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548 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 7-3 Hyperbola 10 DEFINITION Definition of a Hyperbola Drawing a Hyperbola Standard Equations and Their Graphs Applications As before, we start with a coordinate-free definition of a hyperbola. Using this definition, we show how a hyperbola can be drawn and we derive standard equa- tions for hyperbolas specially located in a rectangular coordinate system. Definition of a Hyperbola The following is a coordinate-free definition of a hyperbola: HYPERBOLA A hyperbola is the set of all points P in a plane such that the absolute value of the difference of the distances of P to two fixed points in the plane is a positive con- stant. Each of the fixed points, F' and F, is called a focus. The intersection points V' and V of the line through the foci and the two branches of the hyperbola are called vertices, and each is called a ver- tex. The line segment V' V is called the transverse axis. The midpoint of the transverse axis is the center of the hyperbola. Draunng a Hyperbola d21 Constant p dl Thumbtacks, a straightedge, string, and a pencil are all that are needed to draw a hyperbola (see Fig. 1). Place two thumbtacks in a piece of cardboard—these form the foci of the hyperbola. Rest one corner of the straightedge at the focus F' so that it is free to rotate about this point. Cut a piece of string shorter than the length of the straightedge, and fasten one end to the straightedge corner A and the other end to the thumbtack at F. Now push the string with a pencil up against the straightedge at B. Keeping the string taut, rotate the straightedge about F', keep- ing the corner at F'. The resulting curve will be part of a hyperbola. Other parts of the hyperbola can be drawn by changing the position of the straightedge and string. To see that the resulting curve meets the conditions of the definition, note that the difference of the distances BF' and BF is BF' - BF - BF' + - BF - BA - AE' - (BF + BA) Straightedge length — Constant String length
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7-3 Hyperbola 549 FIGURE I Drawing a hyperbola. FIGURE a Hyperbola with foci on the x axis. B String d2 y) x d21 — — Positive constant Standard Equations and The•r Graphs Using the definition of a hyperbola and the distance-between-two-points formula, we can derive the standard equations for a hyperbola located in a rectangular coor- dinate system. We start by placing a hyperbola in the coordinate system with the foci on the x axis equidistant from the origin at F'(—c, 0) and F(c, 0), c > 0, as in Figure 2. Just as for the ellipse, it is convenient to represent the constant difference by 2a, a > 0. Also, the geometric fact that the difference of two sides of a triangle is always less than the third side can be applied to Figure 2 to derive the fol- lowing useful result: Idl - d21 < 2c 2a < 2c (1) We will use this result in the derivation of the equation of a hyperbola, which we now begin. Referring to Figure 2, the point P(x, y) is on the hyperbola if and only if Idl — 6/21 — 2a Id(P, F') - d(P, - (x + + — (x — + y21 — 2a After eliminating radicals and absolute value signs by appropriate use of squar- ing and simplifying, another good exercise for you, we have (c2 — a2)x2 2 a — a2(c2 — 612) -1 2 2 (2) (3) Dividing both sides of equation (2) by a2(c2 — a2) is permitted, since neither a 2 2 — a2 is 0. From equation (1), a < c; thus, a2 < c2 and c2 - a2 > O. The nor c constant a was chosen positive at the beginning.
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2 FIGURE 3 Hyperbola with foci on the y axis. y) x -c) 550 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY To simplify equation (3) further, we let b2 _ -1 b2 _ 2 a to obtain 2 x 12 = 1 2 a (4) (5) From equation (5) we see that the x intercepts, which are also the vertices, are ±a and there are no y intercepts. To see why there are no y intercepts, let x x — 0 and solve for y: 02 2 a b2 ± —b2 An imaginary number If we start with the foci on the y axis at F' (0, —c) and F(O, c) as in Figure 3, instead of on the x axis as in Figure 2, then, following arguments similar to those used for the first derivation, we obtain 2 b2 a where the relationship among a, b, and c remains the same as before: 2 2 a (6) (7) The center is still at the origin, but the transverse axis is now on the y axis. As an aid to graphing equation (5), we solve the equation for y in terms of x, Idl d21 = Positive constant another good exercise for you, to obtain 2 b a (8) a 2 x As x changes so that IXI becomes larger, the expression 1 — (a2/x2) within the rad- ical approaches 1. Hence, for large values of Ixl, equation (5) behaves very much like the lines b x a (9) These lines are asymptotes for the graph of equation (5). The hyperbola approaches these lines as a point P(x, y) on the hyperbola moves away from the origin (see Fig. 4). An easy way to draw the asymptotes is to first draw the rec- tangle as in Figure 4, then extend the diagonals. We refer to this rectangle as the asymptote rectangle.
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551 FIGURE q Asymptotes. THEOREM - 2b Asymptote b a b —b 7-3 Hyperbola Asymptote b a a x Starting with equation (6) and proceeding as we did for equation (5), we obtain the asymptotes for the graph of equation (6): a x (10) The perpendicular bisector of the transverse axis, extending from one side of the asymptote rectangle to the other, is called the conjugate axis of the hyperbola. Given an equation of the form of equations (5) or (6), how can we find the coor- dinates of the foci without memorizing or looking up the relation b2 2 — 612'? Just as with the ellipse, there is a simple geometric relationship in a hyperbola that enables us to get the same result using the Pythagorean theorem. To see this relationship, we rewrite b2 2 — 612 in the form 2 — 612 + b2 (11) Note in the figures in Theorem I below that the distance from the center to a focus is the same as the distance from the center to a corner of the asymptote rectangle. Stated in another way: A circle, with center at the origin, that passes through all four corners of the asymptote rectangle also passes through all foci of hyperbolas with asymptotes determined by the diagonals of the rectangle. We summarize all the preceding results in Theorem I for convenient reference. STANDARD EOURTIONS OF R HYPERBOLA WITH CENTER RT [O, O] 1. -1 2 b2 a x intercepts: ±a (vertices) y intercepts: none b Foci: F'(-c, 0), F(c, 0) x c 2 — 612 + b2 Transverse axis length Conjugate axis length — 2a
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552 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Foci: F'(0, -c), F(0, c) - 10 THEOREM continued Explore/ Olscuss E HRM PLE Solutions FIGURE S 9x2 16y2 — 144. 5 Y2 x2 2. —1 2 b2 a x intercepts: y intercepts: none ±a (vertices) x 2 — 612 + b2 Transverse axis length = 2a Conjugate axis length = 2b [Note: Both graphs are symmetric with respect to the x axis, y axis, and origin.] The line through a focus F of a hyperbola that is perpendicular to the transverse axis intersects the hyperbola in two points G and H. For each of the two standard equations of a hyperbola with center (0, 0), find an expression in terms of a and b for the distance from G to H. Graphing Hyperbolas Sketch the graph of each equation, find the coordinates of the foci, and find the lengths of the transverse and conjugate axes. Check by graphing on a graphing utility. (A) 9x2 — 16y2 (B) 16y2 - 9x2 - 144 (C) 2x2 - — 144 (A) First, write the equation in standard form by dividing both sides by 144: 9X2 — 16y2 -1 16 9 2 — 16 and b2 —c c x 6 —5 Locate x intercepts, x ±4; there are no y intercepts. Sketch the asymp- totes using the asymptote rectangle, then sketch in the hyperbola (Fig. 5). 2 — 612 + b2 Foci: c — 16 +9 - 25 Thus, the foci are F' (—5, 0) and F(5, 0). Transverse axis length — 2(4) -8 Conjugate axis length — 2(3) 6
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FIGURE 6 (9x2 144)/16; (9x2 — 144)/16. FIGURE ? 16y2 — 144. FIGURE 8 (9X2 + 144)/ 16; (9x2 + 144)/16. FIGURE S = 10. -1 553 7-3 Hyperbola To check the graph on a graphing utility, we solve the original equation for y: 9x2 — 16y2 This produces two functions whose graphs are shown in Figure 6. 16y2 — 9x2 — 144 (B) Y2 x2 — 9 and b2 16 (C) - 16 ± 3; there are no x intercepts. Sketch the asymp- Locate y intercepts, y totes using the asymptote rectangle, then sketch in the hyperbola (Fig. 7). It is important to note that the transverse axis and the foci are on the y axis. Foci: c —9 + 16 - 25 Thus, the foci are F'(O, —5) and F(O, 5). Transverse axis length = 2(3) = 6 Conjugate axis length = 2(4) A check of the graph is shown in Figure 8. - 10 - 10 — 5 and b2 10 F 5 l/ Locate x intercepts, x — ± VS; there are no y intercepts. Sketch the asymp- totes using the asymptote rectangle, then sketch in the hyperbola (Fig. 9). — 612 + b2 Foci: c _ 5 + 10 - 15 15
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554 1— 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY FIGURE 10 - 10. 6 MATCHED PROBLEM CRU TION E HRM PLE 2 Solutions 9 Thus, the foci are 0) and F(V'G, 0). Transverse axis length — 2 VS 4.47 Conjugate axis length — 2 Vfiö 6.32 A check of the graph is shown in Figure 10. Sketch the graph of each equation, find the coordinates of the foci, and find the lengths of the transverse and conjugate axes. Check by graphing on a graphing utility. (A) 16x2 - 25y2 (B) 25y2 — 400 — 16x2 — 400 — 3x2 - 12 Hyperbolas of the form Y2 x2 and -1 are called conjugate hyperbolas. In Example 1 and Matched Problem 1, the hyperbolas in parts A and B are conjugate hyperbolas—they share the same asymptotes. When making a quick sketch of a hyperbola, it is a common error to have the hyperbola opening up and down when it should open left and right, or vice versa. The mistake can be avoided if you first locate the intercepts accurately. Finding the Equation of a Hyperbola Find an equation of a hyperbola in the form 2 x -1 if the center is at the origin, and: (A) Length of transverse axis is 12 Length of conjugate axis is 20 (A) Start with (B) Length of transverse axis is 6 Distance of foci from center is 5 2 a 2 x -1 b2
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7-3 Hyperbola 555 and find a and b: - 10 FIGURE Il Asymptote rectangle. a 12 2 and 20 b 2 Thus, the equation is Y2 x2 -1 36 100 (B) Start with 2 a -1 b2 and find a and b: 6 a 2 To find b, sketch the asymptote rectangle (Fig. Il), label known parts, and use the Pythagorean theorem: 5 —bl —5 5 3 x 1b b2 Thus, 9 — 52 — 32 - 16 the equation is -1 16 MATCHED PROBLEM 2 Explore/ DI scuss Find an equation of a hyperbola in the form x2 Y2 -1 if the center is at the origin, and: (A) Length of transverse axis is 50 Length of conjugate axis is 30 (B) Length of conjugate axis is 12 Distance of foci from center is 9 (A) Does the line with equation y = x intersect the hyperbola with equa- 2 — (Y2/4) tion x — 1? If so, find the coordinates of all intersection points.
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556 1— 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Explore/ Olscuss continued FIGURE 12 Uses of hyperbolic forms. E HRM PLE 3 FIGURE 13 dl - (12 = 50. (B) (C) Does the line with equation y — 3x intersect the hyperbola with 2 — 1? If so, find the coordinates of all intersec- equation x tion points. For which values of m does the line with equation y = mx intersect the hyperbola — 2 a points. Rpphcatlons — l? Find the coordinates of all intersection b2 You may not be aware of the many important uses of hyperbolic forms. They are encountered in the study of comets; the loran system of navigation for pleasure boats, ships, and aircraft; sundials; capillary action; nuclear cooling towers; opti- cal and radiotelescopes; and contemporary architectural structures. The TWA building at Kennedy Airport is a hyperbolic paraboloid, and the St. Louis Sci- ence Center Planetarium is a hyperboloid. With such structures, thin concrete shells can span large spaces [see Fig. 12(a)]. Some comets from outer space occa- sionally enter the sun's gravitational field, follow a hyperbolic path around the sun (with the sun as a focus), and then leave, never to be seen again [Fig. 12(b)]. The next example illustrates the use of hyperbolas in navigation. Comet Sun Comet around sun (b) St. Louis Planetarium (a) Nau•gatlon A ship is traveling on a course parallel to and 60 miles from a straight shore- line. Two transmitting stations, Sl and S2, are located 200 miles apart on the shoreline (see Fig. 13). By timing radio signals from the stations, the ship's navigator determines that the ship is between the two stations and 50 miles closer to S2 than to Sl. Find the distance from the ship to each station. Round answers to one decimal place. 60 miles 200 miles
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Solution FIGURE 200 -1 557 7-3 Hyperbola If dl and 612 are the distances from the ship to Sl and S2, respectively, then — 50 and the ship must be on the hyperbola with foci at Sl and RS2 and dl — 612 fixed difference 50, as illustrated in Figure 14. In the derivation of the equation of a hyperbola, we represented the fixed difference as 2a. Thus, for the hyperbola in Figure 14 we have: - 100 - 25 -100 —1 x2 - 625 c12 — (x, 60) x 100 MATCHED PROBLEM 3 a b 9,375 The equation for this hyperbola is -1 625 9,375 Substitute y — 60 and solve for x (see Fig. 14): 2 x 625 Thus, x - 602 9,375 625 865 3,600 9,375 3,600 + 9,375 9,375 - 865 29.41. (The negative square root is discarded, since the ship is closer to S2 than to LSI.) Distance from ship to Sl (29.41 + 100)2 + 602 20,346.9841 Distance from ship to S2 — + 602 (29.41 8,582.9841 92.6 miles Notice that the difference between these two distances is 50, as it should be. Repeat Example 3 if the ship is 80 miles closer to S2 than to LSi. Example 3 illustrates a simplified form of the loran (Long RAnge Naviga- tion) system. In practice, three transmitting stations are used to send out signals simultaneously (Fig. 15), instead of the two used in Example 3. A computer onboard a ship will record these signals and use them to determine the differences of the distances that the ship is to Sl and S2, and to S2 and S3.
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558 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY FIGURE Loran navigation. -10 —c Ship q2 ql Plotting all points so that these distances remain constant produces two branches, PI and 192, of a hyperbola with foci Sl and S2, and two branches, and 612, of a hyperbola with foci RS2 and S3. It is easy to tell which branches the ship is on by comparing the signals from each station. The intersection of a branch of each hyperbola locates the ship and the computer expresses this in terms of longitude and latitude. Problems Rnsuers to (B) -10 Matched 10 c -10 10 C -10 10 10 25 16 Foci: 0), 0) Transverse axis length = 10 Conjugate axis length = 8 16 25 Foci: F'(O, -uh), F(0, »@1) Transverse axis length = 8 Conjugate axis length = 10 12 4 Foci: F'(O, F(O, 4) Transverse axis length = 2 VT) Conjugate axis length = 4 6.93
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559 - 159.5 miles, 7-3 Hyperbola d2 = 79.5 miles 225 EXERCISE 7-3 625 x2=1 45 36 coordinates of the foci, and find the lengths of the transverse and conjugate axes. Check by graphing on a graphing utility. 13. 3x2 2y2 15. 7y2 - 4x2 form = 12 = 28 14. 3x2 - 4y2 24 16. 3y2 - 2x2 = 24 In Problems 1—4, match each equation with one of graphs In Problems 17—28, find an equation of a hyperbola in the N if the center is at the origin, and M (a) (c) (b) (d) 17. The graph is 10 -10 -10 18. The graph is -10 Sketch a graph of each equation in Problems 5—12, find the coordinates of the foci, and find the lengths of the transverse and conjugate axes. Check by graphing on a graphing utility. 10 -10 19. The graph is 9. 4x2 11. 9y2 = 16 10. 25 4y2 25 x2 — 25x2 = 100 -10 10 -10 10 10 10 Sketch a graph of each equation in Problems 13—16, find the
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21. Transverse axis on x axis Transverse axis length = 14 Conjugate axis length = 10 22. Transverse axis on x axis Transverse axis length = 8 Conjugate axis length = 6 23. Transverse axis on y axis Transverse axis length = 24 Conjugate axis length = 18 24. Transverse axis on y axis Transverse axis length = 16 Conjugate axis length = 22 25. Transverse axis on x axis Transverse axis length = 18 Distance of foci from center 26. Transverse axis on x axis Transverse axis length = 16 Distance of foci from center 27. Conjugate axis on x axis Conjugate axis length = 14 Distance of foci from center 28. Conjugate axis on x axis Conjugate axis length = 10 Distance of foci from center 29. (A) How many hyperbolas have center at (0, 0) and a focus at (1, 0)? Find their equations. (B) How many ellipses have center at (0, 0) and a focus at (1, 0)? Find their equations. (C) How many parabolas have center at (0, 0) and focus at (1, 0)? Find their equations. 30. How many hyperbolas have the lines y = ±2x as asymp- totes? Find their equations. _LL In Problems 31—38, find the coordinates of all points of inter- section. Round any approximate values to three decimal places. 560 1— 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY 20. The graph is 10 x = 11 = 10 25 y 2y Y2 + 2x2 36. 38. 36 16 -10 10 -10 31. 3y2 - 33. 2x2 + 2 2 = 12 24 -12 32. — x 2 34. 2x2 + Y2 = 17 —5 Eccentricity. Problems 39 and 40 below and Problems 39 and 40 in Exercise 7-2 are related to a property of conics called eccentricity, which is denoted by a positive real number E. Parabolas, ellipses, and hyperbolas all can be defined in terms of E, afixed point called a focus, and a fixed line not containing the focus called a directrix as follows: The set of points in a plane each of whose distance from a fixed point is E times its distance from a fixed line is an ellipse if 0 < E < 1, a parabola if E = 1, and a hyperbola if E > 1. 39. Find an equation of the set of points in a plane each of whose distance from (3, 0) is three-halves its distance 4 from the line x — — S. Identify the geometric figure. 40. Find an equation of the set of points in a plane each of whose distance from (0, 4) is four-thirds its distance from the line y = Identify the geometric figure. In Problems 41 —44, find the coordinates of all points of intersection to two decimal places. 41. 2x2- 3y2 = 20, 7x + 15y = 10 42. — 3x2 = 8, x2 — 3 43. 24y2 - 18x2 - 175, 90x2 + 3y2 = 200 44. 8x2 7y2 = 58, - 11x2 = 45 APPLICATIONS 45. Architecture. An architect is interested in designing a thin-shelled dome in the shape of a hyperbolic paraboloid, as shown in Figure (a). Find the equation of the hyperbola located in a coordinate system [Fig. (b)] satisfying the in- dicated conditions. How far is the hyperbola above the vertex 6 feet to the right of the vertex? Compute the an- swer to two decimal places. Hyperbola 1 Parabola Hyperbolic paraboloid (a)
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10 -10 (8, 12) x 10 -1 561 7-3 Hyperbola 47. Space Science. In tracking space probes to the outer plan- ets, NASA uses large parabolic reflectors with diameters equal to two-thirds the length of a football field. Needless to say, many design problems are created by the weight of these reflectors. One weight problem is solved by using a hyperbolic reflector sharing the parabola's focus to reflect the incoming electromagnetic waves to the other focus of the hyperbola where receiving equipment is installed (see Hyperbola part of dome (b) 46. Nuclear Power. A nuclear cooling tower is a hyper- boloid, that is, a hyperbola rotated around its conjugate axis, as shown in Figure (a). The equation of the hyper- bola in Figure (b) used to generate the hyperboloid is 1002 1502 the figure). Common focus Hyperbola focus Receiving cone (a) Radiotelescope (b) Incoming wave Hyperbola Parabola Nuclear cooling tower (a) 500 -500 x 500 -500 Hyperbola part of dome (b) If the tower is 500 feet tall, the top is 150 feet above the center of the hyperbola, and the base is 350 feet below the center, what is the radius of the top and the base? What is the radius of the smallest circular cross section in the tower? Compute answers to 3 significant digits. For the receiving antenna shown in the figure, the com- mon focus F is located 120 feet above the vertex of the parabola, and focus F' (for the hyperbola) is 20 feet above the vertex. The vertex of the reflecting hyperbola is 110 feet above the vertex for the parabola. Introduce a coordi- nate system by using the axis of the parabola as the y axis (up positive), and let the x axis pass through the center of the hyperbola (right positive). What is the equation of the reflecting hyperbola? Write y in terms of x.

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