Conic Sections in Polar CoordinatesMATH 211, Calculus II

J. Robert Buchanan

Department of Mathematics

Spring 2018

Introduction

We have developed the familiar formulas for the parabola,ellipse, and hyperbola from their definitions in rectangularcoordinates.

a(x − b)2 + c = y (parabola)(x − x0)

2

a2 +(y − y0)

2

b2 = 1 (ellipse)

(x − x0)2

a2 − (y − y0)2

b2 = 1 (hyperbola)

Today we will see that polar coordinates unify and simplify thedefinitions of the conic sections.

Definition of the Conic Sections (1 of 2)Consider a fixed point P (called the focus) in the plane and afixed line (called the directrix) not containing P. Let a curve bedefined as the set of all points in the plane whose distance fromthe focus is a constant multiple e (called the eccentricity) oftheir distance to the directrix.

Pdirectrix

d

e*d

Definition of the Conic Sections (2 of 2)

TheoremThe set of all points whose distance to the focus is the productof 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.√

x2 + y2 = e(d − x)

Definition of the Conic Sections (2 of 2)

TheoremThe set of all points whose distance to the focus is the productof 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.√

x2 + y2 = e(d − x)

Polar Coordinate Formulation

Making use of the identities: r =√

x2 + y2 and x = r cos θ wecan develop a single equation for all three conic sections.√

x2 + y2 = e(d − x)

r = e(d − r cos θ)

r =ed

1 + e cos θ

0 < e < 1 : ellipsee = 1 : parabolae > 1 : hyperbola

Different Orientations of the Directrix

TheoremThe conic section with eccentricity e > 0, focus at (0,0) and theindicated directrix has the polar equation

1. r =ed

1 + e cos θ, if the directrix is the line x = d > 0,

2. r =ed

−1 + e cos θ, if the directrix is the line x = d < 0,

3. r =ed

1 + e sin θ, if the directrix is the line y = d > 0,

4. r =ed

−1 + e sin θ, if the directrix is the line y = d < 0.

Examples

Find polar equations for the conic sections with focus (0,0) anddirectrix d = x = −2 and eccentricities

1. e = 1/2,2. e = 1,3. e = 2.

-6 -4 -2 2 4 6x

-4

-2

2

4

y

e=1�2

e=1

e=2

Solution

1. e = 1/2:

r =ed

−1 + e cos θ=

12(−2)

−1 + 12 cos θ

=2

2 − cos θ

2. e = 1:

r =(1)(−2)

−1 + (1) cos θ=

21 − cos θ

3. e = 2:

r =(2)(−2)

−1 + (2) cos θ=

41 − 2 cos θ

Solution

1. e = 1/2:

r =ed

−1 + e cos θ=

12(−2)

−1 + 12 cos θ

=2

2 − cos θ

2. e = 1:

r =(1)(−2)

−1 + (1) cos θ=

21 − cos θ

3. e = 2:

r =(2)(−2)

−1 + (2) cos θ=

41 − 2 cos θ

Solution

1. e = 1/2:

r =ed

−1 + e cos θ=

12(−2)

−1 + 12 cos θ

=2

2 − cos θ

2. e = 1:r =

(1)(−2)−1 + (1) cos θ

=2

1 − cos θ

3. e = 2:

r =(2)(−2)

−1 + (2) cos θ=

41 − 2 cos θ

Solution

1. e = 1/2:

r =ed

−1 + e cos θ=

12(−2)

−1 + 12 cos θ

=2

2 − cos θ

2. e = 1:r =

(1)(−2)−1 + (1) cos θ

=2

1 − cos θ

3. e = 2:r =

(2)(−2)−1 + (2) cos θ

=4

1 − 2 cos θ

Examples

Find polar equations for the conic sections with focus (0,0) andeccentricity e = 1/2 and directrices

1. d = x = 4,2. d = y = 1,3. d = y = −2.

-4 -3 -2 -1 1 2 3 4x

-2

-1

1

2

y

Solution

1. d = x = 4:

r =ed

1 + e cos θ=

12(4)

1 + 12 cos θ

=4

2 + cos θ

2. d = y = 1:

r =ed

1 + e sin θ=

12(1)

1 + 12 sin θ

=1

2 + sin θ

3. d = y = −2:

r =ed

−1 + e sin θ=

12(−2)

−1 + 12 sin θ

=2

2 − sin θ

Solution

1. d = x = 4:

r =ed

1 + e cos θ=

12(4)

1 + 12 cos θ

=4

2 + cos θ

2. d = y = 1:

r =ed

1 + e sin θ=

12(1)

1 + 12 sin θ

=1

2 + sin θ

3. d = y = −2:

r =ed

−1 + e sin θ=

12(−2)

−1 + 12 sin θ

=2

2 − sin θ

Solution

1. d = x = 4:

r =ed

1 + e cos θ=

12(4)

1 + 12 cos θ

=4

2 + cos θ

2. d = y = 1:

r =ed

1 + e sin θ=

12(1)

1 + 12 sin θ

=1

2 + sin θ

3. d = y = −2:

r =ed

−1 + e sin θ=

12(−2)

−1 + 12 sin θ

=2

2 − sin θ

Solution

1. d = x = 4:

r =ed

1 + e cos θ=

12(4)

1 + 12 cos θ

=4

2 + cos θ

2. d = y = 1:

r =ed

1 + e sin θ=

12(1)

1 + 12 sin θ

=1

2 + sin θ

3. d = y = −2:

r =ed

−1 + e sin θ=

12(−2)

−1 + 12 sin θ

=2

2 − sin θ

Parametric Equations for Conic Sections (1 of 2)

ExampleFind parametric equations for the conic section with equation

x2

4+

y2

9= 1.

Use the fundamental trigonometric identity: cos2 t + sin2 t = 1.

x = 2 cos ty = 3 sin t

We can see that

x2

4+

y2

9=

(2 cos t)2

4+

(3 sin t)2

9= 1.

Parametric Equations for Conic Sections (1 of 2)

ExampleFind parametric equations for the conic section with equation

x2

4+

y2

9= 1.

Use the fundamental trigonometric identity: cos2 t + sin2 t = 1.

x = 2 cos ty = 3 sin t

We can see that

x2

4+

y2

9=

(2 cos t)2

4+

(3 sin t)2

9= 1.

Parametric Equations for Conic Sections (1 of 2)

ExampleFind parametric equations for the conic section with equation

x2

4+

y2

9= 1.

Use the fundamental trigonometric identity: cos2 t + sin2 t = 1.

x = 2 cos ty = 3 sin t

We can see that

x2

4+

y2

9=

(2 cos t)2

4+

(3 sin t)2

9= 1.

Parametric Equations for Conic Sections (2 of 2)

ExampleFind parametric equations for the conic section with equation

(x − 2)2

9− (y + 2)2

25= 1.

Use the hyperbolic trigonometric identity: cosh2 t − sinh2 t = 1.

x = 2 + 3 cosh ty = −2 + 5 sinh t

We can see that

(x − 2)2

9−(y + 2)2

25=

(2 + 3 cosh t − 2)2

9−(−2 + 5 sinh t + 2)2

25= 1.

Parametric Equations for Conic Sections (2 of 2)

ExampleFind parametric equations for the conic section with equation

(x − 2)2

9− (y + 2)2

25= 1.

Use the hyperbolic trigonometric identity: cosh2 t − sinh2 t = 1.

x = 2 + 3 cosh ty = −2 + 5 sinh t

We can see that

(x − 2)2

9−(y + 2)2

25=

(2 + 3 cosh t − 2)2

9−(−2 + 5 sinh t + 2)2

25= 1.

Parametric Equations for Conic Sections (2 of 2)

ExampleFind parametric equations for the conic section with equation

(x − 2)2

9− (y + 2)2

25= 1.

Use the hyperbolic trigonometric identity: cosh2 t − sinh2 t = 1.

x = 2 + 3 cosh ty = −2 + 5 sinh t

We can see that

(x − 2)2

9−(y + 2)2

25=

(2 + 3 cosh t − 2)2

9−(−2 + 5 sinh t + 2)2

25= 1.

Kepler’s 2nd Law of Planetary MotionBased on astronomical observations Johannes Keplerhypothesized that the planets move in elliptical orbits with thesun at one focus. His 2nd law of planetary motion states thatthe orbits sweep out equal areas in equal times. This impliesthat the planets speed up when they are close to the sun andslow down when they are further away.

-1 -0.5 0 0.5 1

x

-2

-1.5

-1

-0.5

0

0.5

y

Calculation of Area and Arc Length

Suppose r =2

2 + sin θ.

A[0,π] =12

∫ π

0

(2

2 + sin θ

)2

dθ ≈ 0.9455994

A[3π/2,5.224895] =12

∫ 5.224895

3π/2

(2

2 + sin θ

)2

dθ ≈ 0.9455995

L[0,π] =

∫ π

0

√r2 +

(drdθ

)2

dθ ≈ 2.53

L[3π/2,5.224895] =

∫ 5.224895

3π/2

√r2 +

(drdθ

)2

dθ ≈ 1.02

Homework

I Read Section 9.7I Exercises 1–27 odd.