290
Section 10.6 – Polar Equations of Conics
Objective #1:
Graphing Polar Equations of Conics.
Let's begin by reviewing how to graph a polar equation by looking at an
example.
Graph the following:
2
Ex. 1
r=
1+ cos(θ)
Solution:
First, we apply the tests for symmetry:
Polar Axis (replace θ by – θ)
r=
r=
2
1+ cos(− θ)
2
1+ cos(θ)
(cosine is an even function)
This is an equivalent equation so the graph is symmetric with
respect to the polar axis.
Line θ = π/2 (replace θ by π – θ)
r=
2
1+ cos(π −θ)
r=
2
1+ [cos(!)cos(θ) + sin(!)sin(θ)]
2
1− cos(θ)
r=
(cosine of a difference of angles)
This test fails.
Pole (replace r by – r)
2
–r=
r=
1+ cos(θ)
−2
1+ cos(θ)
This test fails.
Since the function is symmetric with respect the Polar Axis, we need
to just plot the function for values of θ between 0 and π inclusively
and then reflect it across the polar axis.
291
2
1+ cos(θ)
2
2
r=
=
=1
1+ cos(θ)
1+1
2
2
=
≈
π
1+0.886025404
1+ cos( )
r=
θ
0
r=
π
6
1.072
6
r=
π
4
2
1+ cos(
2π
3
r=
3π
4
r=
5π
6
r=
1+
)
1+ cos(
π
2
)
2
1+ cos(
3π
cos(
4
2
1+
2
1+0.5
=
2
2
5π
cos(
6
r=
π
π
3
1+ cos(
r=
π
2
)
2
r=
π
3
π
4
2
1+0.707106781
=
)
=
)
≈ 1.333
=
2
1+0
=
2
1−0.5
=2
=4
2
≈
1− 0.707106781
=
)
2π
3
≈ 1.172
2
1−0.886025404
2
1+ cos(π)
=
2
1−1
6.828
≈ 14.928
= undefined
Now, we will plot the points and then draw a smooth curve through
the points:
π
2
3π
4
3π
4
π
4
€
€
π
2
π
4
€
€
π
€
€
0π
1 2 3 4 5
5π
4
7π
4
1 2 3 4 5
5π
4
7π
4
3π
2
€
3π
2
€
€
0
€
€
€
292
Finally, use symmetry to get the rest of the graph:
π
2
3π
4
π
4
€
€
€
π
0
1 2 3 4 5
5π
4
7π
4
3π
2
€Notice the graph is a parabola.
€ In fact, we can convert the equation into
rectangular form to find all the appropriate information.
€
Convert the following equation into rectangular form:
2
Ex. 2
r=
1+ cos(θ)
Solution:
2
r=
1+ cos(θ)
(multiply both sides by [1 + cos(θ)])
r[1 + cos(θ)] = 2 (distribute)
r + rcos(θ) = 2
x 2 + y2 + x = 2
(replace r by
x 2 + y2 and rcos(θ) by x)
(subtract x from both sides)
(square both sides)
x 2 + y2 = 2 – x
2
2
2
x + y = (2 – x)
(simplify)
2
2
2
x + y = 4 – 4x + x
(subtract x2 from both sides)
y2 = 4 – 4x
(factor out a – 4)
2
y = – 4(x – 1)
This is an equation of a parabola that opens to the left. The vertex is
(1, 0), the axis of symmetry runs along the x-axis, and since a = 1,
293
the focus is (0, 0), and the directrix is x = 2. All of this matches the
graph we obtained in example #1.
We will use the ideas in example #1 and #2 to give an alternate definition of
conics that will tie the parabola, ellipse, and hyperbola together.
Definition of Conics
Let D be a fixed line called the directrix, let F be a fixed point not on line D
called the focus, and let e be a fixed positive number called the
eccentricity. A Conic is collection of all points P such that the distance
from the point P to the focus F divided by the distance from the point P to
the closest point on the directix D is equal to the eccentricity e:
d(F,P)
=e
d(D,P)
If e = 1, then the conic is a parabola.
If e < 1, then the conic is an ellipse. (Note: e = c/a for an ellipse)
If e > 1, then the conic is a hyperbola. (Note: e = c/a for a hyperbola)
Notice that when e = 1, if we multiply both sides d(D, P), the definition
becomes the exact same definition for the parabola as in section 10.2.
When e < 1, the conic is an ellipse with the major axis passing through the
focus F perpendicular to the directrix D. When e > 1, the conic is a
hyperbola with the transverse axis passing through the focus F
perpendicular to the directrix D.
Now, let's derive the equations
d(D, P)
for the conics in polar form when
P = (r, θ)
a focus is at the pole. Suppose
the directrix D is perpendicular to
the polar axis and is p units to the
d(F, P)
rcos(θ)
left of the pole. Let P = (r, θ) be
any point on the graph of the
conic (see figure to the right).
Then we can establish two
p
relationships:
1)
d(F, P) = r
2)
d(D, P) = p + rcos(θ)
The operation in the second one
is plus since rcos(θ) is negative.
Keep in mind that there are two
different p's in the problem; capital P refers to the point P = (r, θ) and
italicize p refers to the shortest distance between the directrix D and the
294
focus F. Going back to the definition of conics, we can replace d(F, P) by r
and d(D, P) by p + rcos(θ)
d(F,P)
=e
d(D,P)
r
=
p+r cos(θ)
(replace d(F, P) by r and d(D, P) by p + rcos(θ))
e
(multiply both sides by p + rcos(θ))
r = e(p + rcos(θ))
r = ep + recos(θ)
r – recos(θ) = ep
r(1 – ecos(θ)) = ep
€
r=
(distribute)
(subtract recos(θ) from both sides)
(factor out r)
(divide both sides by 1 – ecos(θ))
ep
1−ecos(θ)
Even though we have drawn the case for the parabola in the figure
previously, the relationships do hold for the ellipse and hyperbola if the
focus is at the pole and the directrix is to left of the pole perpendicular to
€ polar axis. For the other three cases, the derivation of the formula is
the
very similar to the one presented here.
Polar Equation of a Conic Theorem
Let F be the focus at the pole, let D be the directrix a distance p from the
pole, and let e be the eccentricity. The polar equation of a conic will be in
one of the following forms:
ep
1)
r=
D
F
1+ecos(θ)
D is p units to the right of F and
perpendicular to the polar axis.
€
2)
r=
p
ep
1−ecos(θ)
D F
D is p units to the left of F and
perpendicular to the polar axis.
€
3)
r=
ep
1+esin(θ)
D is p units above F and
parallel to the polar axis.
€
4)
r=
D
p
F
ep
1−esin(θ)
D is p units below F and
parallel to the polar axis.
€
p
F
p
D
295
Analyze and graph the following:
12
Ex. 3
r=
4+8sin(θ)
Solution:
The constant term on the bottom has to be one, so we need to divide
top and bottom by 4:
€
12
3
ep
4
r = 4 + 8 sin(θ
=
The equation matches case 3: r =
)
1+2sin(θ)
1+esin(θ)
4
Since the coefficient of the sine function is e = 2, the conic is a
hyperbola. Also, the numerator is 3 which means ep = 3. Substituting
e = 2,
€ we get 2p = 3 or p = 1.5. Thus, the hyperbola
€ has a focus at
the pole, the directrix is 1.5 units above the pole parallel to the polar
π
axis and the transverse axis runs along the θ =
line. Since the
€
2
vertices are on the transverse axis, we can find them by evaluating
π
3π
the equation for θ =
and θ =
:
2
r=
12
4+8sin(
π
2
)
=
12
4+8
2
€r=
=1
π
2
Thus, the vertices
are €
(1,
€
) and (– 3,
3π
2
12
3π
4+8sin(
2
) = (3,
π
2
)
=
12
4−8
=–3
). The center is
π
the midpoint
between the two vertices, which is €
(2, ) (or (0, 2) in
€
2
€
rectangular form). This means that c, the distance between the center
c
€
and the focus, is€2. Using e = € and plugging
in e = 2 and c = 2, we
€
a
get a = 1. Recall that for a hyperbola, b2 = €
c2 – a2 = 22 – 12, which
implies b = 3 . Thus, in
rectangular form, the€equation is:
(y −2)2
12
€
or
€
–
(x−0)2
( 3 )2
(y – 2)2 –
=1
x2
3
=1
and the oblique asymptotes are:
€
a
y – k = ± (x – h)
b
or
y€– 2 = ±
1
3
x
Now, we can sketch the graph.
€
€
296
Ex. 4
8
2− sin(θ)
r=
Solution:
The constant term on the bottom has to be one, so we need to divide
top and bottom by 2:
€
8
4
ep
2
r = 2 − sin(θ
=
The
equation
matches
case
4:
r
=
)
1−0.5sin(θ)
1−esin(θ)
2
€
Since the coefficient of the sine function is e = 0.5, the conic is an
ellipse. Also, the numerator is 4 which means ep = 4. Substituting
e = 0.5,
€ we get 0.5p = 4 or p = 8. Thus, the ellipse€has a focus at the
pole, the directrix is 8 units below the pole parallel to the polar axis
π
and the major axis runs along the θ =
line. Since the vertices
2
are on the major axis, we can find them by evaluating the equation
π
3π
for θ =
and θ =
:
r=
2
8
2−sin(
2
π
2
)
=
8
2−1
€
=8
π
2
€ Thus, the€vertices are (8,
€
r=
) and (
8
3
,
8
2−sin(
3π
2
3π
2
)
=
8
2+1
8
3
,
) = (–
8
π
2
8
3
=
). The center is
8
π
the midpoint
between the two vertices, which
€
€ is€( 3 , 2 ) (or (0, 3 ) in
€
rectangular form). This means that c, the distance between the center
8
8
€ c and plugging
€ €
and the focus, is€ . Using€e =
in e = 0.5 and c = ,
we get a =
16
3
3
a
2€
. Recall that for an ellipse,€b = a – c =€(
2
2
16 2
)
3
8 3
which implies b =
. Thus, in
3 €
€
rectangular form, the equation is:
8
€
(y − )2
(x−0)2
+ 16 32 = 1
8 3 2
( )
( €)
3
3
x2
or
64
3
+
8 2
)
3
256
9
(y −
=1
€
€ and endpoints
on the minor axis
8 3
8
are: (h ± b, k): (±
, ) in
3
3
€
€
rectangular coordinates.
Now, we can sketch the graph.
€
€
€
€
€
–(
3
8 2
),
3
297
Ex. 5
r=
6sec(θ)
2sec(θ)− 2
Solution:
Since sec(θ) =
1
,
cos(θ)
we need to multiply top and bottom by
€
get the equation in proper form: r =
€
The equation matches case 2: r =
cos(θ )
2
cos(θ )
cos(θ )
2sec(θ)•
− 2•
2
2
6sec(θ)•
ep
1−ecos(θ)
=
cos(θ)
2
to
3
1− cos(θ)
€
Since the coefficient of the cosine function is e = 1,€the conic is a
€ is 3 which means ep = 3. Substituting
parabola. Also, the numerator
e = 1, we get (1)p = 3 or p = 3. Thus, the parabola has a focus at the
€ to the left of the pole perpendicular to the
pole, the directrix is 3 units
polar axis and the axis of symmetry runs along the polar axis. Since
the vertex is on the axis of symmetry, we can find it by evaluating
the equation at θ = 0 or θ = π:
r=
€
3
1− cos(0)
=
3
1−1
= undefined
r=
3
1− cos(π)
=
3
1+1
= 1.5
Thus, the vertex is (1.5, π) = (– 1.5, 0). Since the distance between
the focus and the vertex is 1.5, then a = 1.5. Thus, the equation of the
parabola is:
€(y – 0)2 = 4(1.5)(x – (– 1.5)) or
€
€ + 1.5)
y2 = 6(x
To help graph the parabola,
we will find the points that
determine the latus rectum
by plugging x = h = 0 into
the equation and solving
for y:
y2 = 6((0) + 1.5)
y2 = 9
or y = ± 3
Thus, the points (0, 3)
and (0, – 3) determine the
latus rectum. Now, we can
sketch the graph.
Objective 2: Converting from Polar to Rectangular Coordinates.
We could first convert the equation directly into rectangular form like
we did in example 2 and then sketch the graph. To see another
298
example, let's take the equation from example 3 and convert it
directly.
Convert the following equation into rectangular form:
12
Ex. 6
r=
4+8sin(θ)
Solution:
r=
€
12
4+8sin(θ)
(multiply both sides by [4 + 8sin(θ)])
€ + 8sin(θ)] = 12
r[4
(distribute)
4r + 8rsin(θ) = 12
(replace r by
4 x 2 + y2 + 8y = 12
(divide both sides by 4)
x 2 + y2 + 2y = 3
x 2 + y2 and rsin(θ) by y)
(subtract 2y from both sides)
(square both sides)
x 2 + y2 = 3 – 2y
2
2
2
x + y = (3 – 2y)
(simplify)
2
2
2
x + y = 4y – 12y + 9
(add – 4y2 + 12y to both sides)
x2 – 3y2 + 12y = 9
(factor out – 3 from the y2 and the y-term)
x2 – 3(y2 – 4y) = 9
(complete the square {add – 3(4) to both sides})
2
2
x – 3(y – 4y + 4) = 9 – 3(4)
(simplify)
2
2
x – 3(y – 2) = – 3
(divide both sides by – 3)
–
x2
3
+ (y – 2)2 = 1
(y – 2)2 –
€
x2
3
=1
Notice this is the same result that we had in example 3.
€