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Transforming Equations between Polar and Rectangular Forms
We can now convert coordinates between polar and rectangular form. Converting equations can be moredifficult, but it can be benefic al to be able to convert between the two forms. Since there are a number of polar equations that cannot be expressed clearly in Cartesian form, and vice versa, we can usethe same procedures we used to convert points between the coordinate systems.
Example 6 Write the Cartesian equation^ x^2 +^ y^2 =^ 9 in polar form.
x
y
Solution
Example 7 Rewrite the Cartesian equation x 2 + y 2 = 6 y as a polar equation. Solution
x
y
Example 8 a) Rewrite the Cartesian equation y = 3 x + 2 as a polar equation. b)
Solution
Rewrite the Cartesian equation y 2 = 3 − x 2 in polar form.
Identify and Graph Polar Equations by Converting to Rectangular Equations
We have learned how to convert rectangular coordinates to polar coordinates, and we have seen that the points are indeed the same. We have also transformed polar equations to rectangular equations and vice versa. Now we will demonstrate that their graphs, while drawn on different grids, are identical.
Example 9 Convert the polar equation r = 2sec θ to a rectangular equation, and draw its corresponding graph. Solution
x = 2
x
y
Exercise:^ Convert the polar equation^ r^ =^ 2csc^ θ^ to a rectangular equation, and draw its corresponding graph.
r = 2 sec °
MORE EXERCISES
For the following exercises, convert the given Cartesian equation to a polar equation.
x = 3
y = 4
y = 4 x 2
y = 2 x^4
x 2 + y 2 = 4 y
x 2 + y 2 = 3 x
x 2 − y 2 = x
x 2 − y 2 = 3 y
x 2 + y 2 = 9
x 2 = 9 y
y 2 = 9 x
9 xy = 1
For the following exercises, convert the given polar equation to a Cartesian equation. Write in the standard form of a conic if possible, and identify the conic section represented.
r = 3sin θ
r = 4cos θ
r = __^4 sin θ + 7cos θ
r = __^6 cos θ + 3sin θ
r = 2sec θ
r = 3csc θ
r = √
— rcos θ + 2
r 2 = 4sec θ csc θ
r = 4
r = __^1 4cos θ − 3sin θ
r = __^3 cos θ − 5sin θ
θ = − 2 _π
r = −10sin θ
θ = _π 4
LEARNING OBJECTIVES
In this section, you will:
- Identify a conic in polar form.
- Graph the polar equations of conics.
- Define conics in terms of a focus and a directrix.
9. 3 CONIC SECTIONS IN POLAR COORDINATES
Figure 1 Planets orbiting the sun follow elliptical paths. (credit: NASA Blueshift, Flickr)
Most of us are familiar with orbital motion, such as the motion of a planet around the sun or an electron around an atomic nucleus. Within the planetary system, orbits of planets, asteroids, and comets around a larger celestial body are often elliptical. Comets, however, may take on a parabolic or hyperbolic orbit instead. And, in reality, the characteristics of the planets’ orbits may vary over time. Each orbit is tied to the location of the celestial body being orbited and the distance and direction of the planet or other object from that body. As a result, we tend to use polar coordinates to represent these orbits. In an elliptical orbit, the periapsis is the point at which the two objects are closest, and the apoapsis is the point at which they are farthest apart. Generally, the velocity of the orbiting body tends to increase as it approaches the periapsis and decrease as it approaches the apoapsis. Some objects reach an escape velocity, which results in an infinite orbit. These bodies exhibit either a parabolic or a hyperbolic orbit about a body; the orbiting body breaks free of the celestial body’s gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system.
Identifying a Conic in Polar Form
Any conic may be determined by three characteristics: a single focus, a fixed line called the directrix, and the ratio of the distances of each to a point on the graph. Consider the parabola x = 2 + y^2 shown in Figure below.
Polar axis
x = 2 + y
F, Focus @ pole
r
D
P(r, θ)
Directrix
θ
Example 1 Identifying a Conic Given the Polar Form
For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity.
a. r = 3 +
_________^6
2 sin θ
b. r = 4 +
_________^12
5 cos θ
c. r = 2 −
_________^7
2 sin θ
Solution
d. (^) r(2 − cos θ) = 1
Graphing the Polar Equations of Conics
When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with 1. This enables us to determine e and, therefore, the shape of the curve. The next step is to substitute values for θ and solve for r to plot a few key points. Setting θ equal to 0, π__ 2
, π, and ___^3 π 2
provides the vertices so we can create a rough sketch of the graph.
Example 2 Graphing a Parabola in Polar Form : Graph r = _________^5 3 + 3 cos θ
Solution
A B C D
θ 0 π__ 2
π ___^3 π 2
r = _________^5 3 + 3 cos θ
5 __
≈ 0.83 __^5
≈ 1.67 undefined 5 __ 3
A
Directrix
B
F
D
2
r
3 4 5
x = (^53)
Example 4 Graphing an Ellipse in Polar Form Graph^ r^ =^
__________^10
5 − 4 cos θ.
Solution
A B C D
θ 0 __π 2
π ___^3 π 2
r = _________^10 5 − 4 cos θ 10 2
___^10
2 D
C
B A
r
4 6 8 10 12
x = − (^52) Directrix
Graph r = 4 −
_________^2
cos θ.
Defining Conics in Terms of a Focus and a Directrix
So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse;
we will use information about the origin, eccentricity, and directrix to determine the polar equation.
Example 5 Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix Find the polar form of the conic given a focus at the origin, e = 3 and directrix y = − 2. Solution
Example 6 Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix
Find the polar form of a conic given a focus at the origin, e = _^3 5
, and directrix x = 4. Solution
MORE EXERCISES
For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity.
r = __________^6 1 − 2 cos θ
r = __________^3 4 − 4 sin θ
r = __________^8 4 − 3 cos θ
r = __________^5 1 + 2 sin θ
r =
__________^16
4 + 3 cos θ
r =
____________^3
10 + 10 cos θ
r =
________^2
1 − cos θ
r =
__________^4
7 + 2 cos θ
r(1 − cos θ) = 3
r(3 + 5sin θ) = 11 r(4 − 5sin θ) = 1 r(7 + 8cos θ) = 7
For the following exercises, convert the polar equation of a conic section to a rectangular equation.
r = __________^4 1 + 3 sin θ
r = __________^2 5 − 3 sin θ
r = __________^8 3 − 2 cos θ
r = __________^3 2 + 5 cos θ
r =
__________^4
2 + 2 sin θ
r = __________^3 8 − 8 cos θ
r =
__________^2
6 + 7 cos θ
r =
___________^5
5 − 11 sin θ
r(5 + 2 cos θ) = 6 r(2 − cos θ) = 1 r(2.5 − 2.5 sin θ) = 5 r = ___________6sec^ θ − 2 + 3 sec θ r = __________6csc^ θ 3 + 2 csc θ
For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci.
r =
________^5
2 + cos θ
r =
__________^2
3 + 3 sin θ
r =
__________^10
5 − 4 sin θ
r =
__________^3
1 + 2 cos θ
r = __________^8 4 − 5 cos θ
r = __________^3 4 − 4 cos θ
r = _^2 1 − sin θ
r = __________^6 3 + 2 sin θ r(1 + cos θ) = 5 r(3 − 4sin θ) = 9 r(3 − 2sin θ) = 6 r(6 − 4cos θ) = 5
For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix.
Directrix: x = 4; e = __^1 5
Directrix: x = − 4; e = 5 Directrix: y = 2; e = 2
Directrix: y = − 2; e =
__^1
2 Directrix:^ x^ =^ 1;^ e^ =^1 Directrix:^ x^ =^ −1;^ e^ =^1
Directrix: x = −
__^1
4 ;^ e^ =^
__^7
2 Directrix:^ y^ =^
__^2
5 ;^ e^ =^
__^7
2 Directrix:^ y^ =^ 4;^ e^ =^
__^3
Directrix: x = −2; e = __^8 3
Directrix: x = −5; e = __^34 Directrix: y = 2; e = 2.
Directrix: x = −3; e =
__^1
