lecture 11 (conics in polar coordinates)

7/31/2019 Lecture 11 (Conics in Polar Coordinates)

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Conic Sections in Polar Coordinates

Institute of Mathematics, University of the Philippines Diliman

Mathematics 54 (Elementary Analysis 2)

Polar Curves 1/ 1


2/101

Focus and Directrix of a Parabola

Recall that for a parabola, we have

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3/101



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We define the eccentricity to be e= 1.

Polar Curves 2/ 1


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Foci and Directrices of an Ellipse

For an ellipse, we have

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We define the eccentricity to be e= ca.

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We define the eccentricity to be e= ca.The directrices are of distances ae from the center.

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14/101

Foci and Directrices of a Hyperbola

For a hyperbola, we have

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d f b l


19/101



We define the eccentricity to be e= ca.

Polar Curves 4/ 1

F i d Di i f H b l


20/101



We define the eccentricity to be e= ca.The directrices are of distances ae from the center.

Polar Curves 4/ 1

C i i P l C di t


21/101

Conics in Polar Coordinates

Focus Directrix.

Let F be a fixed point (called focus) and let D be a fixed line, not containingF,

(called directrix). Let e be a positive number (called eccentricity). The set of all

points P in the plane such that

|P F

||P D| = e

Polar Curves 5/ 1

C i i P l C di t


22/101


Focus Directrix.




|P F

||P D| = e:=c

a

is a conic section.

Polar Curves 5/ 1



23/101


Focus Directrix.




|P F

||P D| = e:=c

a

is a conic section.

Moreover, ife< 1, its an ellipse

Polar Curves 5/ 1



24/101


Focus Directrix.




|P F

||P D| = e:=c

a

is a conic section.


e= 1, its a parabola

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25/101


Focus Directrix.




|P F

||P D| = e:=c

a

is a conic section.


e= 1, its a parabolae> 1, its a hyperbola

Polar Curves 5/ 1

Conics in Polar


26/101

Conics in Polar

Consider the following conic section C, with eccentricitye:

Polar Curves 6/ 1

Conics in Polar


27/101

Conics in Polar


one focus F at the pole

Polar Curves 6/ 1

Conics in Polar


28/101

Conics in Polar



corresponding directrixD: x= d

Polar Curves 6/ 1

Conics in Polar


29/101

Conics in Polar



corresponding directrixD: x= dIfP(r,) C, thenP F

=

Polar Curves 6/ 1

Conics in Polar


30/101

C




=r,

and P D=

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Conics in Polar


31/101




=r,

and P D=dF B

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Conics in Polar


32/101




=r,

and P D=dF B= d rcos

Polar Curves 6/ 1

Conics in Polar


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=r,

and P D=dF B= d rcosRecall that

|P F||P D| = e

Polar Curves 6/ 1

Conics in Polar


34/101




=r,


|P F||P D| = e

Hence, rdrcos = e

Polar Curves 6/ 1

Conics in Polar


35/101




=r,


|P F||P D| = e

Hence, rdrcos = e

Solving for r,

r=ed

1+ ecos

Polar Curves 6/ 1

Conics in Polar


36/101

Example.

A hyperbola has one focus at the pole, and corresponding directrix at distance 2

units to the right of the pole. Ife= 32 , find the polar equation of this conic.

Polar Curves 7/ 1

Conics in Polar


37/101

Example.



Solution:

Note: e= 32 and d=

Polar Curves 7/ 1

Conics in Polar


38/101

Example.



Solution:

Note: e= 32 and d= 2 we have

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Conics in Polar


39/101

Example.



Solution:


r = ed1+ecos

Polar Curves 7/ 1

Conics in Polar


40/101

Example.



Solution:


r = ed1+ecos

=

32

(2)

1+

32

cos

Polar Curves 7/ 1

Conics in Polar


41/101

Example.



Solution:


r = ed1+ecos

=

32

(2)

1+

32

cos

= 31+ 32 cos

Polar Curves 7/ 1

Conics in Polar


42/101

Example.



Solution:


r = ed1+ecos

=

32

(2)

1+

32

cos

= 31+ 32 cos

= 62+3cos

Polar Curves 7/ 1

Conics in Polar


43/101

Example.



Solution:


r = ed1+ecos

=

32

(2)

1+

32

cos

=3

1+ 32 cos

= 62+3cos

Polar Curves 7/ 1

Conics in Polar


44/101

Example.



Solution:


r = ed1+ecos

=

32

(2)

1+

32

cos

=3

1+ 32 cos

= 62+3cos some points on the hyperbola

65 , 0

,

3, 2

, (6,),

3, 32 Polar Curves 7/ 1

Conics in Polar


45/101

Example.



Solution:


r = ed1+ecos

=

32

(2)

1+

32

cos

=3

1+ 32 cos

= 62+3cos some points on the hyperbola

65 , 0

,

3, 2

, (6,),

3, 32 Polar Curves 7/ 1

Conics in Polar (General)


46/101

Polar Curves 8/ 1


d


47/101

r= ed1+ecos

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ed


48/101

r= ed1+ecos

(directrix to the right of the pole)

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ed


49/101

r= ed1+ecos


r= ed1ecos

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ed


50/101

r= ed1+ecos


r= ed1ecos

(directrix to the left of the pole)

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ed ed


51/101

r= ed1+ecos


r= ed1ecos


r= ed1+ esin

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ed ed


52/101

r= ed1+ecos


r= ed1ecos


r= ed1+ esin

(directrix above the pole)

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ed ed


53/101

r= ed1+ecos


r= ed1ecos


r= ed1+ esin


r= ed1 esin

Polar Curves 8/ 1


ed ed


54/101

r=1+ecos


r= ed1ecos


r=1+ esin


r= ed1 esin

(directrix below the pole)

Polar Curves 8/ 1

Example


55/101

Identify the following conic section:

13

12sin

22

1+0.5cos

3 51+sin

Polar Curves 9/ 1

Example


56/101


13

12sin

22

1+0.5cos

3 51+sin

Answer:

1 e= 2

Polar Curves 9/ 1

Example


57/101


13

12sin

22

1+0.5cos

3 51+sin

Answer:

1 e= 2 Hyperbola

Polar Curves 9/ 1

Example


58/101


13

12sin

22

1+0.5cos

3 51+sin

Answer:

1 e= 2 Hyperbola

2 e= 0.5

Polar Curves 9/ 1

Example


59/101


13

12sin

22

1+0.5cos

3 51+sin

Answer:

1 e= 2 Hyperbola

2 e= 0.5 Ellipse

Polar Curves 9/ 1

Example


60/101


13

12sin

22

1+0.5cos

3 51+sin

Answer:

1 e= 2 Hyperbola

2 e= 0.5 Ellipse3 e= 1

Polar Curves 9/ 1

Example


61/101


13

12sin

22

1+0.5cos

3 51+sin

Answer:

1 e= 2 Hyperbola

2 e= 0.5 Ellipse3 e= 1 Parabola

Polar Curves 9/ 1

Conics in Polar


62/101

Example.

Identify the graph of the following

r= 12sin .

Polar Curves 10/ 1

Conics in Polar


63/101

Example.


r= 12sin .

Solution:Rewrite the equation above as

Polar Curves 10/ 1

Conics in Polar


64/101

Example.


r= 12sin .


r=

12

(1)

1

12

sin

Polar Curves 10/ 1

Conics in Polar


65/101

Example.


r= 12sin .


r=

12

(1)

1

12

sin= e= 1

2,

Polar Curves 10/ 1

Conics in Polar


66/101

Example.Identify the graph of the following

r= 12sin .

Solution:

Rewrite the equation above as

r=

12

(1)

1

12

sin= e= 1

2, d= 1

Polar Curves 10/ 1

Conics in Polar


67/101

Example.Identify the graph of the following

r= 12sin .

Solution:

Rewrite the equation above as

r=

12

(1)

1

12

sin= e= 1

2, d= 1

Hence, it is an ellipse.

Polar Curves 10/ 1

Conics in Polar


68/101

Example.

Find the polar equation of the following conic.

2x22y2+4x+1= 0

Polar Curves 11/ 1

Conics in Polar


69/101

Example.


2x22y2+4x+1= 0

Solution:

Completing the square, we have

Polar Curves 11/ 1

Conics in Polar

l


70/101

Example.


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =

1

Polar Curves 11/ 1

Conics in Polar

E l


71/101

Example.


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =

1=

c=

2

2

2

+

2

2

2

=1

Polar Curves 11/ 1

Conics in Polar

E l


72/101

Example.


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =

1=

c=

2

2

2

+

2

2

2

=1

Thus, the center is at (1,0)

Polar Curves 11/ 1

Conics in Polar

Example


73/101

Example.


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =

1=

c=

2

2

2

+

2

2

2

=1

Thus, the center is at (1,0) = focus at (0,0).

Polar Curves 11/ 1

Conics in Polar

Example


74/101

Example.


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =

1

=c

=

2

2

2

+

2

2

2

=1

Thus, the center is at (1,0) = focus at (0,0). Thus, e= ca

Polar Curves 11/ 1

Conics in Polar

Example


75/101

Example.


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =

1

=c

=

2

2

2

+

2

2

2

=1

Thus, the center is at (1,0) = focus at (0,0). Thus, e= ca= 122

=

2.

Polar Curves 11/ 1

Conics in Polar

Example


76/101

Example.


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =

1

=c

=

2

2

2

+

2

2

2

=1


=

2.

The corresponding directrix is

Polar Curves 11/ 1

Conics in Polar

Example


77/101

Example.


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =

1

=c

=

2

2

2

+

2

2

2

=1


=

2.

The corresponding directrix is ae to the right of the center.

Polar Curves 11/ 1

Conics in Polar

Example.


78/101

Example.


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =

1

=c

=

2

2

2

+

2

2

2

=1


=

2.

The corresponding directrix is ae to the right of the center. Hence

d=

Polar Curves 11/ 1

Conics in Polar

Example.


79/101

p


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =

1

=c

=

2

2

2

+

2

2

2

=1


=

2.


d=

a

e 1

Polar Curves 11/ 1

Conics in Polar

Example.


80/101

p


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =

1

=c

=

2

2

2

+

2

2

2

=1


=

2.


d=

a

e 1=

2

2 12 1

Polar Curves 11/ 1

Conics in Polar

Example.


81/101

p


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =

1

=c

=

2

2

2

+

2

2

2

=1


=

2.


d=

a

e 1=

2

2 12 1=

1

2

.

Polar Curves 11/ 1

Conics in Polar

Example.


82/101


2x22y2+4x+1= 0

Solution:


(x+1)222

2

y222

2 =1

=c

=

2

2

2

+

2

2

2

=1


=

2.


d=

a

e 1=

2

2 12 1=

1

2

. Thus, we have

Polar Curves 11/ 1

Conics in Polar

Example.


83/101


2x22y2+4x+1= 0

Solution:


(x+1)22

22

y2

222 = 1 = c=

2

2

2

+

2

2

2

=1


=

2.


d=

a

e 1=

2

2 12 1=

1

2

. Thus, we have

r=

2

12

1

2cos

Polar Curves 11/ 1

Conics in Polar

Example.


84/101


2x22y2+4x+1= 0

Solution:


(x+1)22

22

y2

222 = 1 = c=

2

2

2

+

2

2

2

=1


=

2.


d=

a

e 1=

2

2 12 1=

1

2

. Thus, we have

r=

2

12

1

2cos

= 122cos

Polar Curves 11/ 1

Conics in Polar

Example.

Set-up the integral that will give the area of the region common to both


85/101

Set up the integral that will give the area of the region common to both

C1 : r

=6

2

sin

and C2 : r

=3

1

+sin

Polar Curves 12/ 1

Conics in Polar

Example.



86/101


C1 : r

=6

2

sin

and C2 : r

=3

1

+sin

Solution:

Note that C1 is

Polar Curves 12/ 1

Conics in Polar

Example.



87/101


C1 : r= 62

sinand C2 : r= 31

+sin

Solution:

Note that C1 is an ellipse, and C2 is

Polar Curves 12/ 1

Conics in Polar

Example.



88/101

p g g g

C1 : r= 62

sinand C2 : r= 31

+sin

Solution:

Note that C1 is an ellipse, and C2 is a

parabola.

Polar Curves 12/ 1

Conics in Polar

Example.



89/101

p g g g

C1 : r= 62

sinand C2 : r= 31

+sin

Solution:


parabola.

Polar Curves 12/ 1

Conics in Polar

Example.



90/101

C1 : r= 62

sinand C2 : r= 31

+sin

Solution:


parabola.

Solving for the point of

intersection, we get

Polar Curves 12/ 1

Conics in Polar

Example.

Set-up the integral that will give the area of the region common to both6 3


91/101

C1 : r= 62

sinand C2 : r= 31

+sin

Solution:


parabola.



6

2 sin =3

1+ sin

Polar Curves 12/ 1

Conics in Polar

Example.

Set-up the integral that will give the area of the region common to both6 3


92/101

C1 : r= 62

sinand C2 : r= 31

+sin

Solution:


parabola.



6

2 sin =3

1+ sin 9sin = 0

Polar Curves 12/ 1


93/101

Conics in Polar

Example.

Set-up the integral that will give the area of the region common to both6 d 3


94/101

C1 : r= 62

sinand C2 : r= 31

+sin

Solution:


parabola.



6

2 sin =3

1+ sin 9sin = 0 = 0,

Thus,

0

1

2

3

1+sin

2d

Polar Curves 12/ 1

Conics in Polar

Example.


C 6 d C 3


95/101

C1 : r= 62

sinand C2 : r= 31

+sin

Solution:


parabola.



6

2 sin =3

1+ sin 9sin = 0 = 0,

Thus,

0

1

2

3

1+sin

2d+

2

1

2

6

2sin

2d

Polar Curves 12/ 1

Conics in Polar

Example.

Set-up the integral that will give the perimeter of the ellipse r= 42sin .


96/101

Polar Curves 13/ 1

Conics in Polar

Example.



97/101

Solution:

Polar Curves 13/ 1

Conics in Polar

Example.



98/101

Solution:

L

Polar Curves 13/ 1

Conics in Polar

Example.



99/101

Solution:

L = 20

r2+ d r

d

2d

Polar Curves 13/ 1

Conics in Polar

Example.



100/101

Solution:

L = 20

r2+ d r

d

2d

=2

0

4

2 sin

2+ 4cos

(2 sin)22

d

Polar Curves 13/ 1

Exercises

1 Find the eccentricity and identify the ff. conics

15

1cos


101/101

2

1

1+3sin

32

2+cos

44

35cos

5 132sin

2 Write a polar equation for the conic with a focus at the origin and satisifes theff. condition

1 Hyperbola, eccentricity= 74

, directrix: y= 62 Parabola, directrix: x

=3

3 Ellipse, eccentricity= 12

, directrix: x= 54 Parabola, directrix: y= 25 Ellipse, eccentricity= 3

4, directrix: y=4

6 Hyperbola, eccentricity= 3, directrix: x= 9

Polar Curves 14/ 1

lecture 11 (conics in polar coordinates)

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