Chapter 8 Analytic Geometry in Two and Three Dimensions

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1 Section 8. Conic Sections and Parabolas Chapter 8 Analtic Geometr in Two and Three Dimensions Section 8. Conic Sections and Parabolas Eploration. From Figure 8., we see that the ais of the parabola is =0. Thus, we want to find the point along =0 that is equidistant from both (0, ) and the line =. Since the ais is perpendicular to the directri, the point on the directri closest to the parabola is (0, ) and (0, ), it must be located at (0, 0).. Choose an point on the parabola (, ). From figures 8. and 8., we see that the distance from (, ) to the focus is d + - = = and the distance from (, ) to the directri is d + = =. Since d must equal d, we have d + = + - = = d +(-) =(+) + -+= ++ = or = =. From the figure, we see that the first dashed line above =0 is =, and we assume that each subsequent dashed line increases b =. Using the equation above, we solve = r Eploration. to find: 0 b =, =, =, =, =, -,, -,, -,, -,, -,, -,, 0, 0,,,,,,,,,,,, = 0 F(, ). 0 The equation of the ais is =.. 0. Since the focus (h, k+p)=(, ) and the directri =k-p=, we have k+p= and k-p=. Thus, k=, p=. As a result, the focal length p is and the focal width p is.. Since the focal width is, each endpoint of the chord is units awa from the focus (, ) along the line =. The endpoints of the chord, then, are (-, ) and (+, ), or (, ) and (8, ). 0 A(, ). 0 0 F(, ) = V(, ) 0 F(, ) = = V(, ) 0 F(, ) = = = B(8, ) = A(, ) V(, ) 0 F(, ) B(8, ) =

2 Chapter 8 Analtic Geometr in Two and Three Dimensions 7. Downward 8. h=, p=, k=, so (-) = (-) Quick Review = + =. a - + b +. =, =_. =, =_. +7= ( -), +7-= (-), += (-). +=( +), ++ = a + b + = + 7. Verte: (, ). f() can be obtained from g() b stretching b, shifting up units, and shifting right unit. [, ] b [, 0] 8. Verte: (, ). f()= (-) +. f() can be obtained from g() b stretching b, reflecting across the -ais, shifting up units and shifting right units. [, 7] b [ 0, 0]. f()=a(+) +, so =a+, a=, f()= (+) f()=a(-) -, so =a-, a=, f()=(-) - Section 8. Eercises. k=0, h=0, p= Verte: (0, 0), Focus: a 0,, =. b Directri: = - Focal width: p = # =, -8. k=0, h=0, p= Verte: (0, 0), = -. Focus: (, 0), Directri: =, Focal p@ -@ = 8. k=, h=, p= =. Verte: (, ), Focus: (, ), Directri: = -=, Focal =. -. k=, h=, p= Verte: (, ), = -. - Focus: a -,, Directri: = - - a - b = b, Focal width: p = a - b = -. k=0, h=0, p= so p= - Verte: (0, 0),,. Focus: a 0, - Directri: = Focal width: b,, p = a - b =. k=0, h=0, p= so p = Verte: (0, 0),,. Focus: a Directri: = -, 0 b,, Focal width: p = a b = 7. (c) 8. (b). (a) 0. (d) For # 0, recall that the standard form of the parabola is dendent on the verte (h, k), the focal length p, the focal p@, and the direction that the parabola opens.. p= and the parabola opens to the left, so =.. p= and the parabola opens upward, so =8.. p= (so p= ) and the parabola opens downward, so =.. p= (so p=) and the parabola opens to the right, so =8.. p= and the parabola opens upward, so =0.. p= and the parabola opens to the left, so =. 7. h=0, p@ = 8 p = (since it opens to the right): (-0) =8(-0); =8. 8. h=0, p@ = p = - (since it opens to the left): (-0) = (-0); =. h=0, p@ = p = - (since it opens downward): (-0) = (-0); = 0. h=0, p@ = p = (since it opens upward): (-0) =(-0); =. h=, k=, = +p, so p= and the parabola opens to the right; (+) =8(+). h=, k=, +p=, so p= and the parabola opens downward; (+) = (-)

3 Section 8. Conic Sections and Parabolas. Parabola opens upward and verte is halfwa between focus and directri on =h ais, so h= and k = + = ; = - p, so p = = a - b. Parabola opens to the left and verte is halfwa between focus and directri on =k ais, so k= and h = + = 7 so p = - ; = 7 - p,. + = - a - 7 b. h=, k=; =-p, so p= and parabola opens to the left. (-) = 8(-). h=, k=; 7=-p, so p= and the parabola opens downward. (-) = 8(-) 7. h=, k= p@ = p = (since it opens upward): (-) =(+) 8. h=, p@ = 0 p = - (since it opens downward): (+) = 0(-). h=, k= p@ = 0 p = - (since it opens to the left): + = h=, p@ = p = (since it opens to the right): - = [, ] b [, 8] [ 0, 0] b [ 8, ].. [ 8, ] b [, ]

4 Chapter 8 Analtic Geometr in Two and Three Dimensions [, 8] b [, ] 8. [, ] b [ 0, ]. [ 0, ] b [, 7] [ 0, 8] b [ 0, ]. Completing the square produces -=(+).The verte is (h, k)=(, ), so the focus is (h, k+p)= a -, +, and the b = a -, b. [, 8] b [, ] directri is =k-p= - = 7 0. Completing the square produces a - 7 b = -. The verte is (h, k)= a, 7 so the focus is b,... [, ] b [ 0, ] [, ] b [, ] [, ] b [, ] [ 7, 7] b [ 7, ] (h, k+p)= a, 7, and the directri is + b = a, b 7 =k-p= - =.. Completing the square produces 8(-)=(-). The verte is (h, k)=(, ) so the focus is (h+p, k)=(+, )=(, ), and the directri is =h-p=-=0.. Completing the square produces - a - =(-). The verte is b (h, k)= a so the focus is, b (h+p, k)= a, and -, b = a, b the directri is =h-p=. + = 7. h=0, k=, and the parabola opens to the left, so (-) =p(). Using (, ), we find ( -) =p( ) p = - The equation = -. for the parabola is: (-) =. h=, k=, and the parabola opens to the right, so (+) =p(-). Using a we find, 0 b, (0-) =p a p= # The equation - b =. for the parabola is: (+) =(-).. h=, k= and the parabola opens down so (-) =p(+). Using (0, ), we find that (0-) =p( +), so = p and p=. The equation for the parabola is: (-) = (+).

5 Section 8. Conic Sections and Parabolas 7. h=, k= and the parabola opens up so (+) =p(-). Using (, ), we find that (+) =p(-), so =8p and p=. The equation for the parabola is (+) =8(-) 7. One possible answer: If p is rlaced b p in the proof, then the result is = p, which is the correct result. 8. One possible answer: Let P(, ) be a point on the parabola with focus (p, 0) and directri = p. Then - p = distance from (, ) to (p, 0) and - -p + - = distance from (, ) to line = p. Because a point on a parabola is equidistant from the focus and the directri, we can equate these distances. After squaring both sides, we obtain (-p) +(-0) =(-( p)) +(-) -p+p + = +p+p =p.. For the beam to run parallel to the ais of the mirror, the filament should be placed at the focus. As with Eample, we must find p b using the fact that the points (, ) must lie on the parabola. Then, ( ) =p() =8p p= =. cm 8 Because p=. cm, the filament should be placed. cm from the verte along the ais of the mirror. 0. For maimum efficienc, the receiving antenna should be placed at the focus of the reflector. As with Eample, we know that the points (., ) lie on the parabola. Solving for p, we find (.) =p() 8p=. p=0.78 ft The receiving antenna should be placed 0.78 ft, or.7 inches, from the verte along the ais of the reflector.. p=0, so p= and the focus is at (0, p)=(0,.). The electronic receiver is located. units from the verte along the ais of the parabolic microphone.. p=, so p= and the focus is at (0, p)=(0, ). The light bulb should be placed units from the verte along the ais of the headlight.. Consider the roadwa to be the ais. Then, the verte of the parabola is (00, 0) and the points (0, 0) and (00, 0) both lie on it. Using the standard formula, (-00) =p(-0). Solving for p, we have (00-00) =p(0-0), or p=00, so the formula for the parabola is (-00) =00(-0). The length of each cable is the distance from the parabola to the line =0. After solving the equation of the parabola for (= ), we determine that the length of each cable is B - + a b = Starting at the leftmost tower, the lengths of the cables are: {7.,.,,.,.78, 0,.78,.,,., 7.}.. Consider the -ais as a line along the width of the road and the -ais as the line from the middle stripe of the road to the middle of the bridge the verte of the parabola. Since we want a minimum clearance of feet at each side of the road, we know that the points (, ) lie on the parabola. We also know that the points ( 0, 0) lie on the parabola and that the verte occurs at some height k along the line =0, or (0, k). From the standard formula, (-0) =p(-k), or =p(-k). Using the points (, ), and (0, 0), we have: 0 =p(0-k) =p(-k) Solving these two equations gives p=.87 and k.. The maimum clearance must be at least. feet.. False. Ever point on a parabola is the same distance from its focus and its directri.. False. The directri of a parabola is perpendicular to the parabola s ais. 7. The word oval does not denote a mathematicall precise conct. The answer is D. 8. (0) =p(0) is true no matter what p is. The answer is D.. The focus of =p is (p, 0). Here p=, so the answer is B. 70. The verte of a parabola with equation (-k) =p(-h) is (h, k). Here, k= and h=. The answer is D. 7. (a) (c) = l (d) As A moves, P traces out the curve of a parabola. (e) With labels as shown, we can ress the coordinates of P using the point-slope equation of the line PM: - /+c = - b c -/ a - + b b - /+c - b = c -/ c -/a - /+c b = - b This is the equation of a parabola with verte at and focus at a b, /+c + p b where a b, /+c b p = c -/. F(b, c) P(, ) A(, l) Midpoint of AF M a + b l + c, b slope = b c l

6 8 Chapter 8 Analtic Geometr in Two and Three Dimensions 7. (a) (d) (c) Ais = n = (0, ) P(, n ) + ( ) = n Generator (e) A parabola with directri = and focus at (0, ) has equation =. Since P is on the circle +(-) =n and on the line =n-, its -coordinate of P must be = n - n - - = n - n -. Substituting n - n -, n-) into = shows that n - n - =(n-) so P lies on the parabola =. 7. (a) Ais Generator (b) Clinder Circle Two parallel lines Single line (d) Plane Line 7. The point (a, b) is on the parabola = p if and onl if b = a The parabola = and p. p a the line =m(-a)+ intersect in eactl p one point (namel the point a a, a if and onl if p bb the quadratic equation p - m + am - a p = 0 has eactl one solution. This happens if and onl if the discriminant of the quadratic formula is zero. -m - a a baam - = m - am p + a p p b p = a m - a =0 if and onl if m = a p b p. Substituting m = a and =0 into the equation of the p line gives the -interct = a a 0 - a + - a - a = b. p + a p p = p = p 7. (a) The focus of the parabola = is at (0, p) so an line with slope m that passes through the focus must have equation =m+p. The endpoints of a focal chord are the intersection points of the parabola = and the line =m+p. Solving the equation using the p - m - p = 0 quadratic formula, we have = Plane m ; m - a p b-p = m ; m + p a p b p p = pm ; m +.

7 Section 8. Ellipses (b) The -coordinates of the endpoints of a focal chord are = and p pm + m + = p pm - m + p p m + mm + + m + = p p m - mm + + m + = pm + mm + + = pm - mm + + Using the distance formula for pm - m +, pm - mm + + and pm + m +, pm + mm + +, we know that the length of an focal chord is = pm + + mpm + = m p + p + m p + m p = m p + m p + p The quantit under the radical sign is smallest when m=0. Thus the smallest focal chord has length p p@. 7. (a) For the parabola =p, the ais and directri intersect at the point (0, p). since the latus rectum is perpendicular to the ais of smmetr, its slope is 0, and from Eercise we know the endpoints are ( p, p) and (p, p). These points are smmetric about the -ais, so the distance from ( p, p) to (0, p) equals the distance from (p, p) to (0, p). The slope of the line joining (0, p) and (p, p) is -p - p = - and the slope of the line joining 0 - -p -p - p (0, p) and (p, p) is So the lines are 0 - p =. perpendicular, and we know that the three points form a right triangle. (b) B Eercise, the line passing through (p, p) and (0, p) must be tangent to the parabola; similarl for ( p, p) and (0, p). Section 8. Ellipses Eploration. The equations = + cos t and =+7 sin t can be rewritten as cos t = + and sin t = -. 7 Substituting these into the identit cos t+sin t= + - ields the equation + =.. [ 7.,.] b [, ]. Eample : Since, a parametric solution is + = = cos t and = sin t. Eample : Since, a parametric solution is + = = sin t and = cos t. - + Eample : Since + =, a parametric solution is = cos t+ and = sin t-.. [, ] b [, ] [, ] b [, ] [, ] b [, ] Answers ma var. In general, students should find that the eccentricit is equal to the ratio of the distance between foci over distance between vertices.. Eample : The equations = cos t, = sin t can be rewritten as cos t =, sin t =, which using cos t+sin t= ield or + =. + = Eample : The equations = cos t, = sin t can be rewritten as cos t =, sin t =, which using sin t+cos t= ield. + = Eample : B rewriting =+ cos t, = + sin t as cos t = -, sin t = + and using cos t+sin t=, we obtain =.

8 0 Chapter 8 Analtic Geometr in Two and Three Dimensions Eploration Answers will var due to erimental error. The theoretical answers are as follows.. a= cm, b= cm, c= cm, e=/ 0., b/a 0... a=8 cm, b= cm, c= cm, e=/=0., b/a 0.7; a=7 cm, b= 0. cm, c= cm, e=/7 0., b/a 0.0; a= cm, b= 0.7 cm, c= cm, e=/ 0.7, b/a The ratio b/a decreases slowl as e=c/a increases rapidl. The ratio b/a is the height-to-width ratio, which measures the shape of the ellipse when b/a is close to, the ellipse is nearl circular; when b/a is close to 0, the ellipse is elongated. The eccentricit ratio e=c/a measures how off-center the foci are when e is close to 0, the foci are near the center of the ellipse; when e is close to, the foci are far from the center and near the vertices of the ellipse. The foci must be etremel off-center for the ellipse to be significantl elongated.. Quick Review = + =. a b - - = a + + b +. + =, =-, = ;. + =00, =00-, = ; [ 0.,.] b [0,.] b a = a - c a = B - c a = - e [ 0.,.] b [0,.] - = ; - C 00 - = ; - C. +=(0- - 8) += = 0-8 = - 8 =-8 = =8. +=(+ + ) +=(+ + ++) += + += + ++=+ --8=0 (-)(+)=0 = 7. +=(- + ) += = + += -=0 -=0 =, = 8. +8=(8- + ) +8= = =( + ) = ( +) -8 +7=0 =, = ;. a - - =0, so = b 0. (+) 7-7=0, so = ; C Section 8. Eercises. h=0, k=0, a=, b= 7, so c= - 7= Vertices: (, 0), (, 0); Foci: (, 0), (, 0). h=0, k=0, a=, b=, so c= - = Vertices: (0, ), (0, ); Foci: (0, ), (0, ). h=0, k=0, a=, b=, so c= - 7= Vertices: (0, ), (0, ); Foci: (0, ), (0, ). h=0, k=0, a=, b= 7, so c= - 7= Vertices: (, 0), (-, 0); Foci: (, 0), (-, 0). h=0, k=0, a=, b=,so + =. c= - = Vertices: (, 0), (, 0); Foci: (, 0), (, 0). h=0, k=0, a=, b=, so + =. c= - =. Vertices: (0, ), (0, ); Foci: (0, ), (0, )

9 Section 8. Ellipses 7. (d) 8. (c). (a) 0. (b) [.,.] b [.,.] = ; [.7,.7] b [ 8., 8.] = ; - +. [.7,.7] b [.,.]. 0 + = ; - C [, 7] b [, ]. 8 = - ; =. + = 0. c= and a= =, so b= a - c = : + =

10 Chapter 8 Analtic Geometr in Two and Three Dimensions 0. c= and b= =, so a= b - c = = : + =.. + = + = 7. b=; 8. b=;. a=; + = + = + = 0. a=; + =. The center (h, k) is (, ) (the midpoint of the aes); a and b are half the lengths of the aes ( and, - - respectivel): + =. The center (h, k) is (, ) (the midpoint of the aes); a and b are half the lengths of the aes ( and, + - respectivel): + =. The center (h, k) is (, ) (the midpoint of the major ais); a=, half the lengths of the major ais. Since c= (half the distance between the foci), - + b= a - c = : + =. The center (h, k) is (, ) (the midpoint of the major ais); b=, half the lengths of the major ais. Since c= (half the distance between the foci), + - a= b - c = : + =. The center (h, k) is (, ) (the midpoint of the major ais); a and b are half the lengths of the aes ( and, respectivel): =. The center (h, k) is (, ) (the midpoint of the major ais); a and b are half the lengths of the aes ( and, + - respectivel): + = - h - k For #7 0, an ellipse with equation + = a b has center (h, k), vertices (h _a, k), and foci (h_c, k) where c = a - b. 7. Center (, ); Vertices ( _, )=(, ), (, ); Foci ( _, )=(, ), (, ) 8. Center (, ); Vertices: (_, ) (., ), ( 0., ); Foci=(_, )=(, ), (, ). Center (7, ); Vertices: (7, _)=(7, ), (7, ); Foci: (7, _ 7) (7,.), (7, 7.) 0. Center (, ); Vertices: (, _)=(, ), (, ); Foci: (, _)=(, ), (, ). [ 8, 8] b [, ] = cos t, = sin t. [, 0] b [, ] = 0 cos t, = sin t. [ 8, ] b [0, 0] = cos t-, = sin t+. [, 7] b [, ] = cos(t)+, = sin(t)- For # 8, complete the squares in and, then put in standard form. (The first one is done in detail; the others just show the final form.) =0 can be rewritten as ( -)+( +)=. This is equivalent to ( -+)+( ++)=++,or (-) +(+) =. Divide both sides b to - + obtain + =. Vertices: (, ) and (, ) Foci: (, ). Eccentricit: =. Vertices: ( ;, ). Foci: (, ). Eccentricit: = ; C =. Vertices: ( 7, ) and (, ). 7 Foci: ( ; 7, ). Eccentricit:

11 Section 8. Ellipses =. Vertices: (, 0) and (, ). Foci: (, 8 ; ). Eccentricit:. The center (h, k) is (, ) (given); a and b are half the lengths of the aes ( and, respectivel): = 0. The center (h, k) is (, ) (given); a and b are half the lengths of the aes ( and, respectivel): =. Consider Figure 8.(b); call the point (0, c) F, and the point (0, c) F. B the definition of an ellipse, an point P (located at (, )) satisfies the equation PF + PF = a thus, c c = + - c c = a then + - c =a- +(-c) =a -a + + c + + c + +(+c) -c+c =a -a + + c + +c+c a + + c =a +c a + + c =a +c a ( +(+c) )=a +a c+c a +(a -c ) =a (a -c ) a +b =a b b + a = c. Recall that e= means that c=ea, b= a - c and a a celestial object s perihelion occurs at a-c for Pluto, c=ea=(0.8)(00)., so its perhelion is 00-.=,. Gm. For Ntune, c=ea=(0.000)(7)., so its perihelion is 7-.=,7. Gm. As a result of its high b eccentric orbit, Pluto comes over 0 Gm closer to the Sun than Ntune.. Since the Moon is furthest from the Earth at,70 miles and closest at,, we know that a=,70+., or a=7,08.. Since c+,=a, we know c=,. and b= a - c = 7,08. -,.,7. c,. From these, we calculate e= = 0.0. a 7,08. The orbit of the Moon is ver close to a circle, but still takes the shape of an ellipse.. For Mercur, c=ea=(0.0)(7.).0 Gm and its perihelion a-c=7.-.0 Gm. Since the diameter of the sin is. Gm. Mercur gets within. -. Gm of the Sun s surface.. For Saturn, c=ea=(0.00)(,7) 7. Gm. Saturn s perihelion is a-c= Gm and its aphelion is a+c= Gm.. Venus: c=ea=(0.008)(08.) 0.7,so b= L 08..,707. +,70.70 = Mars: c=ea=(0.0)(7.)., so b= L.,8 +,8 = 7. For sungrazers, a-c.(.)=.088. The eccentricit of their ellipses is ver close to a=, b=, c= a - b = 7. Au C a.8 b - a. b 7. thus, e= 8.0 L 0.7. a=8 and b=., so c= a - b =.7. Foci at (.7, 0 L ;7., a= and b=, so c= a - b = Place the source and the patient at opposite foci inches from the center along the major ais.. Substitute =- into the first equation: = + - +(- )= =0 = =, =0 Solution: (, 0), (, 0). Substitute =- into the first equation: - + = -++ = -=0 (-)=0 =0 or = = =0 Solution: (, 0), (0, ). (a) [.7,.7] b [.,.] Approimate solutions: (.0, 0.8), (.7, 0.7) ; - (b) a, - + b, 8 a ; +, b

12 Chapter 8 Analtic Geometr in Two and Three Dimensions. One possibilit: a circle is perfectl centric : it is an ellipse with both foci at the center. As the foci move off the center and toward the vertices, the ellipse becomes more eccentric as measured b the ratio=e=c/a. In everda life, we sa a person is eccentric if he or she deviates from the norm or central tendencies of behavior.. False. The distance is a-c=a(-c/a)=a(-e).. True, because a =b +c in an ellipse =, so c= a - b = - =. The answer is C. 8. The focal ais runs horizontall through (, ). The answer is C.. Completing the square produces =. The answer is B. 70. The two foci are a distance c apart, and the sum of the distances from each of the foci to a point on the ellipse is a. The answer is C. 7. (a) When a=b=r, A=pab=prr=pr and r + rr + r P p(r) a - b r + r =pr a - r b =pr a - r r r b =pr (-)=pr. (b) One possibilit: with A=p and + = P - p.0, and with 00 + = A=0p and P - 0p (a) Answers will var. See Chapter III: The Harmon of Worlds in Cosmos b Carl Sagan, Random House, 80. (b) Drawings will var. Kler s Second Law states that as a planet moves in its orbit around the sun, the line segment from the sun to the planet swes out equal areas in equal times. 7. (a) Graphing in parametric mode with Tst = p. [.7,.7] b [.,.] (b) The equations (t)=+cos(t-) and (t)= sin(t-) can be rewritten as cos (t-)=- and sin(t-)= ->. Substituting these into the identit cos (t-)+sin (t-)= ields the equation > + - =. This is the equation of an ellipse with = as the focal ais. The center of the ellipse is (, 0) and the vertices are (, ) and (, ). The length of the major ais is and the length of the minor ais is. 7. (a) The equations (t)=+ sin a pt + p and b (t)= cos a pt + p can be rewritten as b - sin a pt + p = and cos a pt + p = b b p. Substituting these into the identit cos a pt + p + b sin a pt + p = ields the equation b - + p =. This is the equation of an ellipse. (b) The pendulum begins its swing at t=0 so (0)=+ sin a p =8 ft, which is the maimum b distance awa from the detector. When t=, ()=+ sin a p + p = ft, which is the b minimum distance from the detector. When t=, the pendulum is back to the 8-ft position. As indicated in the table, the maimum velocit (. ft/sec) happens when the pendulum is at the halfwa position of ft from the detector. 7. Write the equation in standard form b completing the squares and then dividing b the constant on the righthand side. D E A +D+ +C +E+ C = D A + E A C - F + D A + D + E A C + E C + C A = AC a D A + E C - F b a + D A b a + E C b CD + AE - ACF + = C A A C A C a * CD + AE - ACF b a + D A b C [ 8, 8] b [ 0, 0] + a + E C b = A

13 Section 8. Hperbolas A C a + D A b AC a + E CD + AE - ACF + C b CD + AE - ACF = Since AC 7 0, A Z 0 and C Z 0 (we are not dividing b zero). Further, AC 7 0 A C 7 0 and AC 7 0 (either A 7 0 and C 7 0, or A 0 and C 0), so the equation rresents an ellipse. 7. Rewrite the equation to a - h b + a - k b = 0 a b Since that a Z 0 and b Z 0 (otherwise the equation is not defined) we see that the onl values of, that satisf the equation are (, )=(h, k). In this case, the degenerate ellipse is simpl a single point (h, k). The semimajor and semiminor aes both equal 0. See Figure = [ 7.,.] b [.,.] = Section 8. Hperbolas Eploration. The equations = +/cos t= + sec t and =+ tan t can be rewritten as + sec t= and tan t = -. Substituting these into the identit sec t-tan t= ields the equation =.. [.,.] b [.,.] [.,.] b [.,.] In Connected graphing mode, pseudo-asmptotes appear because the grapher connects computed points b line segments regardless of whether this makes sense. Using Dot mode with a small Tst will produce the best graphs.. Eample : = >cost, = tant Eample : = tan(t), = /cos (t) Eample : =+/cos(t), = + tan(t) Eample : = +/cos(t), =+7 tan(t). - = [.,.] b [.,.] - =. Eample : The equations = >cos t = sec t, = tan t can be rewritten as sec t =, tan t =, which using the identit sec t-tan t= ield. - = Eample : The equations = tan t, = >cos t = sec t can be rewritten as tan t =, sec t =, which using sec t-tan t= ield. - = Eample : B rewriting =+/cos t, = + tan t as sec t = -, tan t = + and using sec t-tan t=, we obtain =. Eample : B rewriting = +/cos t, =+7 tan t as sec t = +, tan t = - and using sec t-tan t=, we obtain - =. Quick Review 8.. [ 0, 8] b [ 8, 8] = =. b - a + c - - = b - a + c +. - = =+ = ; +. - = = - = ; - [.,.] b [.,.] =; -

14 Chapter 8 Analtic Geometr in Two and Three Dimensions. + = = =0-8 = - 8 no solution. + = = = =+ +-= = 8 - ; = 8 = or =. When = -, = = - 7 C - C = - The onl solution is = =+ + += = + = + -=0 -=0 =, = 8. + = = = + + += =0 ={.,.} (the other solutions are etraneous) a. c=a+, (a+) -a =, a a +a+-a =, a=: a=, c= a 0. c=a+, (a+) -a =, a a +a+-a = : a=, c= Section 8. Eercises For #, recall the Pthagorean relation that c =a +b.. a=, b= 7, c= + 7 = ; Vertices: (, 0); Foci: (, 0). a=, b=, c= + = ; Vertices: (0, ); Foci: (0, ). a=, b=, c= + = 7; Vertices: (0, ); Foci: (0, 7). a=, b=, c= + = ; Vertices: (, 0); Foci: (, 0). ; a=, b=, c= 7; - = Vertices: (, 0); Foci: ( 7, 0). ; a=, b=, c= ; - = Vertices: (, 0); Foci: (, 0) 7. (c) 8. (b). (a) 0. (d). Transverse ais from ( 7, 0) to (7, 0); asmptotes: = ;, 7 = ; 7-0. Transverse ais from (0, 8) to (0, 8); asmptotes: = ; 8, = ; Transverse ais from (0, ) to (0, ); asmptotes: = ;, = ; + 0

15 Section 8. Hperbolas 7. Transverse ais from (, 0) to (, 0); asmptotes: = ;, = ; - 8. [ 8.8, 8.8] b [.,.] = ; The center (h, k) is (, ). Since a = and b =,we have a= and b=. The vertices are at (, ) or ( 7, ) and (, ). [.,.] b [.,.] = ; - 0. [.,.] b [.,.] = ; +. The center (h, k) is (, ). Since a = and b =,we have a= and b=. The vertices are at (, ).. [.,.] b [.,.] = ; [., 7.] b [.,.] [ 8.8, 8.8] b [.,.] = ; - = ; c= and a=, so b= c - a = :. c= and b=, so a= c - b = :. c= and b=, so a= c - b = 0 : - 0 = - = - =

16 8 Chapter 8 Analtic Geometr in Two and Three Dimensions. c= and a=/, so b= c - a = :. - = or.7 > - > = 7. a= and c=ea=0, so b= 00 - = : - 7 = 8. a= and c=ea=, so b= - = : - 0 =. b=, a= c - b = - = : - = c 0. c=, a=, b= c - a = - = : e = - 7 =. The center (h, k) is (, ) (the midpoint of the transverse ais endpoints); a=, half the length of the transverse ais. And b=, half the length of the conjugate ais =. The center (h, k) is (, ) (the midpoint of the transverse ais endpoints); a=, half the length of the transverse ais. And b=, half the length of the conjugate ais =. The center (h, k) is (, ) (the midpoint of the transverse ais); a=, half the length of the transverse ais. - - Since b/a =, b=: - =. The center (h, k) is a-,, the midpoint of the b transverse ais); a=, half the length of the transverse 7 ais. Since a/b =, b= : 8 - > + - 8> 7> =. The center (h, k) is (, ), the midpoint of the transverse ais. a=, half the length of the transverse ais. The center-to-focus distance is c=, so b= c - a + - = : - =. The center (h, k) is a-, -, the midpoint of the b 7 transverse ais. b=, half the length of the transverse ais. The center-to-focus distance is c=, so +. + a= c - b = 8 : - = > 8 7. The center (h, k) is (, ), the midpoint of the transverse ais. a=, half the length of the transverse ais. The center-to-focus distance c=ea =# =0, so b= c - a = 00 - = = 7 8. The center (h, k) is (, ), the midpoint of the transverse ais. c=, the center-to-focus distance c a= b= c - a = - = 7 e = =, = 7 For #, a hperbola with equation - h - k - = has center (h, k) vertices a b (h_a, k), and foci (h_c, k) where c = a + b. - k - h A hperbola with equation - = has a b center (h, k), vertices (h, k_a), and foci (h, k_c) where again c = a + b.. Center (, ); Vertices: ( _, )=(, ), (, ); Foci: ( _, )=(, ), (, ) 0. Center (, ); Vertices: ( _, ); Foci: ( _, )=(, ), (, ). Center (, ); Vertices: (, _8)=(, ), (, ); Foci: (, _ ). Center (, ); Vertices: (, _)=(, ), (, ); Foci: (, _)=(, ), (, 7).. [.,.] b [.,.] =/cos t, = tan t [.,.] b [.,.] = 0/cos t, = tan t. [.,.] b [ 0.,.] = + /cos t, =+ tan t

17 Section 8. Hperbolas. 0. [ 7.,.] b [ 7.,.] = + /cos t, =+ tan t 7. [.,.] b [., 7.] Divide the entire equation b. Vertices: (, ) and (, ), Foci: (, ; ), e=. 8. [.8,.8] b [ 7., 0] Vertices: a and a, Foci:, - b, - b > + > e= > = B + a ;, - b =. For # 0, complete the squares in and, then write the equation in standard form. (The first one is done in detail; the other shows just the final form.) As in the previous problems, the values of h, k, a, and b can be read from the equation - h - k ; < =. The asmptotes are a b -k= ; b (-h). If the term is positive, the transverse a ais endpoints are ( h ; a, k); otherwise the endpoints are (h, k ; b).. [.,.] b [., 7.] =. Vertices: (, ) and (, ). Foci: (,; ), e=. a=, (h, k)=(0, 0) and the hperbola opens to the left and right, so Using (, ): - - b =. b =, b -=b, b =, b = ; - =. a=, (h, k)=(0, 0) and the hperbola opens upward and downward, so Using (, ): - b =. =, b =; - - b =, = b. Consider Figure 8.(b). Label (0, c) as point F, label (0, c) as point F and consider an point P(, ) along the hperbola. B definition, PF -PF = a, with c 7 a c c = a + + c = ;a c + +c+c = a ; a + - c + + -c+c ;a + - c =a -c a ( + -c+c )=a -a c+c a +(c -a ) =a (c -a ) b -a =a b. (a) =0 - = = ; a - b = [.,.] b [.,.] =0 can be rewritten as ( -)-( -)=. This is equivalent to ( -+)-( -+)=+-,or (-) -(-) =. Divide both sides b to - - obtain - =. Vertices: (0, ) and (, ). Foci: (;,), e=

18 0 Chapter 8 Analtic Geometr in Two and Three Dimensions (b) =0 - = = ;. c-a=0, b =0a c -a =b (a+0) -a =0a a +0a+,00-a =0a 0a=,00 a=0 Gm 0 a=0 Gm, b=00 Gm, c=0, e= =. 0 The Sun is centered at focus (c, 0)=(0, 0).. c-a=0, b =0a c -a =b (a+0) -a =0a a +80a+,00-a =0a a=,00 a=.8 a=.8 Gm, b= Gm, c=.8 Gm, e=. 8 The Sun is centered at focus (c, 0)=(7, 0). 7. The Princess Ann is located at the intersection of two hperbolas: one with foci O and R, and the other with foci O and Q. For the first of these, the center is (0, 0), so the center-to-focus distance is c=0 mi. The transverse ais length is b=(.7 Âsec)(80 ft/âsec)=,80. ft 0 mi. Then a 0-0 = 700 mi. For the other hperbola, c=00 mi, a=(. Âsec) (80 ft/âsec)=. ft 0 mi, and b 00-0 = 80 mi. The two equations are therefore and = =. The intersection of the upper branch of the first hperbola and the right branch of the second hperbola (found graphicall) is approimatel (88.7, 0.8). The ship is located about 887 miles east and 0 miles north of point O a bearing and distance of about 0. and 7. miles, respectivel. 8. The gun is located at the intersection of two hperbolas: one with foci A and B, and the other with foci B and C. For the first of these, the center is (0, 000), so the centerto-focus distance is c=000 mi. The transverse ais length is b=( sec)(00 ft/sec)=00 ft. Then a = 007 ft. For the other hperbola, c=00 ft, a=( sec)(00 ft/sec)= 00 ft, and b = 007 ft. The two equations are therefore and 00,70,000 = ,0,000 = The intersection of the upper branch of the first hperbola and the right branch of the second hperbola (found graphicall) is approimatel (,7., 7.). The gun is located about,7 ft (. mi) east and 7 ft (.8 mi) north of point B a bearing and distance of about 0. and,8. ft (.8 mi), respectivel.. = - - = Solve the second equation for and substitute into the first equation. = - a - b - = a b - = - = 0 - = 0 =0 or = Solutions: (, 0), (, ) 0. Add: [.,.] b [.,.] = - + = =0 =8 = ; + = 8+ = = ; [.,.] b [.,.] There are four solutions: (, )

19 Section 8. Hperbolas. (a) [.,.] b [.,.] There are four solutions: (.,.8) (b) The eact solutions are a ;0 B, ;0 B b.. One possibilit: Escape speed is the minimum speed one object needs to achieve in order to break awa from the gravit of another object. For eample, for a NASA space probe to break awa from the Earth s gravit is must meet or eceed the escape speed for Earth E= GM>r L,00 m/s. If this escape speed is eceeded, the probe will follow a hperbolic path.. True. The distance is c-a=a(c/a-)=a(e-).. True. For an ellipse, b +c =a.., so c= + and the foci are each - = units awa horizontall from (0, 0). The answer is B.. The focal ais passes horizontall through the center, (, ). The answer is E. 7. Completing the square twice, and dividing to obtain on the right, turns the equation into =. The answer is B. 8. a=, b=, and the slopes are b/a. The answer is C.. (a d) (e) a=, c=, b=; /- /= 70. Assume that the focus for the primar parabolic mirror occurs at F P and the foci for the hperbolic mirror occur at F H and F H. Assume also that the -ais etends from the ee piece to the right most F H, and that the -ais is perpendicular through the -ais 0 cm from the ee piece. Then, the center (h, k) of the hperbolic mirror is (0, 0), the foci ( c, 0)=( 0, 0) and the vertices ( a, 0)=( 0, 0). Since a=0, c=0, b =c -a =000. The equation for the hperbolic mirror is =. 7. From Section 8., Question #7, we have A +C +D+E+F=0 becomes A C a + D A b AC a + F C b CD + AE - ACF + CD + AE - ACF = Since AC 0 means that either (A 0 and C 7 0) or (A 7 0 and C 0), either (A C 0 and AC 7 0), or (A C 7 0 and AC 0). In the equation above, that means that the + sign will become a ( ) sign once all the values A, B, C, D, E, and F are determined, which is eactl the equation of the hperbola. Note that if A 7 0 and C 0, the equation becomes: AC a + E C A C@a + D A b CD + AE - ACF - CD + AE - ACF = If A 0 and C 7 0, the equation becomes: A C a + D A + E C b CD + AE - ACF - CD + AE - ACF = 7. With a Z 0 and b Z 0, we have Then a - h a a - h a b = a - k b b b = -a - k b a - h b a or = a - k b. b b. Solving these two equations, we find that = ; b - h + k. The graph consists of a two intersecting slanted lines through (h, k). Its smmetr is like that of a hperbola. Figure 8. shows the relationship between an ordinar hperbola and two intersecting lines. 7. The asmptotes of the first hperbola are = ; b - h + k and the asmptotes of the second a hperbola are = ; b - h + k; the are the same. a [Note that in the second equation, the standard usage of a+b has been revised.] The conjugate ais for hperbola is b, which is the same as the transverse ais for hperbola. The conjugate ais for hperbola is a, which is the same as the transverse ais of hperbola.

20 Chapter 8 Analtic Geometr in Two and Three Dimensions c a - b 7. When =c, = c b -a =a b a =b (c -a ) b =c -a = = ; b a One possible answer: Draw the points and a c, -b on a cop of figure 8.(a). Clearl the points a b a c, ; b on the hperbola are the endpoints of a a b segment perpendicular to the -ais through the focus (c, 0). Since this is the definition of the focal width used in the construction of a parabola, appling it to the hperbola also makes sense. 7. The standard forms involved multiples of,,, and, as well as constants; therefore the can be rewritten in the general form A +C +D+E+F=0 (none of the standard forms we have seen require a B term). For eample, rewrite =a as a -=0; this is the general form with A=a and E=, and all others 0. - Similarl, the hperbola can be put in b a = standard form with A= - C=, F=, and a, b B=D=E=0. Section 8. Translation and Rotation of Aes Quick Review 8.. cos Å= 8. cos Å= 7. cos Å=. cos Å= p p. Å= so Å=,. Å=sin p a so Å= b = p, 7. cos Å= cos Å-=, cos 8 Å=, cos Å=, cos Å= b a a c, b a b 8. cos Å= cos Å-=, cos 7 Å=, cos 7 Å=, cos Å= = = # B 8 cos Å=. cos Å=- sin Å=, sin Å= - sin Å= sin Å= sin Å= B 0. cos Å=- sin Å=, sin 8 Å= sin 8 Å= 0 sin Å= Section 8. Eercises. Use the quadratic formula with a=, b=0, and c= -+8. Then b -ac=(0) -( - +8)= ++8=( ++7), and = -0 ; = - ; [.,.] b [.,.]. Use the quadratic formula with a=, b=, and c= ++. Then b -ac= ( ) -( ++)= --80= ( --), and = ; = ; [.,.] b [.,.]. Use the quadratic formula with a=, b= 8, and c= 8+8. Then b -ac= ( 8) -( 8+8)=+=(+), and 8 ; + = = ; + [.8, 7.8] b [ 8.,.]

21 Section 8. Translation and Rotation of Aes. Use the quadratic formula with a=, b= 0, and c= ++. Then b -ac= ( 0) -( )( ++)= ++0 =( ++), and = 0 ; = - ; + + [.7,.7] b [.,.] 0. Use the quadratic formula with a=, b=-0, and c= --0. Then b -ac=(-0) ( --0)= +0+0, and = ; [ 7., 7.] b [.8,.8]. +=0 = =/ [ 0, 0] b [ 8, 8] [.,.] b [.,.]. + = = /. Use the quadratic formula with a=8, b=-, and c= -0-. Then b -ac=(-) -( -0-) = =8( ++), and - ; = = - ; [.,.] b [.,.] =0 (-)=8 =8/(-) [ 0, ] b [, ] 8. -+=0 (-)= = /(-) [, 8] b [, ]. Use the quadratic formula with a=, b=-, and c= --. Then b -ac=(-) 8( --)=-8=(-), and - ; - = = - ; -. [,.] b [ 0., 0.8]. Use the quadratic formula with a=, b=-, and c= --. Then b -ac =(-) -( --)= +8+88, and = - ; [ 0, 0] b [, 0]. h=0, k=0 and the parabola opens downward, so p= ). Using (, ): p=, p=. The standard form is =.. h=0, k=0 and the parabola opens to the right, so p= (p>0). Using (, ): 8p=, p=. The standard form is =8.

22 Chapter 8 Analtic Geometr in Two and Three Dimensions. h=0, k=0 and the hperbola opens to the right and left, so a=, and b=. The standard form is. - =. h=0, k=0, and the -ais is the focal ais, so a= and b=. The standard form is. + = For #7 0, recall that =-h and =-k. 7. (, )=(, ) 8. (, )=(, ). (, )=(, - 0. (, )= - -, -. ( -)-( +)=, so (-) -(+) =+-=. Then =. This is a hperbola, with a=, b=, and c=. - =. ' 8 '. a - 7 =(-), a parabola. The verte is b (h, k)= a, 7 so =( ). b, 8. ( -)+( +)=, so (-) +(+) =++=. Then =. This is an ellipse, with a=, b=, and c=. Foci:, - ;. Center (, ), so + =. ' ' ' '. ( +)+( -8)= 0, so (+) +(- )= 0+8+8=. Then =. This is an ellipse with a= and c=. +, b =, =. ' '. ( -)-( +)=7, so (-) - (+) =7+-=. - + Then - =. This is a hperbola, with a=, b=8, and c= 8 = 7. Foci: Center (, ), so - ; 7, -. =. ' 0. -=(+), a parabola. The verte is (h, k)=(, ), so =( ). ' ' '

23 Section 8. Translation and Rotation of Aes 7. 8(-)=(-), a parabola. The verte is (h, k)=(, ), so 8 =( ). 8 ' 8. ( -)+( -)=, so (-) +(-) - - =++=0. Then =. This is an ellipse, with a= 0, b = 0 =, and c = 0. Foci: Center (, ), so +, ; = '. ( +)- =, so (+) - = + + =, Then - =. This is a hperbola, with a =, b =, and c =. Foci: Center (, 0), so - -, ;. = ' '. The horizontal distance from O to P is =h+ = +h, and the vertical distance from O to P is =k+ = +k.. Given = +h, subtract h from both sides: -h= or =-h. And given = +k, subtract k from both sides: -k= or =-k. For #, recall that = cos Å+ sin Å and = sin Å+ cos Å.. (, )= a - cos p + sin p, sin p + cos p b =. (, )= a cos p - sin p, - sin p - cos p b = a -, - - b = a -, - - b L (0.0,.70). Å.0, (, )=( cos (.0)- sin (.0), sin (.0)- cos (.0)) (.,.8) p. Å (, ), = a cos p + sin p, - sin p + cos p b = For #7 0, use the discriminant B -AC to determine the tpe of conic. Then use the relationship of cot a = A - C B to determine the angle of rotation. 7. B -AC=>0, hperbola; cot Å=0, so a = p. Translating, a, b - a, 7 b a - ba + b = 8, =, =; (-.)=(-), a parabola. The verte is (h, k)=(., ), so =( ) ' 8. B -AC=>0, hperbola; cot Å=0, so a = p. Translating, 0 [.,.] b [.,.] - 0 a - ba + b + = 0, =, =; + 0 ' [.,.] b [.,.]

24 Chapter 8 Analtic Geometr in Two and Three Dimensions. B -AC=-()()= <0, ellipse; cot a = Translating,, a = p. = cos p, so the - sin p, = sin p + cos p equation becomes + = 0, + = 0 [.,.] b [.,.] 0. B -AC=-()()=0, parabola; cot Å=, p p Å= Translating, = cos sin p,. - p = sin so ( ) =, =; + cos p,. This is a degenerate form consisting of onl two parallel lines.. B -AC= 7<0, ellipse. Use the quadratic formula with a=, b= 0, and c= -0. Then b -ac=( 0) -()( -0) = 7 +0=( +0), and = 0 ; = [.7,.7] b [.,.] 0 ; 0 - [.,.] b [.,.] cot Å= - 7, Å 0..º 0. B -AC=>0, hperbola. Use the quadratic formula with a=, b= +0, and c= --. Then b -ac=( +0) -()( --) = -+8=( -+7), and = - 0 ; ; = [ 8, ] b [, ] cot Å= -, Å 0..º. B -AC=-()(0)= <0; ellipse. B -AC=-()(0)=>0; hperbola. B -AC=-()()=0; parabola. B -AC=-(0)()=>0; hperbola 7. B -AC=-(8)()= 8<0; ellipse 8. B -AC=-()()=>0; hperbola. B -AC=0-()( )=>0; hperbola 0. B -AC=-()()= <0; ellipse. B -AC=-()()= <0; ellipse. B -AC=-()()= 00<0; ellipse. In the new coordinate sstem, the center (, )= (0, 0), the vertices occur at (, 0) and the foci are located at ;, 0. We use = cos p - sin p, p p = sin + cos to translate back. Under the old coordinate sstem, the center (, )=(0, 0), the vertices occured at a and a -., -, b b and the foci are located at (, ) and (, ).. (a) Reversing the translation and rotation of the parabola, we see that the verte in the (, ) coordinate sstem is V(0, 0), with h= and, k = This means that the verte of the 0. parabola in the (, ) coordinate sstem is ( +h, +k)= a 0 + Since, b. cos Å= and sin Å=, rotating back into the (, ) coordinate sstem gives (, )=( cos Å- sin Å, sin Å+ cos Å) a # # + # - # 0, 0 =(., 8.7). (b) See (a).. Answers will var. One possible answer: Using the geometric relationships illustrated, it is clear that = cos Å- cos a p - a b = cos a - sin a and that = cos a p - a b + cos a = sin Å+ cos Å.

25 Section 8. Translation and Rotation of Aes 7. = cos Å+ sin Å = sin Å+ cos Å cos Å= cos Å+ sin Å cos Å sin Å= sin Å+ cos Å sin Å cos Å- sin Å= cos Å+ sin Å cos Å+ sin Å- sin Å cos Å cos Å- sin Å= cos Å+ sin Å cos Å- sin Å= Similarl, =cos Å+ sin Å = sin Å+ cos Å sin Å= cos Å sin Å+ sin Å cos Å= sin Å cos Å+ cos Å sin Å+ cos Å= cos Å sin Å+ sin Å - sin Å cos Å+ cos Å sin Å+ cos Å=(sin Å+cos Å) sin Å+ cos Å= 7. True. The B term is missing and so the rotation angle Å is zero. 8. True. Because the and terms have the same coefficient (namel ), completing the square to put the equation in standard form will produce the same denominator under (-k) as under (-h).. Eliminating the cross-product term requires rotation, not translation. The answer is B. 0. Moving the center or verte to the origin is done through translation, not rotation. The answer is C.. Completing the square twice, and dividing to obtain on the right, turns the equation into = The vertices lie units to the left and right of center (, ). The answer is A.. The equation is equivalent to =/. The answer is E.. (a) The rotated aes pass through the old origin with slopes of, so the equations are =_. (b) The location of, = 0, 0 in the sstem can be found b reversing the transformations. In the sstem,, = 0, 0 has coordinates h, k = a.the coordinates of this point, 0 b in the sstem are then given b the second set of rotation formulas; with cos Å =, sin Å = : = a b - 0 a b = 8 = a b + 0 a b = 87 0 The aes pass through the point (, ) > = a 8 with slopes of, 87 0 b > = and its negative reciprocal, -. Using this information z to write linear equations in point-slope form, and then converting to slope-interct form, we obtain. (a) If the translation on =-h and =-k is applied to the equation, we have: A( ) +B +C( ) +D +E +F=0, so A(-h) +B(-h)(-k)+C(-k) +D(-h)+E(-k)+F=0, which becomes A +B+C +(D-Bk-Ah)+ (E-ck-Bh)+(Ah +Ck -Ek-Dh) +Bhk+F=0 The discriminants are eactl the same; the coefficients of the,, and terms do not change (no sign change). (b) If the equation is multiplied b some constant k, we have ka +kb+kc +kd+ke+kf=0, so the discriminant of the new equation becomes (kb) -(ka)(kc)=k B -k AC=k (B - AC). Since k >0 for k Z 0, no sign change occurs.. First, consider the linear terms: D+E=D( cos Å- sin Å) +E( sin Å+ cos Å) =(D cos Å+E sin Å) +(E cos Å-D sin Å) This shows that D+E=D +E, where D =D cos Å+E sin Å and E =E cos Å-D sin Å. Now, consider the quadratic terms: A +B+C =A( cos Å- sin Å) + B( cos Å- sin Å)( sin Å+ cos Å)+ C( sin Å+ cos Å) =A( cos Å- cos Å sin Å+ sin Å) +B( cos Å sin Å+ cos Å- sin Å - sin Å cos Å)+C( sin Å+ sin Å cos Å + cos Å) =(A cos Å+B cos Å sin Å+C sin Å) +[B(cos Å-sin Å) +(C-A)(sin Å cos Å)] +(C cos Å-B cos Å sin Å+A sin Å) =(A cos Å+B cos Å sin Å+C sin Å) +[B cos Å+(C-A) sin Å] +(C cos Å-B cos Å sin Å+A sin Å) This shows that A +B+C =A +B +C, where A =A cos Å+B cos Å sin Å+C sin Å, B =B cos Å+(C-A) sin Å, and C =C cos Å- B cos Å sin Å+A sin Å. The results above impl that if the formulas for A, B, C, D, and F are applied, then A +B +C +D +E +F =0 is equivalent to A +B+C +D+E+F=0. Therefore, the formulas are correct.. This equation is simpl a special case of the equation we have used throughout the chapter, where B=0. The discriminant B -AC, then, reduces simpl to AC. If AC>0, we have a hperbola; AC=0, we have a parabola; AC<0, we have a ellipse. More simpl: a hperbola if AC<0; a parabola if AC=0; an ellipse if AC>0. = + = - +

26 8 Chapter 8 Analtic Geometr in Two and Three Dimensions 7. Making the substitutions = cos Å- sin Å and = sin Å+ cos Å, we find that: B =(B cos Å-B sin Å+C sin Å cos Å -A sin Å cos Å) A =(A cos Å+B sin Å cos Å+C sin Å)( ) C =(A sin Å+C cos Å-B cos Å sin Å)( ) B -A C =(B cos (Å)-(A-C)sin (Å)) -(A cos Å+B sin Å cos Å+C sin Å) (A sin Å-B sin Å cos Å+C cos Å) = B cos (Å)+ B +BC sin (Å)-BA sin (Å) + C - C cos (Å)-CA+CA cos (Å) + A - A cos (Å)- a A cos (Å)+ A + B sin (Å)+ C- C cos (Å) a A - A cos Å + C cos Å + C - sin Å B = B cos (Å)+ B +BC sin (Å)-BA sin (Å) + C - C cos (Å)-CA+CA cos (Å)+ A - A cos (Å)-BC sin (Å)+BA sin (Å)-AC - C - A + A cos (Å)+ B - B cos (Å) + C cos (Å)-AC cos (Å) =B -AC. 8. When the rotation is made to the (, ) coordinate sstem, the coefficients A, B, C, D, E, and F become: A B A = (+cos (Å)+ sin (Å) C + (-cos (Å)) B =B cos (Å)-(A-C) sin (Å) A B C = (-cos (Å))- sin (Å) C + (cos (Å)+) D =D cos Å+E sin Å E = D sin Å+E cos Å F =F (a) Since F =F, F is invariant under rotation. A (b) Since A +C = [+cos (Å)+-cos (Å)] B C + [sin (Å)-sin (Å)]+ [-cos (Å) +cos (Å)+]=A+C, A+C is invariant under rotation. (c) Since D +E =(D cos Å+E sin Å) +( D sin Å+E cos Å) =D cos Å+DE cos Å sin Å+E sin Å +D sin Å-DE cos Å sin Å+E cos Å =D (cos Å+sin Å)+E (sin Å+cos Å) =D +E, D +E is invariant under rotation.. Intersecting lines: + = 0 can be rewritten as = 0 (the -ais) and = - [.7,.7] b [.,.] A plane containing the ais of a cone intersects the cone. Parallel lines: = can be rewritten as = ; (a pair of vertical lines) [.7,.7] b [.,.] A degenerate cone is created b a generator that is parallel to the ais, producing a clinder. A plane parallel to a generator of the clinder intersects the clinder and its interior. One line: = 0 can be rewritten as = 0 (the -ais). [.7,.7] b [.,.] A plane containing a generator of a cone intersects the cone. No graph: = - [.7,.7] b [.,.] A plane parallel to a generator of a clinder fails to intersect the clinder. Circle: + = [.7,.7] b [.,.] A plane perpendicular to the ais of a cone intersects the cone but not its verte.

27 Section 8. Polar Equations of Conics Point: + = 0, the point (0, 0). Section 8. Eercises. r= a parabola. - cos [.7,.7] b [.,.] A plane perpendicular to the ais of a cone intersects the verte of the cone. No graph: + = - [ 0, 0] b [ 0, 0] 0. r= a hperbola. + a = b cos + cos [.7,.7] b [.,.] A degenerate cone is created b a generator that is perpendicular to the ais, producing a plane. A second plane perpendicular to the ais of this degenerate cone fails to intersect it. Section 8. Polar Equations of Conics Eploration For e=0.7 and e=0.8, an ellipse; for e=, a parabola; for e=. and e=, a hperbola. [ 0, 0] b [ 0, 0]. r= an ellipse. + a = b sin + sin [ 7., 7.] b [ 7, ] [, ] b [, ] The five graphs all have a common focus, the pole (0, 0), and a common directri, the line =. As the eccentricit e increases, the graphs move awa from the focus and toward the directri. Quick Review 8.. r=. r= 7p. = or - p p. = - p or. h=0, k=0, p=, so p= The focus is (0, ) and the directri is =.. h=0, k=0, p=, so p= The focus is (, 0) and the directri is =. 7. a=, b=, c= ; Foci: ;, 0; Vertices: (, 0) 8. a=, b=, c=; Foci: (0, ); Vertices: (0, ). a=, b=, c=; Foci: (, 0); Vertices: (, 0) 0. a=, b=, c= ; Foci: 0, ;; Vertices: (0, ). r= a parabola. + sin [, ] b [, ] 7 7. r= a hperbola. - a 7 = b sin - 7 sin [, ] b [, ]

28 0 Chapter 8 Analtic Geometr in Two and Three Dimensions 0 0. r= an ellipse. - a = b cos - cos [, ] b [, ] 7. Parabola with e= and directri =. 8. Hperbola with e= and directri =.. Divide numerator and denominator b. Parabola with e= and directri = - = Divide numerator and denominator b. Ellipse with e= and directri =. = 0.. Divide numerator and denominator b. Ellipse with e= and directri =.. Divide numerator and denominator b. 7 Hperbola with e= and directri =. =.. Divide numerator and denominator b. Ellipse with e= and directri =. = 0.. Divide numerator and denominator b. Hperbola with e= and directri =. =.. (b) [, ] b [ 0, 0]. (d) [, ] b [, ] 7. (f) [, ] b [, ] 8. (e) [, ] b [, ]. (c) [ 0, 0] b [, 0] 0. (a) [, ] b [, ] For # 8, one must solve two equations a= and + e b= for e and p (given two constants a and b). The - e b - a ab general solution to this is e= and p=. b + a b - a. The directri must be =p 7 0, since the right majorais endpoint is closer to (0, 0) than the left one, so the equation has the form r=. Then + e cos.= and = + e cos 0 = + e + e cos p = (so a=. and b=). Therefore e= =0. - e. and p=, so r= + > cos = + cos.. The directri must be = p 0, since the left majorais endpoint is closer to (0, 0) than the right one, so the equation has the form r=. Then - e cos.= and = - e cos 0 = - e - e cos p = (so a= and b=.). Therefore e= =0. + e and p= (the directri is = ),so. r= - > cos = - cos.. The directri must be =p 7 0, since the upper majorais endpoint is closer to (0, 0) than the lower one, so the equation has the form r=. Then + e cos = = and = + e sinp> + e + e sinp> = (so a= and b=). Therefore e= =0. - e. and p=, so r= + > sin = + sin.. The directri must be = p 0, since the lower majorais endpoint is closer to (0, 0) than the upper one, so the equation has the form r=. Then - e cos = = and = - e sin p> - e - e sinp> = (so a= and b=). Therefore e= =0. + e and p= (the directri is = ),so. r= - > sin = - sin.. The directri must be =p 7 0, since both transverseais endpoints have positive coordinates, so the equation has the form r=. Then + e cos = and = + e cos 0 = + e + e cos p = (so a= and b= ). Therefore e= - e 7. =. and p=, so r= + > cos = + cos.. The directri must be = p 0, since both transverseais endpoints have negative coordinates, so the equation has the form r=. Then - e cos = and.= - e cos 0 = - e - e cos p = (so a=. and b= ). Therefore e= + e and p= (the directri is = ), so r= - cos.

29 Section 8. Polar Equations of Conics 7. The directri must be =p 7 0, since both transverseais endpoints have positive coordinates, so the equation has the form r=. Then. + e cos = = and = + e sinp> + e + e sinp> = (so a=. and b= ). Therefore e= - e =. and p=, so r= + > sin = + sin. 8. The directri must be = p 0, since both transverseais endpoints have negative coordinates, so the equation has the form r=. Then - e cos = = and = - e sinp> - e - e sinp> = (so a= and b= ). Therefore e= + e and p= (the directri is = ), so r= - sin.. The directri must be =p 7 0, so the equation has the form r=. Then 0.7= + e cos 0 = + e cos + e and = = (so a=0.7 and b=). + e cos p - e Therefore e= =0. and p=, so. r= + > cos = + cos. 0. Since this is a parabola, e=, and with =p 7 0 as the p directri, the equation has the form r= Then + sin. p p = = p=, and therefore + sinp> +, r=. Alternativel, for a parabola, the distance + sin from the focus to the verte is the same as the distance from the verte to the directri (the same is true for all points on the parabola). This distance is unit, so we again conclude that the directri is =.. r=, so e=0.. The vertices - cos = cos are (7, 0) and (, p), so a=0, a=, c=ae =(0.)()=, so b= a - c = - =. are a, p and a, p, so a=, a=. b b c=ae= # =, so b= a - c = - =. [, 0] b [, ] e=, a=, b=, c=. r=, so e=. The vertices + sin = + > sin p p are (, ) and (, ), so a=, a=8. c=ae = # 8 =, so b= a - c = - =. [, ] b [, ] e=, a=8, b=, c=. r=, so e=. The + cos = > + > cos vertices are (, 0) and (8, p), so a=0, a=, c=ae = a =, so b= a - c = - =. b [ 0, ] b [, ] e=0., a=, b=, c=. r=, so e=. The + cos = > + > cos vertices are (, 0) and ( 8, p), so a=, a=, c=ae = # = and b= c - a = - =. [, ] b [ 7, ] e=0., a=, b=, c=. r=, so e=. The vertices - sin = > - > sin [, ] b [, ] e=, a=, b=, c=