Explanation Results used: Polar equations of conic sections : Let d > 0. The conic section with a focus at the origin and eccentricity e has the polar equation r = e d 1 + e cos θ or r = e d 1 − e cos θ . The conic section with focus at the origin and eccentricity e has the polar equation r = e d 1 + e sin θ or r = e d 1 − e sin θ . If 0 < e <1, the conic section is an ellipse; if e = 1, it is a ellipse; and if e > 1 it is a hyperbola. The curves are defined over any interval in θ of length 2 π . Calculation: The given equation of the curve is r = 1 2 − cos θ . Rewrite the equation as r = 1 2 1 − 1 2 cos θ . Compare with r = e d 1 − e cos θ and obtain that e d = 1 2 and e = 1 2 . Therefore, obtain that d = 1 . As the term cos θ is in the equation the directrix are vertical. As e = 1 2 the curve is an ellipse. As the equation has negative sign, one of the the directrix is x = − 1 . As the directrix is vertical the vertex lies on the line which is parallel to the x -axis. So, to calculate the vertex set θ = 0 and θ = π in r = 1 2 1 − 1 2 cos θ . r = 1 2 1 − 1 2 cos 0 r = 1 2 1 − 1 2 r = 1 2 1 2 r = 1 Similarly, r = 1 2 1 − 1 2 cos π r = 1 2 1 + 1 2 r = 1 2 3 2 r = 1 3 Therefore, the vertices are ( r 1 , θ 1 ) = ( 1 , 0 ) and ( r 2 , θ 2 ) = ( 1 3 , π ) In Cartesian coordinates the coordinates of vertices are ( r 1 cos θ 1 , r 1 sin θ 1 ) = ( 1 , 0 ) and ( r 2 cos θ 2 , r 2 sin θ 2 ) = ( − 1 3 , 0 ) . One of the foci is at origin and other focus will be at the distance which is same as the distance between one focus and the nearest vertex

Calculus: Early Transcendentals (3rd Edition)

3rd Edition

ISBN: 9780134763644

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

See similar textbooks

Question

Chapter 12.4, Problem 59E

To determine

To graph: The curve r=12−cosθ and mention the vertices, foci, directrix and asymptotes ( if exists ) on the graph; verify the answer by using graphing utility.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 12.4, Problem 58E

Chapter 12.4, Problem 60E

Students have asked these similar questions

Q1/Details of square footing are as follows: DL = 800 KN, LL = 500 kN, Fy=414 MPa, Fc = 20 MPa Footing, qa = 120 kPa, Column (400x400) mm. Determine the dimensions of footing and thickness? Q2/ For the footing system shown in Figure below, find the suitable size (BxL) for: 1. Non uniform pressure, 2. Uniform pressure, 3.Uniform pressure with moment in clockwise direction. (Use qmax=qall =200kPa). Property, line M=200KN.m 1m P-1000KN

Q2/ Determine the size of square footing to carry net allowable load of 400 kN. FS-3. Use Terzaghi equation assuming general shear failure. 400KN 1 m += 35" C=0.0 Ya = 18.15 kN/m³ +=25" C=50 kN/m² Ya 20 kN/m³

4 x+3 and g(x)=x2-9 4X-10 2X --13) The domain of rational expression A) 1R. {-2,-8} AB -14) Let f(x) = B) 1R. {2,-4,-8} 4X-12 x² +6x-16 X3+7X²+12X ? C) 1R \ {-4,-3,0} then f(x) + g(x) is equal ro D) IR 2 A) B) c) D) x²-9 x2-9 x²-9 x+4 DB 5x-4 A B If + then the value of B is equal to X+1 A) 4 B) 2 C) 5 D) 3 4X 4x+4 С.В.... x2+5X+6 x2 (x-2)(x+1) X-2 AC 16 The solution set of the equation A){4} B) {-3} C){ 1} 17 The solution set of the equation A) (-3,-2) B) [-3,0) C)[-3,-2] D). [-2,0) BA -18) Which one of the following is proper fraction? 2x+4 ≤0 入×1 x+2x+4 (x+1)(x+2) 2x+4x+2 = 4 X+1 is equal to D). {-5} ≤0 A) x6 +4 2x+12 2X x +4 B) c) x2-9 AL 2x+12 D) x+4 14) let g(x) = [x-3],then g(-2) is equal to A) -5 B)-6 C)-3 D) 3 Part III work out (show every step cleary) (2pt) 20. E9) Find the solution set of the equation 2x+4 x+1 ≤0 P(x) (a) P(x) =≤0 2x+4 50 x+1 x+1≤ 2x+4 (x-1)(x-2) x= 1 or x=2 solution is {1.2} x-1=0 of x-2=0 x = 1 or = 2