Conic Sections in Polar Coordinates MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2018
Introduction We have develop the familiar formulas for the parabola, ellipse, and hyperbola from their definitions in rectangular coordinates. a(x b) 2 + c = y (parabola) (x x 0 ) 2 a 2 + (y y 0) 2 b 2 = 1 (ellipse) (x x 0 ) 2 a 2 (y y 0) 2 b 2 = 1 (hyperbola) Today we will see that polar coordinates unify and simplify the definitions of the conic sections.
Definition of the Conic Sections (1 of 2) Consider a fix point P (call the focus) in the plane and a fix line (call the directrix) not containing P. Let a curve be defin as the set of all points in the plane whose distance from the focus is a constant multiple e (call the eccentricity) of their distance to the directrix. d e d directrix P
Definition of the Conic Sections (2 of 2) Theorem The set of all points whose distance to the focus is the product of the eccentricity e > 0 and the distance to the directrix is 1. an ellipse if 0 < e < 1, 2. a parabola if e = 1, 3. a hyperbola if e > 1.
Definition of the Conic Sections (2 of 2) Theorem The set of all points whose distance to the focus is the product of the eccentricity e > 0 and the distance to the directrix is 1. an ellipse if 0 < e < 1, 2. a parabola if e = 1, 3. a hyperbola if e > 1. Proof. Assume the focus is at (0, 0) and the directrix is x = d > 0. x 2 + y 2 = e(d x)
Polar Coordinate Formulation Making use of the identities: r = x 2 + y 2 and x = r cos θ we can develop a single equation for all three conic sections. x 2 + y 2 = e(d x) r = e(d r cos θ) r = 1 + e cos θ 0 < e < 1 : ellipse e = 1 : parabola e > 1 : hyperbola
Different Orientations of the Directrix Theorem The conic section with eccentricity e > 0, focus at (0, 0) and the indicat directrix has the polar equation 1. r =, if the directrix is the line x = d > 0, 1 + e cos θ 2. r =, if the directrix is the line x = d < 0, 1 + e cos θ 3. r =, if the directrix is the line y = d > 0, 1 + e sin θ 4. r =, if the directrix is the line y = d < 0. 1 + e sin θ
Examples Find polar equations for the conic sections with focus (0, 0) and directrix d = x = 2 and eccentricities 1. e = 1/2, 2. e = 1, 3. e = 2. y e 2 4 2 e 1 e 1 2-6 -4-2 2 4 6 x -2-4
Solution 1. e = 1/2: 2. e = 1: 3. e = 2:
Solution 1. e = 1/2: r = 1 1 + e cos θ = 2 ( 2) 1 + 1 2 cos θ = 2 2 cos θ 2. e = 1: 3. e = 2:
Solution 1. e = 1/2: r = 1 1 + e cos θ = 2 ( 2) 1 + 1 2 cos θ = 2 2 cos θ 2. e = 1: r = (1)( 2) 1 + (1) cos θ = 2 1 cos θ 3. e = 2:
Solution 1. e = 1/2: r = 1 1 + e cos θ = 2 ( 2) 1 + 1 2 cos θ = 2 2 cos θ 2. e = 1: 3. e = 2: r = r = (1)( 2) 1 + (1) cos θ = 2 1 cos θ (2)( 2) 1 + (2) cos θ = 4 1 2 cos θ
Examples Find polar equations for the conic sections with focus (0, 0) and eccentricity e = 1/2 and directrices 1. d = x = 4, 2. d = y = 1, 3. d = y = 2. y 2 1-4 -3-2 -1 1 2 3 4 x -1-2
Solution 1. d = x = 4: 2. d = y = 1: 3. d = y = 2:
Solution 1. d = x = 4: r = 1 1 + e cos θ = 2 (4) 1 + 1 2 cos θ = 4 2 + cos θ 2. d = y = 1: 3. d = y = 2:
Solution 1. d = x = 4: r = 1 1 + e cos θ = 2 (4) 1 + 1 2 cos θ = 4 2 + cos θ 2. d = y = 1: r = 1 1 + e sin θ = 2 (1) 1 + 1 2 sin θ = 1 2 + sin θ 3. d = y = 2:
Solution 1. d = x = 4: r = 1 1 + e cos θ = 2 (4) 1 + 1 2 cos θ = 4 2 + cos θ 2. d = y = 1: r = 1 1 + e sin θ = 2 (1) 1 + 1 2 sin θ = 1 2 + sin θ 3. d = y = 2: r = 1 1 + e sin θ = 2 ( 2) 1 + 1 2 sin θ = 2 2 sin θ
Parametric Equations for Conic Sections (1 of 2) Example Find parametric equations for the conic section with equation x 2 4 + y 2 9 = 1.
Parametric Equations for Conic Sections (1 of 2) Example Find parametric equations for the conic section with equation x 2 4 + y 2 9 = 1. Use the fundamental trigonometric identity: cos 2 t + sin 2 t = 1. x = 2 cos t y = 3 sin t
Parametric Equations for Conic Sections (1 of 2) Example Find parametric equations for the conic section with equation x 2 4 + y 2 9 = 1. Use the fundamental trigonometric identity: cos 2 t + sin 2 t = 1. x = 2 cos t y = 3 sin t We can see that x 2 4 + y 2 9 = (2 cos t)2 4 + (3 sin t)2 9 = 1.
Parametric Equations for Conic Sections (2 of 2) Example Find parametric equations for the conic section with equation (x 2) 2 9 (y + 2)2 25 = 1.
Parametric Equations for Conic Sections (2 of 2) Example Find parametric equations for the conic section with equation (x 2) 2 9 (y + 2)2 25 = 1. Use the hyperbolic trigonometric identity: cosh 2 t sinh 2 t = 1. x = 2 + 3 cosh t y = 2 + 5 sinh t
Parametric Equations for Conic Sections (2 of 2) Example Find parametric equations for the conic section with equation (x 2) 2 9 (y + 2)2 25 = 1. Use the hyperbolic trigonometric identity: cosh 2 t sinh 2 t = 1. x = 2 + 3 cosh t y = 2 + 5 sinh t We can see that (x 2) 2 9 (y + 2)2 25 = (2 + 3 cosh t 2)2 ( 2 + 5 sinh t + 2)2 9 25 = 1.
Kepler s 2nd Law of Planetary Motion Bas on astronomical observations Johannes Kepler hypothesiz that the planets move in elliptical orbits with the sun at one focus. His 2nd law of planetary motion states that the orbits sweep out equal areas in equal times. This implies that the planets spe up when they are close to the sun and slow down when they are further away. 0.5 0-0.5 y -1-1.5-2 -1-0.5 0 0.5 1 x
Calculation of Area and Arc Length Suppose r = 2 2 + sin θ. A [0,π] = 1 2 A [3π/2,5.224895] = 1 2 L [0,π] = L [3π/2,5.224895] = π ( ) 2 2 dθ 0.9455994 2 + sin θ ( ) 2 2 dθ 0.9455995 3π/2 2 + sin θ ( ) dr 2 r 2 + dθ 2.53 dθ ( ) dr 2 r 2 + dθ 1.02 dθ π 0 5.224895 0 5.224895 3π/2
Homework Read Section 9.7 Exercises 1 27 odd.