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Study Guides > College Algebra

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 ee and, therefore, the shape of the curve. The next step is to substitute values for θ\theta and solve for rr to plot a few key points. Setting θ\theta equal to 0,π2,π0,\frac{\pi }{2},\pi , and 3π2\frac{3\pi }{2} provides the vertices so we can create a rough sketch of the graph.

Example 2: Graphing a Parabola in Polar Form

Graph r=53+3 cos θr=\frac{5}{3+3\text{ }\cos \text{ }\theta }.

Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is 13\frac{1}{3}.
r=53+3 cos θ=5(13)3(13)+3(13)cos θr=531+cos θ\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{5}{3+3\text{ }\cos \text{ }\theta }=\frac{5\left(\frac{1}{3}\right)}{3\left(\frac{1}{3}\right)+3\left(\frac{1}{3}\right)\cos \text{ }\theta }\hfill \end{array}\hfill \\ r=\frac{\frac{5}{3}}{1+\cos \text{ }\theta }\hfill \end{array}
Because e=1e=1, we will graph a parabola with a focus at the origin. The function has a cos θ \cos \text{ }\theta , and there is an addition sign in the denominator, so the directrix is x=px=p.
53=ep53=(1)p53=p\begin{array}{l}\frac{5}{3}=ep\\ \frac{5}{3}=\left(1\right)p\\ \frac{5}{3}=p\end{array}
The directrix is x=53x=\frac{5}{3}. Plotting a few key points as in the table below will enable us to see the vertices.
A B C D
θ\theta 00 π2\frac{\pi }{2} π\pi 3π2\frac{3\pi }{2}
r=53+3 cos θr=\frac{5}{3+3\text{ }\cos \text{ }\theta } 560.83\frac{5}{6}\approx 0.83 531.67\frac{5}{3}\approx 1.67 undefined 531.67\frac{5}{3}\approx 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=823 sin θr=\frac{8}{2 - 3\text{ }\sin \text{ }\theta }.

Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is 12\frac{1}{2}.
r=823sin θ=8(12)2(12)3(12)sin θr=4132 sin θ\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{8}{2 - 3\sin \text{ }\theta }=\frac{8\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)-3\left(\frac{1}{2}\right)\sin \text{ }\theta }\hfill \end{array}\hfill \\ r=\frac{4}{1-\frac{3}{2}\text{ }\sin \text{ }\theta }\hfill \end{array}
Because e=32,e>1e=\frac{3}{2},e>1, so we will graph a hyperbola with a focus at the origin. The function has a sin θ\sin \text{ }\theta term and there is a subtraction sign in the denominator, so the directrix is y=py=-p.
 4=ep 4=(32)p4(23)=p 83=p\begin{array}{l}\text{ }4=ep\hfill \\ \text{ }4=\left(\frac{3}{2}\right)p\hfill \\ 4\left(\frac{2}{3}\right)=p\hfill \\ \text{ }\frac{8}{3}=p\hfill \end{array}
The directrix is y=83y=-\frac{8}{3}. Plotting a few key points as in the table below will enable us to see the vertices.
A B C D
θ\theta 00 π2\frac{\pi }{2} π\pi 3π2\frac{3\pi }{2}
r=823sinθr=\frac{8}{2 - 3\sin \theta } 44 8-8 44 85=1.6\frac{8}{5}=1.6
Figure 5

Example 4: Graphing an Ellipse in Polar Form

Graph r=1054 cos θr=\frac{10}{5 - 4\text{ }\cos \text{ }\theta }.

Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is 15\frac{1}{5}.
r=1054cos θ=10(15)5(15)4(15)cos θr=2145 cos θ\begin{array}{l}\hfill \\ \begin{array}{l}r=\frac{10}{5 - 4\cos \text{ }\theta }=\frac{10\left(\frac{1}{5}\right)}{5\left(\frac{1}{5}\right)-4\left(\frac{1}{5}\right)\cos \text{ }\theta }\hfill \\ r=\frac{2}{1-\frac{4}{5}\text{ }\cos \text{ }\theta }\hfill \end{array}\hfill \end{array}
Because e=45,e<1e=\frac{4}{5},e<1, so we will graph an ellipse with a focus at the origin. The function has a cosθ\text{cos}\theta , and there is a subtraction sign in the denominator, so the directrix is x=px=-p.
 2=ep 2=(45)p2(54)=p 52=p\begin{array}{l}\text{ }2=ep\hfill \\ \text{ }2=\left(\frac{4}{5}\right)p\hfill \\ 2\left(\frac{5}{4}\right)=p\hfill \\ \text{ }\frac{5}{2}=p\hfill \end{array}
The directrix is x=52x=-\frac{5}{2}. Plotting a few key points as in the table below will enable us to see the vertices.
A B C D
θ\theta 00 π2\frac{\pi }{2} π\pi 3π2\frac{3\pi }{2}
r=1054 cos θr=\frac{10}{5 - 4\text{ }\cos \text{ }\theta } 1010 22 1091.1\frac{10}{9}\approx 1.1 22
Figure 6

Analysis of the Solution

We can check our result using a graphing utility.
Figure 7. r=1054 cos θr=\frac{10}{5 - 4\text{ }\cos \text{ }\theta } graphed on a viewing window of [3,12,1]\left[-3,12,1\right] by [4,4,1],θmin =0\left[-4,4,1\right],\theta \text{min =}0 and θmax =2π\theta \text{max =}2\pi .

Try It 2

Graph r=24cos θr=\frac{2}{4-\cos \text{ }\theta }. Solution

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  • Precalculus. Provided by: OpenStax Authored by: OpenStax College. Located at: https://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1/Preface. License: CC BY: Attribution.