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Conic Sections in Polar Coordinates MATH 211, Calculus II J. Robert Buchanan Department of Mathematics Spring 2016 Introduction We have developed the familiar formulas for the parabola, ellipse, and hyperbola from their definitions in rectangular coordinates. a(x − b)2 + c = y (x − x0 )2 (y − y0 )2 + = 1 a2 b2 (x − x0 )2 (y − y0 )2 − = 1 a2 b2 (parabola) (ellipse) (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 fixed point P (called the focus) in the plane and a fixed line (called the directrix) not containing P. Let a curve be defined as the set of all points in the plane whose distance from the focus is a constant multiple e (called 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. q x 2 + y 2 = e(d − x) Polar Coordinate Formulation p 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. q x 2 + y 2 = e(d − x) r r = e(d − r cos θ) = ed 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 indicated directrix has the polar equation ed 1. r = , if the directrix is the line x = d > 0, 1 + e cos θ ed 2. r = , if the directrix is the line x = d < 0, −1 + e cos θ ed 3. r = , if the directrix is the line y = d > 0, 1 + e sin θ ed , if the directrix is the line y = d < 0. 4. r = −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 4 e=1 e=2 2 e=12 -6 -4 -2 2 -2 -4 4 6 x Solution 1. e = 1/2: 2. e = 1: 3. e = 2: Solution 1. e = 1/2: 1 ed 2 2 (−2) r= = = 1 −1 + e cos θ 2 − cos θ −1 + 2 cos θ 2. e = 1: 3. e = 2: Solution 1. e = 1/2: 1 ed 2 2 (−2) r= = = 1 −1 + e cos θ 2 − cos θ −1 + 2 cos θ 2. e = 1: r= 3. e = 2: (1)(−2) 2 = −1 + (1) cos θ 1 − cos θ Solution 1. e = 1/2: 1 ed 2 2 (−2) r= = = 1 −1 + e cos θ 2 − cos θ −1 + 2 cos θ 2. e = 1: r= 3. e = 2: r= (1)(−2) 2 = −1 + (1) cos θ 1 − cos θ (2)(−2) 4 = −1 + (2) cos θ 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 -1 -2 2 3 4 x Solution 1. d = x = 4: 2. d = y = 1: 3. d = y = −2: Solution 1. d = x = 4: r= 2. d = y = 1: 3. d = y = −2: 1 ed 4 2 (4) = = 1 1 + e cos θ 2 + cos θ 1 + 2 cos θ Solution 1. d = x = 4: r= 1 ed 4 2 (4) = = 1 1 + e cos θ 2 + cos θ 1 + 2 cos θ 2. d = y = 1: r= 3. d = y = −2: 1 ed 1 2 (1) = = 1 1 + e sin θ 2 + sin θ 1 + 2 sin θ Solution 1. d = x = 4: 1 ed 4 2 (4) = = 1 1 + e cos θ 2 + cos θ 1 + 2 cos θ r= 2. d = y = 1: r= 1 ed 1 2 (1) = = 1 1 + e sin θ 2 + sin θ 1 + 2 sin θ 3. d = y = −2: r= 1 ed 2 2 (−2) = = 1 −1 + e sin θ 2 − sin θ −1 + 2 sin θ Parametric Equations for Conic Sections (1 of 2) Example Find parametric equations for the conic section with equation x2 y2 + = 1. 4 9 Parametric Equations for Conic Sections (1 of 2) Example Find parametric equations for the conic section with equation x2 y2 + = 1. 4 9 Use the fundamental trigonometric identity: cos2 t + sin2 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 x2 y2 + = 1. 4 9 Use the fundamental trigonometric identity: cos2 t + sin2 t = 1. x = 2 cos t y = 3 sin t We can see that x2 y2 (2 cos t)2 (3 sin t)2 + = + = 1. 4 9 4 9 Parametric Equations for Conic Sections (2 of 2) Example Find parametric equations for the conic section with equation (x − 2)2 (y + 2)2 − = 1. 9 25 Parametric Equations for Conic Sections (2 of 2) Example Find parametric equations for the conic section with equation (x − 2)2 (y + 2)2 − = 1. 9 25 Use the hyperbolic trigonometric identity: cosh2 t − sinh2 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 (y + 2)2 − = 1. 9 25 Use the hyperbolic trigonometric identity: cosh2 t − sinh2 t = 1. x = 2 + 3 cosh t y = −2 + 5 sinh t We can see that (2 + 3 cosh t − 2)2 (−2 + 5 sinh t + 2)2 (x − 2)2 (y + 2)2 − = − = 1. 9 25 9 25 Kepler’s 2nd Law of Planetary Motion Based on astronomical observations Johannes Kepler hypothesized 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 speed up when they are close to the sun and slow down when they are further away. 0.5 0 y -0.5 -1 -1.5 -2 -1 -0.5 0 x 0.5 1 Calculation of Area and Arc Length 2 . 2 + sin θ 2 Z 1 π 2 A[0,π] = dθ ≈ 0.9455994 2 0 2 + sin θ 2 Z 1 5.224895 2 A[3π/2,5.224895] = dθ ≈ 0.9455995 2 3π/2 2 + sin θ s 2 Z π dr 2 L[0,π] = r + dθ ≈ 2.53 dθ 0 s 2 Z 5.224895 dr L[3π/2,5.224895] = r2 + dθ ≈ 1.02 dθ 3π/2 Suppose r = Homework I Read Section 9.7 I Exercises 1–27 odd.

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