 1) 2 − (y − 3) 2 = 1 Therefore, this conic is the hyperbola 2x 2 − y 2 = 1 (a hyperbola with x-intercepts at ± 1 √ 2 ) shifted left 1 unit and up 3 units. ◆ In addition to shifting the conic Ax 2 + Cy 2 = P horizontally and vertically, we can rotate the conic. The resulting algebraic equation is of the form Ax 2 + Bxy +Cy 2 +Dx +Ey + F = 0, where A, B, C, D, E, and F are constants. It can be shown 1 that this equation gives a parabola when B 2 − 4AC = 0, an ellipse when B 2 − 4AC < 0 (a circle if, in addition, A = C), a hyperbola when B 2 − 4AC > 0. Notice that the sign of the expression B 2 − 4AC discriminates between the three types of conics. This is not coincidental In addition to having fascinating geometric characteristics, conic sections have many practical applications. 1 This can be shown most elegantly using linear algebra; therefore the more accessible but less appealing argument is not given here. Both arguments involve introducing a new, rotated coordinate system. 1106 APPENDIX E Conic Sections E.3 THE PRACTICAL IMPORTANCE OF CONIC SECTIONS Reflection Properties The reflective properties of parabolas, hyperbolas, and ellipses are most useful when considering surfaces formed by rotating the two-dimensional figures around the line through the two foci in the latter two cases and through the focus and perpendicular to the directrix in the former. Reflecting properties of parabolas: A light source placed at the focus will be reflected parallel to the axis of symmetry of the parabola. Conversely, sound or light waves coming in parallel to the axis of symmetry of the parabola are reflected through the focus, thereby concentrating them there. This property makes parabolas a useful shape in the design of headlights and search- lights, reflecting mirrors in telescopes, satellite dishes, radio antennas, and micro- phones designed to pick up a conversation far away. F Reflective property of a parabola Parabolic headlight with the bulb at the focus Figure E.11 Reflecting property of ellipses: Sound or light emanating from one focus and reflecting off the ellipse will pass through the other focus. F 1 F 2 Reflection property of an ellipse Figure E.12 There are what are known as “whispering galleries” under the elliptic dome in the Capitol building of the United States as well as in the Mormon Tabernacle in Salt Lake City, Utah. A person standing at one focus and whispering to his neighbor can be heard quite clearly by an individual located at the other focus. As you can imagine, this can be a potential problem for those unaware of the reflection properties of the ellipse. In 1980 an ingenious medical treatment for kidney stones, called lithotripsy (from the Greek word for “stone breaking”), was introduced. The treatment is based on the reflective properties of the ellipse. Intense sound waves generated outside the body are focused on the kidney stone, bombarding and thereby destroying it without invasive E.3 The Practical Importance of Conic Sections 1107 surgery. Over 80% of patients with kidney stones can be treated with lithotripsy; the recovery time is substantially less and the mortality rate much lower than with the traditional surgery . 2 Reflecting property of hyperbolas: Light traveling along a line through one focus of a hyperbola will be reflected off the surface of the hyperbola along the line through the point of reflection and the other focus. F 1 F 2 Reflection property of a hyperbola Figure E.13 The reflective property of hyperbolas is used in the construction of camera and telescope lenses as well as radio navigation systems, such as the long-range navigation system (LORAN). Conic Sections and Astronomy The path of any projectile under the force of gravity is a parabolic arc. Around 1600 Kepler found that planets have elliptical orbits around the sun. This discovery involved painstaking calculations using measurements collected by Tycho Brahe in the late 1500s over a period of more than two decades. Kepler’s conclusions (and the accuracy of Tycho Brahe’s measurements) are rather astounding, given that the foci of most of the planets are very close together, making their paths very nearly circular. 3 Newton later put Kepler’s observations on solid theoretical grounds. Comets can travel in elliptical orbits (like Halley’s comet) or in parabolic orbits. The moon has an elliptical orbit with the earth at one focus. PROBLEMS FOR APPENDIX E Parabolas 1. Begin with the geometric characterization of a parabola. Suppose that the focus F is at (0, c) and the directrix L is at y =−c.You will show that the set of points equidistant from F and L is the parabola y = 1 4c x 2 . (a) Use the fact that the axis of symmetry is perpendicular to the directrix and passes through the focus to deduce that the vertex of the parabola is at (0, 0). 2 The story of the development of this treatment is fascinating. It came out of scientists looking at aircraft. You can get more in- formation on the scientific aspects at http://www.hvlitho.com/lithotripsy and on the mathematics of it at http://www.math.iupui.edu: edu:80/m261vis/litho. 3 For more information on Kepler, consult a history of mathematics book, such as A History of Mathematics An Introduction by Victor Katz, Addison Wesley 1998, pp. 409–416. 1108 APPENDIX E Conic Sections 5 2 y x (2.5, 2) (b) Let (x, y) be a point on the parabola described above. Equating the distance between the point (x, y) and F with the distance between the point (x, y) and L gives the equation  x 2 + (y − c) 2 =  (y + c) 2 . Explain. (c) Square both sides of the equation in part (b) to solve for y. 2. In Problem 1 you showed that a parabola with focus F at (0, c) and the directrix L at y =−ccan be written in the form y = 1 4c x 2 . (a) Find the equation of the parabola with vertex (0, 0) and focus (0, 1). Sketch its graph. In the sketch show the location of the focus and directrix. (b) Find the equation of the parabola with vertex (0, 0) and directrix y = 2. Sketch its graph. In the sketch show the location of the focus and directrix. 3. In Problem 1 you showed that a parabola with focus F at (0, c) and the directrix L at y =−ccan be written in the form y = 1 4c x 2 . Find the focus and directrix of the parabola y = 2x 2 . 4. If the focus and directrix of a parabola are moved farther apart, does the parabola get narrower or does it get wider? Explain. 5. Find the equation of a parabola with its vertex at the origin and its focus at (0, 6). 6. Find the equation of a parabola with vertex at (0, 0) and directrix y = 5. 7. A headlight is made of reflective material in the shape of a parabolic dish. In order to take advantage of the fact that light emanating from the focus will be reflected by the parabolic bowl in the direction of the axis of symmetry, where should the light source be placed if the dish is 5 inches in diameter and 2 inches in depth? (Hint: Look at a cross section of the reflector and introduce a coordinate system. Let the y-axis lie along the axis of symmetry of the reflector and let the origin lie at the vertex of the reflector.) 8. The light filament in the bulb of a floodlight is 1.5 inches from the vertex of the paraboloid reflector. (The cross sections of the reflector taken through the vertex are all identical parabolas.) (a) Find the equation of the parabolic cross section. (b) If the reflector dish is 10 inches in diameter, how deep is it? E.3 The Practical Importance of Conic Sections 1109 Ellipses 9. Let (x, y) be a point on the ellipse with foci at (0, c) and (0, −c). You will arrive at the formula x 2 a 2 + y 2 b 2 = 1, where a 2 − C 2 = b 2 . (a) Denote the sum of the distances from any point (x, y) on the ellipse to the foci by 2a. (This distance is some positive number; denoting it by 2a is convenient.) Using the distance formula we know that  (x + c) 2 + y 2 +  (x − c) 2 + y 2 = 2a. Isolate  (x + c) 2 + y 2 and square both sides of the equation to eliminate it. After multiplying out you should be able to simplify the result to a  (x + c) 2 + y 2 = a 2 + cx. (b) Square both sides of the equation a  (x + c) 2 + y 2 = a 2 + cx to eliminate the radical. Show that the result can be expressed in the form (a 2 − c 2 )x 2 + a 2 y 2 = a 2 (a 2 − c 2 ). (E.1) (c) The distance between the two foci is 2c. This must be less than the sum of the distance between (x, y) and the foci. Thus 2a>2c,ora>c. Consequently, (a 2 − c 2 )>0 and we can divide both sides of Equation E.1 by a 2 (a 2 − c 2 ) to get x 2 a 2 + y 2 a 2 − c 2 = 1. Denote a 2 − c 2 by b 2 to get the desired result. 10. (a) In Problem 9 you showed that the ellipse with foci at (0, c) and (0, −c) is given algebraically by x 2 a 2 + y 2 b 2 = 1. Explain from an algebraic point of view how you know that b<a. (b) What is the geometric significance of the constants a and b? Hyperbolas 11. Let (x, y) be a point on the hyperbola with foci at (0, c) and (0, −c). Denote the magnitude (absolute value) of the differences of the distances from any point on the hyperbola to the foci by 2a. (a) Using the distance formula show that  (x + c) 2 + y 2 −  (x − c) 2 + y 2 =±2a. 1110 APPENDIX E Conic Sections (b) Carry on as in Problem 9. Isolate one radical and square both sides to eliminate it. Then isolate the other radical and square both sides to eliminate the second radical. Show that you obtain (c 2 − a 2 )x 2 − a 2 y 2 = a 2 (c 2 − a 2 ). (c) Argue that c 2 − a 2 > 0. Once this is done, we can set b 2 = c 2 − a 2 . Simplify to obtain x 2 a 2 − y 2 b 2 = 1. . (c,0 )and (−c,0 )and the magnitude of the difference of the distances from any point on the hyperbola to the foci is denoted by 2a, then the hyperbola will have x-intercepts of (a,0 )and (−a,0 )and. of a circle if A = C, and the signs of A, C, and P agree; ii. the equation of an ellipse if A, C, and P have the same signs; iii. the equation of a hyperbola if A and C have opposite signs (to. has foci at (c,0 )and (−c,0 )and the sum of the distances from any point on the ellipse to the foci is denoted by 2a, then the ellipse will have x-intercepts of (a,0 )and (−a,0 )and be given by the