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**Unformatted text preview: **4.4 Conic sections By the end of this subchapter, you will learn
a What IS a ‘conic’ or ‘conic section’
How to deﬁne the following conic sections. Parabolas; ellipses; hyperbolas
How to identify a conic from an equation
How to‘derive the equation of a conic from the given conditions
What are the popular terms used in conic, such as ‘Vertex’, ‘focus/foci’, ‘directrix’,
‘asyrnptotes’, ‘major axis’, ‘minor axis’...
I How to ﬁnd the vertex, focus, and directrix of a parabola
0 How to sketch a parabola
o How to ﬁnd vertices, foci, major axis, and minor axis of an ellipse
- How to sketch an ellipse
- How to ﬁnd vertices, foci, and asymptotes of a hyperbole
-_ How to sketch a hyperbola Notes 4.4 Please refer to Appendix A (extracted from Calculus (2010) by S. T. Tan published by
Brooks/Cole, Canada) Tutorial 4.4 Please refer to your textbook, Calculus 7th Edition (2012) by J. Stewart, for the following:
Exercise 10.5 — Questions 5,7, 11,15, 19, 23, 25, 27, 29, 33, 37, 47 4.5 Conic sections and polar coordinates By the end of this subchapter, you will learn
- What is ‘eccentricity’
o How to identify a conic from its eccentricity o How to deﬁne the polar equation for three types of conics: Parabolas; ellipses;
hyperbolas- o How to derive a polar equation for a conic with the focus at the origin and the given
data G How to ﬁnd the eccentricity, identify the conic, locate the directrix, and sketch the
conic from the given polar equation Notes 4.5 Please refer to Appendix B (extracted from Calculus (2010) by S. T. Tan published by
Brooks/Cole, Canada) Tutorial 4.5 -
Please refer to your textbook, Calculus 7’1Ediz‘icn (2012) by J. Stewart, for the following:
Exercise 10. 6— Questions 1, 3, 5, 7, 9, 11, 13, 15 a radio telescope. The shape of the surface of the reﬁe
lane curve called a parabola about its axis of symm‘ Figure 1 shows the reflector of tor is obtained by revolving a p try. (See Figure 2a.) Figure 2b depicts the
d an ellipse. Figure 2c depicts the trajectory of an incoming alpha 9 curve is calle
cle heading toward and then repulsed by a massive atomic an F. The trajectory is one of two branches of a layer-bola. Axis Ficus: 1 Tim reflector Of a mdw telescope (b) The orbit of a planet around (c) The trajectory of an :11 (a) The cross section of a radio
the sun is an ellipse. tetescope is part of a
parabola. FIGURE 2 orbit of a planet P around the son, 5', T cleus located at the p x P (planet) ' \\
\
\ hF (nucleus) ‘5.
‘ l Alpha pa 3. particle in a Rutherf
scattering is part of a
of a hyperbolic s—parabolas, ellipses, and hyperbolas—are called com’ ‘ _ These curve more simply, conics because they result from the intersection of a plane mapped cone, as shown'in Figure 3. FIGURE 3 The comic sections to) Ellipse (c) Hyperbnla (a) Parabola In this section we give the geometric deﬁnition of each con
an equation for describing each conic section algebraically. E Parabola We ﬁrst consider a conic section called a parabola. primates sagas f
" F bolais‘jthe‘ set'of' all]
' (salad assassins ‘4 'siufs'ifii anesthesia; at: 7
ﬁxed line (Called the direc' . FIGURE 4 The distance between a point P on a
parabola and its focus F is the same
as the distance between P and the
directrix I of the parabola. By deﬁniti ' thalfwa ,
s called the ver _ ”WWW-m Haw-M Thispoint Vi tax of the _=9.W ic section, and; parabolafﬂ'lhgiihempaSSIng thI; m I
> 7760115 5
The parabola with focus F (0, p) and
directrix y = —p, where” p > O FIGURE 6
130132 I = 4p); opens upward
_0 and downward if p < 0. "l
l 10.] Conic Sections 329 and perpendicular to the directIix 1s called the axis of the parabola Observe that the
We Wit respEEtTo Its—8561's. ”””””” 'ﬁ """" ' “WP—"Ru .mdw—h as- 3 M—ﬁHHw—wﬂwwmﬁi mug“ M“ To ﬁnd an equation of a parabola, suppose that the parabola is placed so that its
vertex is at the origin and its axis is along the ,y-axis, as shownin Figure 5. Further,
suppose that its focus F is at (O, p), and its directrix is the line with equation y = —p.
If P(x, )1) IS any point on the parabola, then the distance between P and F 15 d(P,F)— VIE +(y— P)‘ whereas the distance between P and the directrix 1s [ y + pl. By deﬁnition these dis— tances are equal, so
Vx + (11-11) =|y + pi
Squaring both sides and simplifying, we obtain
x2+ (y -p)2:|y+1vl2 =01 +13)2
x2 + y2 — 211711 +132 =32 + 2P)” +172 : 4103’ ‘ “4.5%; {ﬂig’x i s
r" . _—__—__.....L.._._,m.____w—..._._—-_—{
f Standard Equation of a Parabola :
Ii131: equation of the parabola with Eggs (2,5Land d1reetr11ci;:£ E E
é 2:. rum“? g
i [3% 1 if} C. 7'" 5 .“yhn Emm‘ipij m}; Phi-aim i (1) i nu.» WW WWW“ If we write a = 1/ (4p), then Equation (1) becomes y = (11:2. Observe that the
qungs downward, if p < 0. (See Figure 6.) Also,
the parabola is symmetn'c with respect to themii-axis (that is the axis of the parabola co—
incides with the y- axis) since Equation (1) remains unchanged if we replace .7: by -x. i?" P" 1
l
l
J ham)- (1—7 7" 1 830 Chapter 10 Conic Sections. Plane Curves, and Polar Coordinates interchanging x and y in Equation (1) gives “W“? ‘ l” ; a) which is an equation of the parabola with focus Fiﬁ. 0) and directrix x — 2125,3313 WWW: parabola opens to the right if g? (l and opens to the left If Hp < 0. (See Figure 7.) In -_-.._-.m.......7. ...77. .7-7..7.,._-.1 Hm. both cases the axis of the parabola coincides with the x—axis. FIGURE 7
The parabola y2 é, 4px opens to the
right ifp > 0 and to the ieftifp < O. or y—axis is said to be in standard position. (See Figures 6 and 7.) “—.,..__,, _________...._.ﬁm..___._ sketch of the parabola. Solution Rewriting the given equation in the form y2 = -6x and comparing it'-
Equation (2), we see that 4p = —-36 or p 2 +3. Therefore, the focus of the parabo
F( ”g, 0), and its directrix 13 x — %.The parabola is sketched in Figure 8. ' FIGURE 3
The parabola y2 + 6x = 0 equation -4— — 111(3): giving a — —- Therefore, an equation of the parabolal ., = 1%,:
To ﬁnd the focus of the parabola observe that it has the form F (0, ,0). NOW:
1 1 9 P Eli—3):? Therefore, the focus is F(0, _]9§). Its directn'x is y = "p = —('**9*
graph of the parabola is sketched in Figure 9.
1 'r FIGURE 9 The parabola has many applications. For example, the cables of aerial
The parabola y = 43‘1" bridges assume ‘shapes that are parabolic. 7 FIGURE is
A bridge of length Zn
suspended by a ﬂexible cable ' . 10.1 Conic Sections _ 831 _,_ .7 Suspension Bridge Cables Figure 10 depicts a bridge suspended by a
ilexihle cable. If we assume that the weight of the cable 15 negligible in comparison to
the weight of the bridge, then it can be shown that the shape of the cable is described
by the equation ’3 WI” J1=2H where W is the Weightof the bridge in pounds per foot and H is the tension at the low-
est'point of the cable in pounds (the origin). (See Exercise 89.) Suppose that the span
of the cable is 2:; ft and the sag is 11 ft. * 3. Find an equation describing the shape assumed by the cable in terms of a and h.
b. Find the length of the cable if the span of the cable is 400 ft and the sag is
80 ft. Solution
2. We can write the given equation in the form y = kxz, where k = W/(ZH). Since
the point (a, It) lies on the parabola y = kxz, we have h=kaz or k = h/al, so the required equation is y = [Hz/a2.
b- With a ‘= 209 and h = 80 an equation that describes the shape of the cable is- 803:2 J:—2
2002 500 y =
Next, thelength of the cable is given by
zoo
s 2 2i V1+ (y’)2dx
0 But 32’ = .x/ZSO, so 200' I 'J 1 200 ‘
s22] 1+(——) dx=mf V2503+x3dx
O 250 125 0 ' _ The easiest way to evaluate this integral is to use Formula 37 from the Table of
integrals found on the reference pages of this book: 77
- ., u " n
[We +1qu=§m+gmlu +mi + C 832 Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates If we let a = 250 audit =.x, then 2 200
5::[3251250‘irx2 25 2502+x2]
125 2 , _ 0 125 _1—[100'V62500 + 40000 + 3212501an00 + V62500 + 40000 I — 312501n 250} 2 +\/ 25
oo 10_OO):439 5 4V 102500 501D( 250 or 439 ft. Other applications of the parabola include the trajectory of a projectile 1n the absenc
of air resistance. ‘. % Reflective Property of the ‘Parabola Exercise 105.) As _was hteno'oned earlier, the reﬂector a radio telescope has a shape that
obtained by rerolviug a parabola about it axis. Figure 12a shows a cross secti ._
such a reﬂector. A radio wave coming iIKfrorn a great distance may be assumed
FIGURE 11' parallel to the axis of the parabola x‘I‘his wave will strike the surface of the ref],
The reﬂective, property states that and be reﬂected toward the focus/F, where a collector' 1s located. (The angle of
at = ,8. . deuce is equal to the angle off/reﬂection.) _______ «(H.— —.._:.. H
________ _(_ _ _ ; i
_/ _______ .4 _____ " .
___ 71-9%.;05 ________ [has
, f. _________ _< _____ '\‘
———————— —<————— \I
{If _______ _< _____ i a.
FIGURE 12
Applications of the reﬂective ”/1
property of a parabola” (a) A cross section of a radio telescope (bj'A cross section of a headlight
a" The reflective property of the parabola is also used in the design of he
/ automobiles, Here, a light bulb is placed at the focus of the parabola A re
“,2”! emanating from the light bulb wili strike the surface of the reflector and I? outward along a direction parallel to the axis of the parabola (see Figure, l' Elli‘pses . Next, we consider a conic section called an ellipse. r/wau , FIGURE 13“
' An ellipse with foci F, and F2. A
point P(x, y) is on the ellipse if
and only if d; + d: = a constant. :0: ling an ellipse on paper using § - packs, a string, and a pencil 10.1 Conic Sections 833 I, “125?,‘iﬁhé; f' p,
tau ﬁtted pants tcaucdee‘ Figure. 13 shows an ellipse with foci F1 and F2. The line passing through the foci
intersects the ellipse "rat two points, V1 and V;, called the vertices of the ellipse. The
chord joining the vertices is called the major axis, and its midpoint is called the C811!
ter of the ellipse. The chord passing through the center of the ellipse and perpendicu-
lar to the major axis is called the minor axis of the ellipse. ________ -‘n— — V2 (vertex) F1 (focus) Note We can construct an ellipse on paper in the following way: Place a piece of
paper on a flat wooden board. Next, secure the ends of a piece 'of string to two points
(the foci of the ellipse) with thumbtacks. Then trace the required ellipse with a pencil
pushed against the string, as shown in Figure 14, making sure that the string is kept
taut at all times. g To ﬁnd an equation for an ellipse, suppose that the ellipse is placed so that its major
axis lies along the x-axis and its center is at the origin, as shown in Figure 15. Then
its foci F l and F2 are at the points (we, 0) and (c, 0), respectively. Let the sum of the
distances between any point P(x, 3:) on the ellipse and its foci be 2:: > 2c > 0. Then,
by the deﬁnition of an ellipse we have d(P, 3,) + d(P,F2) = 2:17
that is, . ‘
“(X+C)2+J’2+ (I—C)2+y2=2a
or i
m = 2a _ m
Squaring both sides of this equation, we obtain ‘
x2”Zen-t-c2+yz=4nz~4avm+x2+2cx+cé+y3
or, upon simpliﬁcation, - .
am = a2 + C):
Squaring both sides again, We have - i n2(x2 + 2cx + c2 + 322) = n4 + 2a2cx + :3ch2 334 Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates which yields
(a2 — elk: + aZ-y2 = a2(c12 — c2) Recall that o: > c, so a2 —* c2 > 0. Let b?" 2' a1 — a:2 with b > 0. Then the equa—
tion of the ellipse becomes - 52x2 + alyg = a2b2
or, upon dividing both sides by (125.2, we obtain
2 2
i: + .3.)— : 1
a2 b” By setting y .= O, we obtain x = :a, which gives (—a, (l) and (a, O) as the vet.
tices of the ellipse. Similarly, by setting x =1 0, we see that the ellipse intersects the
y—axis at the points (0, '—b) and (O, 5). Since the equation remains unchanged if x is
replaced by —.r and y is replaced by‘—y, we see that the ellipse is symmetric with
respect to both axes. Observe, too, that b < :1, since ' 7 'J ')
-b‘=a2—c'<a‘ origin leads to an equation in which the roles of x and y are reversed. To summariz
we have the following. - Vi? lit 1é\ . mm Standard Equation of an Ellipse ‘
An equation of the ellipse with foci (ic, 0) and vertices (in, 0) is J
'F
l {\{J/ i 2 ’/l ‘ c" if] -_
me" : —+y—Q=1- aab>0 W v' .. (3,)
l l b is cQW’JA ”if” '
n. if" I . . '. . . . ’
" NT?“ 1 and an equation of the ellipse With feel (0, :c) and verttces (0, ta) is 5‘- KIJ '
r RDA; E 2 2 .
{an E3; 11' x_. + 97—. -—- :2 _ i
U i “6—: a! 1 (I b > 0 u! “‘3 {”7
Him-M whet r52; -— a! — 132. (See Figure 16.) 3*?" ’“ “Fwy?”
k/U L l . .11
FIGURE 16 ‘ Two ellipses in standard position
/ with center atrthe origin FIGURE 17 _ x" y
The ellipseig + K = 1 FIGURE 13 The reﬂective property
states that CE = E. 10.1 Conic Sections 835 Note An ellipse with center at the origin and foci lying along the x—axis or the y-axis
is said to be in standard position. (See Figure 16.) E” 7 7 .. Sketch the ellipse % + )1? z 1. What are the foci and vertices? Solution Here, a2 = 16 and b2 = 9, so a = 4 and b : .3- Settingy : 0 and x = 0
in succession gives the x— and y-intercepts as :4 and :3, respectively. Also, from _ czzcﬁ—bzzis—927 We obtain c = W and conclude that the foci of the ellipse are (ix/77, 0}. Its vertices
are ($4, 0). The ellipse is sketched in Figure 17. . E Find an equation of the ellipse with foei (0, :2) and vertices (0, :4). Solution Since the foci and therefore the major axis of the ellipse lie along the y-axis,
we use Equation (4). Here, c = 2 and a = 4, so
bi=a24c3= 16—4: 12 Therefore, the standard form of the equation for the ellipse is 2 2
ii. 2. = 1
12 16
Of
4x2 + 3y2 2: 48 2% E Ref etive Property of the Ellipse The ellipse, like\the parabola, has a reﬂective preperty To describe this property, con—
sider an ellipse with foci F 1 and F; as shown in Figure 18. Let P be a,p“oint on the
ellipse, and let I be t atangent line to the ellipse at P. Then the angled between the
line segment FIP and l ' equal to the angle ,6 between the line-Sbgment FZP and l.
‘ You will be asked to establish this property in Exercise 106/” - 2‘ / sf“ ‘\ . The reﬁectivepiioperty of the ellipse is used to design whispering galleries—rooms
with elliptica/Irshaped ceilings, in which a person standing at one theirs can hear the
whisper offanother person standing at the other focus. A whispering \galleiy can be
foundjj-n’rthe rotunda of the Capitol Building in Washington, DC. Also, Paris subway
t3nn‘éls are almost elliptical, and because of the reﬂective property of the ellipse, whis— pering on one platform can be heard on the other. (See Figure 19.) .1.
1"; 2’ ‘ 836 Chapter 10 Conic Sections. Plane Curves, and Polar Coordinates FIGURE 19
A cross section of a Paris subway
tunnel is almost eiliptica]- / Yet another application of the?eﬂectivd'property of the ellipse can be found in the '
ﬁeld of medicine in a p’roédure for removing kidney stones called shock wav
lithom‘psy. In this progedue an ellipsoidal reﬂector istpositioned so that a transducer" 1
at one focus and alodney stone is at the other focus. Shock waves emanating from th
transducer are/reﬂected according to the reﬂective propertycf the ellipse onto the kid
ney stone (pulverizing it. This procedure obviates the necessity for surgery. /’
//e Eccentricity of an Ellipse To measure the ovalness of an ellipse, we introduce the notion of eccentricity.
a not“ arts-ma % Hyperbolas WEL— between the foci and a point on an ellipse is ﬁxed whereas the diﬁererice of thes
tances is ﬁxed for a hyperbola. [ l
I foci intersects the hyperbole at two points, V1 and Vi, called the ver‘tices bofthb FiGURE 20 . bola The line segment joining the vertiees is called the transVerse axis of Eh
A hyperbola with focj Ft and F2. A bola and the midpoint of the transverse axisis calie- e center of the by point Pot, y) is on the hyperbole if and Observe that a hyperbola, in centrast to a parabola or an “ellipse, has two.-
only if IdI — d1} is a constant. branches. ‘ _..._. FIGURE 21 An equation of the hyperbola with
0) and foci (h c, O) and (c, ‘0) center (0, . 12 3” _ IS? — [2—3 g 1
'I-K‘ig/ /) / [ ‘~ rent 32 a- la. 10.? Conic Sections 837 The derivation of an equation of a hyperbole is similar to that of an ellipse. Consider,
for example, the hyperbole. with center at thigiginwzniifociﬁ 1_(:_g, W, O)
ogﬁenaxisthee Figure 21.) Using the condition (KP, Fl) — (KP, F3) 2 2a, wlﬁie
a is a positive constant, it can be shown that if P(x, )2) is any point on the hyperbola,
then an equation of the hyperbola is (t2 "b- -whereb = Veg—o2 orc = Va2 + 192. Observe Lhat the x—intercepts of the hyperbole are x 2 in, giving (—a, 0) and (a, O)
as its vertices. But there are no y—intercepts, since setting x = 0 gives y2 = -b2, which
has no real solution. Also, obsérve that the hyperbola is symmetric‘with respect to both
axes. ' If we solve‘the equation 3’2
__2=l gel“... CI" for y, We obtain 13
y=i— x —a
(1 Id 2 Since x2 - a2 2 0 or, equivalently, x E “a or x 2 a, we see that the hyperbole actu-’
ally consists of two separate branches, as was noted earlier. Also, observe that if x is
large in magnitude, then x2 — n2 = x2, 50‘}: = jib/cox. This heuristic argument sug—
gests that both branches of the hyperbole approach the slant asymptotes y = i (b/a)x
as 2: increases or decreases without bound. (See Figure 22.) You will be asked in Exer~
cise 101 to demonstrate that this is true. Finally, if the foci of a hyperbole are on the y—axis, then by reversing the roles of .x and y, we obtain 2 2
3’ x 1
a2 , b2
as an equation of the hyperbole. , ..,.._._...s.a..ww--~-~r W ....
, _ 1 t2 ,
= 1 (The transverse axis is along thex-axis.) (b) l, -— ‘r—,, = l (The transverse axis is along the y—axis.)
_ "aw—vmhfea-wnvupewa-‘r’mwr-Wuﬁa;ﬁ_ :hmw—m M_ __,,.,.a~e---r“*‘" ‘ Two hyperbolas in standard position with center at the origin 838 Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates 5 ______.n
i Standard Equation of a Hyperbola
An equation of the hyperbole with foci (ic, 0) and vertices (+a 0) IS ‘ x2 yz. 5
EHﬁzQ) <5) i
l l l E i‘ where c = V a2 + 122. The hyperbole has asymptotes y = jib/(Ox. An equation
l l t l l l 31/” of the hyperbole with foci (O, :c) and vertices (O, in) is
2 ’I -.
3’ x- '\
g—e—g @- where e = Va2 + bz. The hyperbole has asymptotes y z i(a/b)x.
LEN” . l
r r l
itaéqw m».n_. ! The line segment of length 2!; Joining the points (0, mb) and (0 b) or (— —l; 0) an
(E7, 0) 15 called the conjugate axis of the hyperbole.
. WWW Find the fool, vertices, and asytnptotes of die hyperbole 4x2 ~ '9 Solution Dividing both sides of the given equation by 36 leads to the standard equat v 2
1—3’ = Veg + b2 = V 13, and conclude that the foei of the hyperbole eze' (iVlS
Finally, the asymptotes of the hyperbole are b 2
y x i—x = :"x
a 3
FIGURE 23 ' .
The graph of the hyperbgla When you sketch this hyperbole, draw the asymptotes ﬁrst so that you can then
4x27— 9f = 36 ‘ them as guides for sketching the hyperbole itself. (See Figure 23 .) , Find an equation of the hyperbole. What are its foci and esyﬁeptotes? Solution Here, the foci lie along the y—axis, so the standard equation of the hill): has the form
3|,2 x2
3—;=1 Notethetn= 3
To determine b we use the condition that the hyperbole passes throu Oh the 01
to write
25 4
__ __ T = 1 y
9 b"
, ./ .
i _ e: _ 1 _ 1_6
b2 9 10.] Conic Sections 839 or E)2 u 2. Therefore, a required equation of the hyperbola is ”1'51
9 r or, equivalently, y2 — 4x2 :3 9. To ﬁnd the foci of the hyperbola, we compute 45
cg=ag+bz=9+2:— _ 4 4
or c : i 45/ = :3V3/2, from which we see thatche foci are (0, 13¢}. Finally, the asymptotes are obtained by substituting a S 3 and b = % into the equations y— — i(n/b)x, giving y = 221:. The graph of the hyperbola is shown in Figure 24. incoming alpha particles is located at the point (— 2, 0), as shown 1n Figure 25. Sup-
pose that an alpha. particle approaching the nucleus has a trajectory that 15 a bran...

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- Spring '16
- DR ADYDA IBRAHIM
- Conic Sections, Polar Coordinates, Conic section