Polar Equations of Conic Sections

Sometimes it is useful to write or identify the equation of a conic section in polar form. To do this, we need the concept of the focal parameter. The focal parameter of a conic section p is defined as the distance from a focus to the nearest directrix. The following table gives the focal parameters for the different types of conics, where a is the length of the semi-major axis (i.e., half the length of the major axis), c is the distance from the origin to the focus, and e is the eccentricity. In the case of a parabola, a represents the distance from the vertex to the focus.

Using the definitions of the focal parameter and eccentricity of the conic section, we can derive an equation for any conic section in polar coordinates. In particular, we assume that one of the foci of a given conic section lies at the pole. Then using the definition of the various conic sections in terms of distances, it is possible to prove the following theorem.

In the equation on the left, the major axis of the conic section is horizontal, and in the equation on the right, the major axis is vertical. To work with a conic section written in polar form, first make the constant term in the denominator equal to 1. This can be done by dividing both the numerator and the denominator of the fraction by the constant that appears in front of the plus or minus in the denominator. Then the coefficient of the sine or cosine in the denominator is the eccentricity. This value identifies the conic. If cosine appears in the denominator, then the conic is horizontal. If sine appears, then the conic is vertical. If both appear then the axes are rotated. The center of the conic is not necessarily at the origin. The center is at the origin only if the conic is a circle (i.e., [latex]e=0[/latex]).

Example: Graphing a Conic Section in Polar Coordinates

Identify and create a graph of the conic section described by the equation

[latex]r=\dfrac{3}{1+2\cos\theta }[/latex].

 

Show Solution

The constant term in the denominator is 1, so the eccentricity of the conic is 2. This is a hyperbola. The focal parameter p can be calculated by using the equation [latex]ep=3[/latex]. Since [latex]e=2[/latex], this gives [latex]p=\frac{3}{2}[/latex]. The cosine function appears in the denominator, so the hyperbola is horizontal. Pick a few values for [latex]\theta[/latex] and create a table of values. Then we can graph the hyperbola (Figure 17).

[latex]\theta[/latex]	[latex]r[/latex]	[latex]\theta[/latex]	[latex]r[/latex]
0	1	[latex]\pi[/latex]	−3
[latex]\frac{\pi }{4}[/latex]	[latex]\frac{3}{1+\sqrt{2}}\approx 1.2426[/latex]	[latex]\frac{5\pi }{4}[/latex]	[latex]\frac{3}{1-\sqrt{2}}\approx -7.2426[/latex]
[latex]\frac{\pi }{2}[/latex]	3	[latex]\frac{3\pi }{2}[/latex]	3
[latex]\frac{3\pi }{4}[/latex]	[latex]\frac{3}{1-\sqrt{2}}\approx -7.2426[/latex]	[latex]\frac{7\pi }{4}[/latex]	[latex]\frac{3}{1+\sqrt{2}}\approx 1.2426[/latex]

Figure 17. Graph of the hyperbola described in [link].

Watch the following video to see the worked solution to Example: Graphing a Conic Section in Polar Coordinates.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “7.5 Conic Sections” here (opens in new window).

try it

Identify and create a graph of the conic section described by the equation

[latex]r=\frac{4}{1 - 0.8\sin\theta }[/latex].

 

Hint

First find the values of e and p, and then create a table of values.

Show Solution

Here [latex]e=0.8[/latex] and [latex]p=5[/latex]. This conic section is an ellipse.

Figure 18.

General Equations of Degree Two

A general equation of degree two can be written in the form

The graph of an equation of this form is a conic section. If [latex]B\ne 0[/latex] then the coordinate axes are rotated. To identify the conic section, we use the discriminant of the conic section [latex]4AC-{B}^{2}[/latex]. One of the following cases must be true:

1. [latex]4AC-{B}^{2}>0[/latex]. If so, the graph is an ellipse.
2. [latex]4AC-{B}^{2}=0[/latex]. If so, the graph is a parabola.
3. [latex]4AC-{B}^{2}<0[/latex]. If so, the graph is a hyperbola.

The simplest example of a second-degree equation involving a cross term is [latex]xy=1[/latex]. This equation can be solved for y to obtain [latex]y=\frac{1}{x}[/latex]. The graph of this function is called a rectangular hyperbola as shown.

Figure 19. Graph of the equation [latex]xy=1[/latex]; The red lines indicate the rotated axes.

The asymptotes of this hyperbola are the x and y coordinate axes. To determine the angle [latex]\theta[/latex] of rotation of the conic section, we use the formula [latex]\cot{2\theta}=\frac{A-C}{B}[/latex]. In this case [latex]A=C=0[/latex] and [latex]B=1[/latex], so [latex]\cot{2\theta}=\frac{\left(0 - 0\right)}{1}=0[/latex] and [latex]\theta =45^\circ[/latex]. The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled [latex]{A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime },{E}^{\prime },\text{ and }{F}^{\prime }[/latex], and are given by the formulas

The procedure for graphing a rotated conic is the following:

1. Identify the conic section using the discriminant [latex]4AC-{B}^{2}[/latex].
2. Determine [latex]\theta[/latex] using the formula [latex]\cot2\theta =\frac{A-C}{B}[/latex].
3. Calculate [latex]{A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime },{E}^{\prime },\text{and}{F}^{\prime }[/latex].
4. Rewrite the original equation using [latex]{A}^{\prime },{B}^{\prime },{C}^{\prime },{D}^{\prime },{E}^{\prime },\text{and}{F}^{\prime }[/latex].
5. Draw a graph using the rotated equation.

Example: Identifying a Rotated Conic

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

[latex]13{x}^{2}-6\sqrt{3}xy+7{y}^{2}-256=0[/latex].

 

Show Solution

In this equation, [latex]A=13,B=-6\sqrt{3},C=7,D=0,E=0[/latex], and [latex]F=-256[/latex]. The discriminant of this equation is [latex]4AC-{B}^{2}=4\left(13\right)\left(7\right)-{\left(-6\sqrt{3}\right)}^{2}=364 - 108=256[/latex]. Therefore this conic is an ellipse. To calculate the angle of rotation of the axes, use [latex]\cot2\theta =\frac{A-C}{B}[/latex]. This gives

[latex]\begin{array}{cc}\hfill \cot2\theta & =\frac{A-C}{B}\hfill \\ & =\frac{13 - 7}{-6\sqrt{3}}\hfill \\ & =-\frac{\sqrt{3}}{3}.\hfill \end{array}[/latex]

 

Therefore [latex]2 \theta = {120}^{\circ}[/latex] and [latex]\theta ={60}^{\circ}[/latex], which is the angle of the rotation of the axes.

To determine the rotated coefficients, use the formulas given above:

[latex]\begin{array}{ccc}\hfill {A}^{\prime }& =\hfill & A{\cos}^{2}\theta +B\cos\theta \sin\theta +C{\sin}^{2}\theta \hfill \\ & =\hfill & 13{\cos}^{2}60+\left(-6\sqrt{3}\right)\cos60\sin60+7{\sin}^{2}60\hfill \\ & =\hfill & 13{\left(\frac{1}{2}\right)}^{2}-6\sqrt{3}\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+7{\left(\frac{\sqrt{3}}{2}\right)}^{2}\hfill \\ & =\hfill & 4,\hfill \\ \hfill {B}^{\prime }& =\hfill & 0,\hfill \\ \hfill {C}^{\prime }& =\hfill & A{\sin}^{2}\theta -B\sin\theta \cos\theta +C{\cos}^{2}\theta \hfill \\ & =\hfill & 13{\sin}^{2}60+\left(-6\sqrt{3}\right)\sin60\cos60=7{\cos}^{2}60\hfill \\ \hfill & =\hfill & {\left(\frac{\sqrt{3}}{2}\right)}^{2}+6\sqrt{3}\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)+7{\left(\frac{1}{2}\right)}^{2}\hfill \\ & =\hfill & 16,\hfill \\ \hfill {D}^{\prime }& =\hfill & D\cos\theta +E\sin\theta \hfill \\ & =\hfill & \left(0\right)\cos60+\left(0\right)\sin60\hfill \\ & =\hfill & 0,\hfill \\ \hfill {E}^{\prime }& =\hfill & -D\sin\theta +E\cos\theta \hfill \\ & =\hfill & -\left(0\right)\sin60+\left(0\right)\cos60\hfill \\ & =\hfill & 0,\hfill \\ \hfill {F}^{\prime }& =\hfill & F\hfill \\ & =\hfill & -256.\hfill \end{array}[/latex]

 

The equation of the conic in the rotated coordinate system becomes

[latex]\begin{array}{}\\ \hfill 4{\left({x}^{\prime }\right)}^{2}+16{\left({y}^{\prime }\right)}^{2}& =\hfill & 256\hfill \\ \hfill \frac{{\left({x}^{\prime }\right)}^{2}}{64}+\frac{{\left({y}^{\prime }\right)}^{2}}{16}& =\hfill & 1.\hfill \end{array}[/latex]

 

A graph of this conic section appears as follows.

Figure 20. Graph of the ellipse described by the equation [latex]13{x}^{2}-6\sqrt{3}xy+7{y}^{2}-256=0[/latex]. The axes are rotated [latex]60^\circ[/latex]. The red dashed lines indicate the rotated axes.

Watch the following video to see the worked solution to Example: Identifying a Rotated Conic.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “7.5 Conic Sections” here (opens in new window).