Section 10.4 Hyperbolas

Transcription

1 66 Section 10.4 Hyperbols Objective : Definition of hyperbol & hyperbols centered t (0, 0). The third type of conic we will study is the hyperbol. It is defined in the sme mnner tht we defined the prbol nd the ellipse. Definition: A Hyperbol is the set of ll points P in plne such tht the bsolute vlue of the difference of the distnce from one fixed point F 1 to the point P nd the distnce from second fixed point F to the point P is constnt. The two fixed points F 1 nd F re clled the foci (plurl of focus) of conjugte xis d(f 1, P) P d(f, P) F 1 V 1 ϑ V F trnverse xis the hyperbol. Like n ellipse, hyperbol hs two xes of symmetry, the conjugte xis nd trnsverse xis. The trnsverse xis contins the two foci nd the midpoint between the two foci is the center ϑ of the hyperbol. The points of the hyperbol tht intersect the trnsverse xis re the vertices V 1 & V (plurl of vertex) of the hyperbol. The conjugte xis psses d(f 1, P) d(f, P) = constnt through the center of the hyperbol nd is perpendiculr to the trnsverse xis. The hyperbol itself consists of two seprte curves clled brnches. In the illustrtion bove, the trnsverse xis runs prllel to the x-xis. We could hve just s esily drwn hyperbol with the trnsverse xis running prllel to y-xis nd ll of the sme definitions would still pply. If we let be the distnce between the center of the hyperbol nd one of the vertices nd let c be the distnce between the center of the hyperbol nd one of the foci. Then the distnce between V 1 nd F 1 will be c nd the distnce between F nd V will be c s well. Notice tht this is different compred to n ellipse nd it is importnt to keep this reltionship

2 67 distinct from the reltionship with n ellipse. We will then define b such tht b = c. Agin, this is different then how b ws defined for the ellipse. c We will lter see wht the geometric reltionship b hs with F 1 V 1 ϑ V F the hyperbol. For now, when we think of b, remember the b is equl to the longer of nd c squred minus the shorter of nd c squred. Now, we wnt to estblish tht: d(f 1, P) d(f, P) = Proof: 1) Assume d(f 1, P) > d(f, P). Recll tht d(f 1, P) d(f, P) = constnt for ny point P on the hyperbol. c c If we let P = V, then d(f 1, P) = d(f 1, ϑ) + d(ϑ, V ) F 1 ϑ P = V F = c + Also, d(f, P) = c Thus, d(f 1, P) d(f, P) = c + (c ) = c + c + = But, d(f 1, P) d(f, P) = constnt for ny point on the hyperbol, so d(f 1, P) d(f, P) = Now, we re redy to derive the formul for the hyperbol. For our derivtion, we will use hyperbol P: (x, y) centered t (0, 0) nd with the trnsverse xis prllel to the x-xis. The derivtion for the formul for F 1 : ( c, 0) F : (c, 0) the hyperbol with the trnsverse xis prllel to the y-xis works the sme wy, just with the roles of x nd y switched just like with the ellipse.

3 68 Let F 1 : ( c, 0) nd F : (c, 0) be the foci nd P: (x, y) be ny point on the hyperbol. We just showed tht: d[f 1, P] d[f, P] = (pply the definition of the bsolute vlue) d[f 1, P] d[f, P] = ± d[( c, 0), (x, y)] d[(c, 0), (x, y)] = ± (plug in the points) (pply the distnce formul) (x ( c)) +(y 0) (x c) +(y 0) = ± (simplify) (x+c) + y (x c) + y = ± (dd (x c) + y to both sides) (x+c) + y = ± + (x c) + y (squre both sides) ( (x+c) + y ) = ( ± + (x c) + y ) (expnd nd simplify) (x + c) + y = (± ) + (± ) (x c) + y + (x c) + y (expnd) x + cx + c + y = 4 ± 4 (x c) + y + x cx + c + y (dd x + cx c y to both sides) 4cx = 4 ± 4 (x c) + y (subtrct 4 from both sides) 4cx 4 = ± 4 (x c) + y (divide both sides by 4) cx = ± (x c) + y (cx ) = ( ± (x c) + y ) (squre both sides) (expnd nd simplify) c x cx + 4 = ((x c) + y ) c x cx + 4 = (x cx + c + y ) c x cx + 4 = x cx + c + y (dd cx 4 x y to both sides) (expnd nd simplify) (distribute) c x x y = c 4 (fctor x from c x x & from c 4 ) (c )x y = (c ) (replce c by b ) b x y = b (divided both sides by b )

4 69 x y = 1 b If the trnsverse xis is long to the y-xis, then the eqution of the hyperbol is y x b = 1. To see the geometric reltionship between b nd the hyperbol, we will strt with the eqution of the hyperbol with the trnsverse xis long the x-xis. x y b y b y = 1 (subtrct x from both sides nd divide by 1) b = x = x 1 (write the right sides s single frction) y b = ± x y = ± b x (use the squre root property nd simplify) (multiply both sides by b) Since is fixed, then s x, the vlue of x becomes very lrge compred to, mening tht x x. Thus, s x, x x = x. So, for lrge vlues of x, y = ± b x = ± b x. This mens tht the hyperbol will hve two oblique symptotes. Similrly, for hyperbol with the trnsverse xis long the y-xis, we cn find the oblique symptotes: y y y x = x b = x b = 1 (dd x b to both sides) + 1 (write the right side s single frction) +b b y = ± x +b b y = ± x b b (use the squre root property nd simplify) (multiply both sides by ) As x, x b x = x. Thus, the oblique symptotes re y = ± b x.

5 Eqution of n Hyperbol Centered t the Origin 1) The Hyperbol with center ) The Hyperbol with center (0, 0) nd eqution (0, 0) nd eqution x y y b b = 1 hs vertices of (±, 0), foci of (± c, 0), nd (0, ± ), foci of (0, ± c), nd oblique symptotes of oblique symptotes of y = ± b x where b = c, y = ± b x where b = c, b, & c re positive. The, b, & c re positive. The trnsverse xis is long the trnsverse xis is long the x-xis. y-xis. conjugte xis trnsverse xis y = b x c 70 y = trnsverse conjugte xis c 0 c xis 0 y = b x c y = b x Find nd grph the eqution of the following hyperbol: Ex. 1 Hyperbol centered t the origin with focus t ( 5, 0) nd vertex t (3, 0). Solution: Since the focus nd the vertex lie on the x-xis, then the trnsverse xis lies long the x-xis nd the brnches of the hyperbol opens left nd right. Thus, the eqution hs the form: x y = 1 b Since the focus is five units wy from the center, then c = 5 nd since the vertex is three units from the center, then = 3. Using b = c, we cn find b: b = 5 3 = 16 (tke the principl squre root since b > 0) b = 4 b x

6 Hence, the equtions is: x 9 y 16 = 1 The oblique symptotes re y = ± b x = ± 4 3 x. To help us drw the symptotes, we cn construct wht we cll the "symptote rectngle." If we plot the vertices (±, 0) = (± 3, 0) nd then plot the points (0, ± b) = (0, ± 4), we cn use these four points to form the symptote rectngle. The digonls of the rectngle will hve slope of ± b = ± 4. We cn then extend the digonls to get the 3 oblique symptotes. 71 Ex. Hyperbol with vertices (0, ) nd (0, ) nd oblique symptotes y = ± 5 x. Solution: Since the vertices lie on the y-xis, the trnsverse xis runs long the y-xis, mening the hyperbol opens up nd down. Also, the center of the hyperbol is the midpoint between the vertices which is the point (0, 0). So, the form of the eqution needed is: y x = 1 b The vertices re ech two units wy from the center, then =. The slope of the oblique symptotes is ± b = ± b = ± 5 b = 5. Hence, the eqution is: which mens

7 7 y 4 x 5 = 1 To find the foci, plug in = nd b = 5 into b = c nd solve for c: 5 = c 4 9 = c (tke the principl squre root since c > 0) c = 9 Thus, the foci re (0, ± 9 ) Plot the vertices (0, ± ) = (0, ± ) nd then plot the points (± b, 0) = (± 5, 0). Use these points to construct the symptote rectngle nd extend the digonls of the rectngle to get the oblique symptotes. (0, 9 ) (0, 9 ) Anlyze the following eqution: Ex. 3 6x 9y = 54 Solution: To nlyze the eqution, we will find the center, the foci, the vertices, the oblique symptotes nd then sketch the grph of the hyperbol. First, we will need to get the eqution in the pproprite form: 6x 9y = 54 (divide both sides by 54) x 9 y 6 = 1 Since the x term is positive, the trnsverse xis runs long the x-xis nd the hyperbol opens left nd right. Thus, = 9 = 3 nd b = 6. Using b = c, we cn find c: 6 = c 9 (dd 9 to both sides)

8 15 = c (tke the principl squre root since c > 0) c = 15 Thus, foci re (± c, 0) = (± 15, 0). Since (0, 0) is the midpoint of the foci, then the center is t (0, 0). The vertices re (±, 0) = (± 3, 0) nd the oblique symptotes re y = ± b x = ± 6 x. To sketch the grph, 3 73 we will find the symptote rectngle by plotting the vertices (± 3, 0) nd the points (0, ± b) = (0, ± 6 ) nd extending the digonls to get the oblique symptotes. Then, we will complete the grph. ( 15, 0) ( 15, 0) Ex. 4 5y 36x = 900 Solution: First, we will need to get the eqution in the pproprite form: 5y 36x = 900 (divide both sides by 900) y 36 x 5 = 1 Since the y term is positive, the trnsverse xis runs long the y-xis nd the hyperbol opens up nd down. Thus, = 36 = 6 nd b = 5 = 5. Using b = c, we cn find c: 5 = c 36 (dd 36 to both sides) 61 = c (tke the principl squre root since c > 0) c = 61 Thus, foci re (0, ± c) = (0, ± 61 ). Since (0, 0) is the midpoint of the foci, then the center is t (0, 0). The vertices re (0, ± ) = (0, ± 6) nd

9 74 the oblique symptotes re y = ± b x = ± 6 x. To sketch the grph, 5 we will find the symptote rectngle by plotting the vertices (0, ± 6) nd the points (± b, 0) = (± 5, 0) nd extending the digonls to get the oblique symptotes. Then, we will complete the grph. Objective b: Hyperbol with center (h, k). We will now consider hyperbols centered t (h, k). Such hyperbol is shifted h units horizontl nd k units verticl, so we will hve (x h) nd (y k) in plce of x nd y respectively in the formuls for the hyperbol nd the formuls for the oblique symptotes. The vertices nd foci will hve h dded to ll the x-vlues nd k dded to ll the y-vlues. Eqution of n Hyperbol Centered t (h, k) 1) The Hyperbol with center ) The Hyperbol with center (h, k) nd eqution (h, k) nd eqution (x h) (y k) b = 1 hs vertices of (h ±, k), foci of (h ± c, k), nd oblique symptotes of (y k) (x h) b = 1 hs vertices of (h, k ± ), foci of (h, k ± c), nd oblique symptotes of y k = ± b (x h) where b y k = ± (x h) where b b = c,, b, & c re positive. = c., b, & c re positive. The trnsverse xis is prllel The trnsverse xis is prllel to the x-xis. to the y-xis.

10 75 conjugte xis y k = b (x h) (h, k) trnsverse xis (h, k + c) trnsverse xis (h c, k) (h, k + ) (h + c, k) conjugte (h, k) xis (h, k) (h, k ) (h +, k) (h, k c) y k = b (x h) y k = (x h) b y k = (x h) b Find nd grph the eqution of the following hyperbol: Ex. 5 A hyperbol with foci of (, ) nd (, 4) nd vertex of (, 1). Solution: Since the foci lie on the line x =, then the trnsverse xis is prllel to the y-xis. Thus, the form of the eqution needed is: (y k) (x h) = 1 b The center (h, k) is the midpoint between the foci (, ) nd (, 4) which is the point (, 1). Since the distnce between the foci nd the center is 3, then c = 3. Likewise, since the distnce between the vertex (, 1) nd the center (, 1) is, then =. Using b = c, we cn find b: b = 3 = 9 4 = 5 (find the principl squre root since b > 0) b = 5 Plugging into the eqution, we get: (y 1) 4 (x+) 5 = 1 The vertices re (h, k ± ) = (, 1) nd (, 3) nd the oblique symptotes re y 1 = ± 5 (x+ ) or ± 5 5 (x + ). To sketch the grph, we will find the symptote rectngle by plotting the vertices (, 1), (, 3) nd the points (h ± b, k) = ( ± 5, 1) ( 4.4, 1) nd (0.4, 1) nd extending the digonls to get the oblique symptotes. Then, we will complete the grph.

11 76 ( 5, 1) ( + 5, 1) Notice how much esier it is to drw the symptote rectngle thn trying to grph the equtions of the symptotes in the lst exmple. Anlyze the following eqution: Ex. 6 3y + 1y 3x + 18x + 1 = 0 Solution: First, we will need to get the eqution in the pproprite form: 3y + 1y 3x + 18x + 1 = 0 (subtrct 1 from both sides) 3y + 1y 3x + 18x = 1 (fctor out 3 nd 3 respectively) 3(y + 4y) 3(x 6x) = 1 (complete the squre) 3(y + 4y + 4) 3(x 6x + 9) = 1 + 3(4) 3(9) (simplify) 3(y + 4y + 4) 3(x 6x + 9) = 7 (write s perfect squre) 3(y + ) 3(x 3) = 7 (divide both sides by 7) (x 3) 9 (y+) 9 = 1 So, the center (h, k) is (3, ). Since the (x 3) term is positive, the trnsverse xis runs prllel to the x-xis nd the hyperbol opens left nd right. Thus, = 9 = 3 nd b = 9 = 3. Using b = c, we cn find c: 9 = c 9 (dd 9 to both sides) 18 = c (tke the principl squre root since c > 0) c = 18 = 3 The vertices re (h ±, k) = (3 ± 3, ) = (0, ) & (6, ) nd the foci re (h ± c, k) = (3 ± 3, ) ( 1.4, ) & (7.4, ). The

12 77 oblique symptotes re y + = ± 3 (x 3) or ± (x 3). To sketch the 3 grph, we will find the symptote rectngle by plotting the vertices (0, ), (6, ) nd the points (h, k ± b) = (3, ± 3) = (3, 5) nd (3, 1) nd extending the digonls to get the oblique symptotes. Then, we will complete the grph. In this lst exmple, it my hve been esier to grph the symptotes insted of constructing the symptote rectngle since the equtions for the oblique symptotes were so esy to grph. Ex. 7 f(x) = 16+4x Solution: Since f(x) is negtive, we will only hve the "hlf" of the grph of the hyperbol. First, we will need to get the eqution in the pproprite form: f(x) = y = 16+4x (squre both sides, noting y 0) y = x (subtrct 4x from both sides) y 4x = 16 (divide both sides by 16) y 16 x 4 = 1 Since the y term is positive nd the center (h, k) is (0, 0), the trnsverse xis runs long the y-xis nd the hyperbol opens up nd down. However, y 0, so we will only get the brnch below the x-xis s our grph. Thus, = 16 = 4 nd b = 4 =. Using b = c, we cn find c:

13 4 = c 16 (dd 16 to both sides) 0 = c (tke the principl squre root since c > 0) c = 0 = 5 Thus, foci re (0, ± c) = (0, ± 5 ). The vertices re (0, ± ) = (0, ± 4) 78 nd the oblique symptotes re y = ± b x = ± 4 x = ± x. To sketch the grph, we will find the symptote rectngle by plotting the vertices (0, ± 4) nd the points (± b, 0) = (±, 0) nd extending the digonls to get the oblique symptotes. Then, we will complete only the lower brnch of the grph since y 0. Suppose in the lst, the restriction ws x 0 insted of y 0. The process would be exctly the sme except we would hve drwn the left side of the grph nd not the right side:

14 79 Objective c: Applictions of Hyperbols. Hyperbols re used in mny different pplictions. If you turn on lmp with lmp shpe, the light will cst hyperbolic shdow on the wlls. If plne is going fster thn the speed of sound, the sonic boom tht it produces is cone shped. When the cone hits the ground it produces curve in the shpe of one brnch of hyperbol. Anyone stnding on tht curve t tht instnt will her the sonic boom t tht instnt. The structure of the most cooling towers of nucler power plnt re in the shpe of hyperboloid ( hyperbol rotted bout the conjugte xis) since it is the strongest structure tht cn be designed using the lest mount of mteril. In nvigtion, the LORAN nvigtion system is used to locted ships nd plnes. The bsic ide is to hve two trnsmitters tht re fixed distnce prt send out signl t certin time. A nvigtor on the ship or plne then mesures the difference in rrivl times between the signls nd uses specil mps locte the crft on the corresponding hyperbol. In this wy, the nvigtor cn determine how fr the crft is from ech trnsmitter. In ctul prctice, using two trnsmitters would give the nvigtor two possible loctions. crft OR trnsmitter 1 trnsmitter trnsmitter 1 trnsmitter crft

15 80 Thus, the LORAN system ctully uses three trnsmitters in tringulr fshion to ccurtely locte ships nd plnes. Solve the following: Ex. 7 A nucler power plnt is in shpe of hyperboloid hs bse dimeter of 600 feet. The dimeter t its nrrowest point is 300 feet nd is locted 540 feet bove the ground. If the dimeter of the top of the tower is 450 feet, find the height of the cooling tower. Solution: First, plce coordinte xis on the cooling tower such tht the point (0, 0) is in the middle of the nrrowest of the cooling tower. Since the dimeter t tht point is 300 feet, then the distnce between the center nd the wll of the cooling tower is 150 feet. Since this is the nrrowest point, then ( 150, 0) nd (150, 0) re the vertices nd hence, = 150. The dimeter of the bse is 600 feet, so the wll is 300 feet from the conjugte xis. This is lso 540 feet below the center of the hyperbol, thus (300, 540) is point on the hyperbol. Since the trnsverse xis is long the x-xis, the eqution is in the form: x y b 300 ( 540) 150 = 1 (plug in = 150 nd the point (300, 540)) b = 1 (simplify) = 1 b (subtrct 4 from both sides) = 3 b (multiply both sides b ) 3b = (divide both sides by 3) b = 9700 b = Thus, the eqution of the hyperbol is: x 500 y 9700 = 1 ( 150, 0) (tke the principl squre root) (5, y) (0, 0) (150, 0) (300, 540)

16 81 The dimeter t the top of the tower is 450 feet, then the distnce from the conjugte xis to the side of the tower is 5 ft. Thus, the endpoint is in the form of (5, y) where y > 0. Plug this point into the eqution nd solve for y: (5) 500 y 9700 = 1 (simplify).5 y 9700 y 9700 = 1 (subtrct.5 from both sides) = 1.5 (multiply both sides by 9700) y = (tke the principl squre root) y = ft Thus, the top of the tower is feet bove the center nd the bse is 540 feet below the center, so the totl height of the tower is = feet tll.