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 0,2π,π, and 23π provides the vertices so we can create a rough sketch of the graph.
Example 2: Graphing a Parabola in Polar Form
Graph
r=3+3 cos θ5.
Solution
First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is
31.
r=3+3 cos θ5=3(31)+3(31)cos θ5(31)r=1+cos θ35
Because
e=1, we will graph a
parabola with a focus at the origin. The function has a
cos θ, and there is an addition sign in the denominator, so the directrix is
x=p.
35=ep35=(1)p35=p
The directrix is
x=35.
Plotting a few key points as in the table below will enable us to see the vertices.
|
A |
B |
C |
D |
---|
θ |
0 |
2π |
π |
23π |
r=3+3 cos θ5 |
65≈0.83 |
35≈1.67 |
undefined |
35≈1.67 |
Figure 3
Analysis of the Solution
We can check our result with a graphing utility.
Figure 4
Example 3: Graphing a Hyperbola in Polar Form
Graph
r=2−3 sin θ8.
Solution
First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is
21.
r=2−3sin θ8=2(21)−3(21)sin θ8(21)r=1−23 sin θ4
Because
e=23,e>1, so we will graph a
hyperbola with a focus at the origin. The function has a
sin θ term and there is a subtraction sign in the denominator, so the directrix is
y=−p.
4=ep 4=(23)p4(32)=p 38=p
The directrix is
y=−38.
Plotting a few key points as in the table below will enable us to see the vertices.
|
A |
B |
C |
D |
---|
θ |
0 |
2π |
π |
23π |
r=2−3sinθ8 |
4 |
−8 |
4 |
58=1.6 |
Figure 5
Example 4: Graphing an Ellipse in Polar Form
Graph
r=5−4 cos θ10.
Solution
First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is
51.
r=5−4cos θ10=5(51)−4(51)cos θ10(51)r=1−54 cos θ2
Because
e=54,e<1, so we will graph an
ellipse with a
focus at the origin. The function has a
cosθ, and there is a subtraction sign in the denominator, so the
directrix is
x=−p.
2=ep 2=(54)p2(45)=p 25=p
The directrix is
x=−25.
Plotting a few key points as in the table below will enable us to see the vertices.
|
A |
B |
C |
D |
---|
θ |
0 |
2π |
π |
23π |
r=5−4 cos θ10 |
10 |
2 |
910≈1.1 |
2 |
Figure 6
Analysis of the Solution
We can check our result using a graphing utility.
Figure 7. r=5−4 cos θ10 graphed on a viewing window of
[−3,12,1] by
[−4,4,1],θmin =0 and
θmax =2π.
Try It 2
Graph
r=4−cos θ2.
Solution
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