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The General Polar Equation of Conic Sections

Introduction

The polar equation of any conic section is \(r(\theta) = \frac{ed}{1-e\sin\theta}\), where d is the distance to the directrix from the focus and e is the eccentricity.

If we place the focus at the origin, we get a very simple equation of a conic section. In part 2, we will make the directrix cross the pole, which results in a much more complicated equation.

The Focus on the Focus

Figure 1 below shows the pieces needed to derive the general equation of a conic.

In the image, the directrix creates an angle φ with the positive x-axis.

The distance of the directrix from the focus is d. This is also equal to BD where we have drawn the perpendicular FB to PD.

The focus, F, is located at the pole.

A point P on the conic creates an angle θ with the positive x-axis.

Angle BFP is equal to \((\theta + \pi - \phi)\).

By definition of conics, then, the ratio of FP, which is equal to r, to PD is equal to the eccentricity or: \(\frac{r}{\text{PD}} = e\). We will solve for r in terms of θ in the steps below to derive the general polar equation of conics.

(i) \( e = \dfrac{r}{\text{PD}} \)

From the figure above, \(\text{PD} = d + \sin(\theta + \pi - \phi) \).

(ii) \( r = e(d + r\sin(\theta - \phi + \pi)) \)

Making the substitution \(\sin(\theta - \phi + \pi) = -\sin(\theta - \phi) \)

(iii) \( r = e(d - r\sin(\theta - \phi)) \)

(iv) \( r = ed - e\sin(\theta - \phi)\cdot r \)

(v) \( (1+e\sin(\theta-\phi))r = ed \)

(vi) \( r = \dfrac{ed}{1+e\sin(\theta-\phi)} \)

If θ = 0, we essentially get the familiar equation of conics: \( r(\theta) = \frac{ed}{1+e\sin\theta} \). So, essentially, the angle θ was replaced with (θ – φ). I believe to rotate curves in polar, we simply replace θ with (θ – φ).

The Vertices

We can figure out the location of both vertices of the ellipse and hyperbola by noting that the foci are located at an angle that is \(\frac{\pi}{2} \) and \(\frac{3\pi}{2} \) more than the angle created by the directrix, which is φ. We can substitute these values into the equation of the conic.

The distance of the first vertex from the pole is:

(1) \( r(\phi + \frac{\pi}{2}) = \frac{ed}{1+e\sin(\phi + \frac{\pi}{2} - \phi)} = \frac{ed}{1+e} \)

The distance of the second vertex from the pole is:

(2) \( r(\phi + \frac{3\pi}{2}) = \frac{ed}{1+e\sin(\phi + \frac{3\pi}{2} - \phi)} = \frac{ed}{1-e} \)

When e = 1, the conic is a parabola and there is only 1 vertex. The other vertex becomes undefined.

The second vertex can also be written as: \( \left( \frac{ed}{e-1}, \, \phi + \frac{\pi}{2} \right) \) if we want to keep the angle the same. In order to locate the second focus, we need this form so we can add the radii of the two vertices.

The Foci

We placed the first focus point at the pole. The second focus of the ellipse or hyperbola is simply the addition of the radii of the two vertices and the angle is the same. That is a convenience of the polar coordinate system.

The second focus can we rewritten as \( \left(\frac{2e^2d}{e^2-1}, \, \phi + \frac{\pi}{2} \right)\).

The benefit of polar coordinates in terms of e is that we do not have different formulas for the locations of the vertices and the foci. One formula covers all 3 - the parabola, ellipse and hyperbola - because the eccentricity variable takes care of it all.

The Center

Parabolas do not have a center but the ellipse and the hyperbola have a center. The center is the point of symmetry. We can rotate the curves 180° and it will be the same as the original.

The center is quite easy. It is the midpoint of the two vertices and also the foci. Since one focus is at the origin, we can simply divide the other focus in half to find the radius of the center. The angle will be the same, of course.

Lines of Symmetry

One of the lines of symmetry is the obvious one. It passess through the pole and has an angle of \(\phi - \frac{\pi}{2}\). Its equation is \( y = \tan(\phi-\frac{\pi}{2})x \), which can be written as \( y = -(\cot\theta)x \).

The equation of the second line of symmetry that passes through the center of the conic is \(y = (\tan\phi)x + \frac{e^2 d}{e^2-1}\sec\phi \) and in polar form: \( r(\theta) = \frac{\frac{e^2 d}{e^2-1}\sec\phi}{\sin\theta - \tan\phi\cos\theta} \).

a and b

The rectangular form of an ellipse or hyperbola is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) or \( \pm\frac{x^2}{a^2} \mp \frac{y^2}{b^2} = 1\). The constants a and b allow us to find the coordinates of the major and minor vertices.

We want to do the same for the ellipse and hyperbola. Based on the radial distances of the vertices and the foci, we have the following values for a and c.

(1) \( a = \dfrac{ed}{1-e^2}\)

(2) \(c = \dfrac{e^2 d}{e^2-1}\)

Using a and c, we can find b. For an ellipse:

(3) \( b = \sqrt{a^2 - c^2} = \dfrac{ed}{\sqrt{1-e^2}}\)

And for a hyperbola:

(4) \( b = \sqrt{c^2 - a^2} = \dfrac{ed}{\sqrt{e^2-1}}\)

The value of b differs because of the order of subtraction.

The Minor Vertices of an Ellipse

The ellipse has minor vertices. Knowing b from above and using a right triangle as shown in Figure 2, the distance, m, of the minor vertices from the center is \(\sqrt{b^2 + c^2} \) or \( \sqrt{ \frac{e^2d^2}{1-e^2} + \frac{e^4d^2}{(1-e^2)^2}} = \) \( \frac{ed}{e^2-1} \).

Note that the length m is equal to a in an ellipse.

If we draw a circle with this radius, then it will intersect both minor vertices, V₃ and V₄, as shown in Figure 2.

The angle, α, between the center and the minor vertices is \(\alpha = \arccos(\frac{c}{m}) = \arccos(e) \). Therefore, the location of the minor vertices is as follows:

Again, since m = a in an ellipse, \(\alpha = \arccos(\frac{c}{a}) = \arccos(e) \) because \( \frac{c}{a} = e \).

The Asymptotes of a Hyperbola

The hyperbola does not have minor vertices but has asymptotes. We can find the slope of the asymptotes using either sine, cosine, or tangent. If we use tangent, the slope ratio is \( m = \frac{b}{a} = \frac{\frac{ed}{\sqrt{e^2-1}}}{\frac{ed}{1-e^2}} = \) \( -\sqrt{e^2-1} \). Therefore, the equation of the asymptotes in rectangular coordinates are: \( y = \tan\left( \tan^{-1}(\pm\sqrt{e^2-1}) + \phi + \frac{\pi}{2} \right)x + \) \( \tan\left( \tan^{-1}(\pm\sqrt{e^2-1}) + \phi + \frac{\pi}{2} \right)\frac{e^2d}{e^2-1}\sin\phi + \) \( \frac{e^2d}{e^2-1}\cos\phi \). A complex formula.

If we use the arccosine function, we get a nicer value for the angle: \( y = \tan\left( \pm\cos^{-1}(\frac{1}{e}) + \phi + \frac{\pi}{2} \right)x + \) \( \tan\left( \pm\cos^{-1}(\frac{1}{e}) + \phi + \frac{\pi}{2} \right)\frac{e^2d}{e^2-1}\sin\phi + \frac{e^2d}{e^2-1}\cos\phi \).

We can simplify it a bit by using the identity \( \tan(\alpha + \frac{\pi}{2}) = -\cot\alpha\).

\( y = -\cot\left( \pm\cos^{-1}(\frac{1}{e}) + \phi \right)x - \) \( \cot\left( \pm\cos^{-1}(\frac{1}{e}) + \phi \right)\frac{e^2d}{e^2-1}\sin\phi + \frac{e^2d}{e^2-1}\cos\phi \)

The value \(\cot(\pm\cos^{-1}(\frac{1}{e}) - \phi)\) can be expanded using the sum property of cotangent: \( \frac{\cot(\pm\cos^{-1}(\frac{1}{e}))\cot\phi + 1}{\cot\phi - \cot(\pm\cos^{-1}(\frac{1}{e}))} \) = \( \frac{\pm\cot\phi + \sqrt{e^2-1}}{\sqrt{e^2-1}\cot\phi \mp 1} \).

Examples

It is easier to learn with examples, so we will resolve an ellipse and a hyperbola, meaning, we will find all the critical points and the lines.

Example 1: An Ellipse

Let’s resolve an ellipse with an eccentricity of 2/3, directrix distance of 2 from the focus, and the directrix creates an angle of \(\frac{4\pi}{5}\) with the x-axis.

The equation of the ellipse is: \( r(\theta) = \frac{4}{1 + \frac{2}{3}\sin(\theta + \frac{4\pi}{5})} \) = \( r(\theta) = \frac{12}{3 + 2\sin(\theta + \frac{4\pi}{5})} \).

Center: \( \left( \frac{e^2d}{e^2-1}, \phi + \frac{\pi}{2} \right) = \) \( \left( \frac{\frac{2^2}{3^2}\cdot 2}{\frac{2^2}{3^2}-1}, \frac{4\pi}{5} + \frac{\pi}{2} \right) = \) \( \left(-\frac{8}{5}, \frac{13\pi}{10}\right) \).

Since we have a negative radius, we can rotate the coordinate by π and take the positive radius. That will give the same coordinate. Therefore, the center is located in the first quadrant at \( \left(\frac{8}{5}, \frac{3\pi}{10}\right) \) or \( \left(\frac{8}{5}, 54^{\circ}\right) \).

The Foci: One focus point is always at the pole or (0, 0°). The second focus is at \( \left( \frac{ed}{1+e} - \frac{ed}{1-e}, \, \phi + \frac{\pi}{2} \right) \) = \( \left( \frac{\frac{2}{3}\cdot 2}{1+\frac{2}{3}} - \frac{\frac{2}{3}\cdot 2}{1-\frac{2}{3}}, \, \frac{13\pi}{10} \right) \) = \( \left( -\frac{16}{5},\frac{13\pi}{10} \right) \).

Again, since the radius is negative, we can rotate this point by π and take the positive radius: \( \left( \frac{16}{5},\frac{3\pi}{10} \right)\) or \( \left( \frac{16}{5}, 54^{\circ} \right) \). The second focus is also in the first quadrant.

The Major Vertices: The first vertex is at \( \left( \frac{ed}{1+e}, \, \phi + \frac{\pi}{2} \right) \) = \( \left( \frac{\frac{2}{3}\cdot 2}{1+\frac{2}{3}}, \, \frac{13\pi}{10} \right) \) = \( \left( \frac{\frac{2}{3}\cdot 2}{1+\frac{2}{3}}, \, \frac{13\pi}{10} \right) \) = \( \left( \frac{4}{5}, \frac{13\pi}{10} \right) \) or \( \left( \frac{4}{5}, 234^{\circ} \right) \). This vertex is in the 3rd quadrant.

The second vertex is at \( \left( \frac{ed}{1-e}, \, \phi + \frac{3\pi}{2} \right) \) = \( \left( \frac{\frac{2}{3}\cdot 2}{1-\frac{2}{3}}, \, \frac{23\pi}{10} \right) \) = \( \left( 4, \, \frac{23\pi}{10} \right) \). The second vertex angle makes a full 360° circle. We can subtract 2π from that value and simplify the coordinates: \( \left( 4, \, \frac{3\pi}{10} \right) \).

We expected the second major vertex to be in the first quadrant since the first was in the third quadrant.

The Minor Vertices: The first minor vertex is located at \( \left( \frac{ed}{e^2-1}, \phi + \frac{\pi}{2} + \cos^{-1}(e) \right) \) = \( \left( \frac{\frac{2}{3}\cdot 2}{\frac{2^2}{3^2}-1}, 234^{\circ} + 48.19^{\circ} \right) \) = \( \left( -\frac{12}{5}, 282.19^{\circ} \right) \). We can subtract 180° and take the positive radius: \( \left( \frac{12}{5}, 102.19^{\circ} \right) \). This vertex is in the second quadrant.

The second minor vertex is located at \( \left( \frac{ed}{e^2-1}, \phi + \frac{\pi}{2} - \cos^{-1}(e) \right) \) = \( \left( \frac{\frac{2}{3}\cdot 2}{\frac{2^2}{3^2}-1}, 234^{\circ} - 48.19^{\circ} \right) \) = \( \left( -\frac{12}{5}, 185.81^{\circ} \right) \). We can subtract 180° and take the positive radius: \( \left( \frac{12}{5}, 5.81^{\circ} \right) \). This vertex is in the first quadrant.

The Directrix: The equation of the directrix is \(y = (\tan\phi)x + 2\sec\phi \) or \(y = (\tan\frac{4\pi}{5})x + 2\sec\frac{4\pi}{5} \). This is approximately \(y = -0.727x - 2.4721 \).

We can find the exact trigonometric values of the angle \(\frac{4\pi}{5}\). The equation of the directrix is \(y = -(\sqrt{5-2\sqrt{5}})x + 2-2\sqrt{5} \).

Figure 3 shows all of our critical points and lines.

The Second Directrix: The ellipse (and the hyperbola) has a second directrix on the side of V₂ that is parallel to the first directrix. The distance is the same. Since V₁ is located at a radius of 4/5 and the distance of the directrix from the pole is 2, the distance of the directrix from V₁ is \(2 - \frac{4}{5} = \frac{6}{5}\). The second vertex V₂ has a radius of 4 so the directrix will pass through the polar coordinate \( \left( \frac{26}{5}, \frac{3\pi}{10} \right) \).

Converted to rectagular units, the coordinate is \(\left(\frac{26}{5}\cos\frac{3\pi}{10}, \frac{26}{5}\sin\frac{3\pi}{10}\right)\). Using the point-slope formula and some trigonometric identities, the equation of the second directrix is \(y = (\tan\frac{4\pi}{5})x - \frac{26}{5}\sec\frac{4\pi}{5} \). This is approximately \(y = -0.727x + 6.428 \).

As a matter of fact, we can write the general equation of the second directrix in terms of the variables e, d, and φ that we have been using: \( y = (\tan\phi)x - \frac{d(1+e^2)}{1-e^2}\sec(\phi) \).

Figure 4 shows the second directrix. You can see the eccentricity equals 0.667 or 2/3.

Example 2: The Hyperbola

We will resolve the hyperbola with an odd angle in degrees this time. Let e = 2, d = 1.5, and φ = 111°.

The equation of the hyperbola is \( r(\theta) = \frac{2\cdot 1.5}{1+2\sin(\theta - 111)}\) = \( r(\theta) = \frac{3}{1+2\sin(\theta - 111)}\).

Center: \( \left( \frac{e^2d}{e^2-1}, \phi + 90^{\circ} \right) \) = \( \left( \frac{2^2\cdot 1.5}{2^2-1}, \, 111^{\circ} + 90^{\circ} \right) \) = \( \left(2, \, 201^{\circ}\right) \).

The center is in the third quadrant.

The Foci: One focus point is always at the pole or (0, 0°). The second focus is at \( \left( \frac{ed}{1+e} - \frac{ed}{1-e}, \, \phi + 90^{\circ} \right) \) = \( \left( \frac{2\cdot 1.5}{1+2} - \frac{2\cdot 1.5}{1-2}, \, 201^{\circ} \right) \) = \( \left( 4, \, 201^{\circ} \right) \).

The second focus is in the third quadrant. Now, we have an idea of the hyperbola’s orientation.

The Vertices: The hyperbola does not have major and minor vertices like an ellipse. It has just two vertices on the transverse axis.

The first vertex is at \( \left( \frac{ed}{1+e}, \, \phi + 90^{\circ} \right) \) = \( \left( \frac{2\cdot 1.5}{1+2}, \, 201^{\circ} \right) \) = \( \left( 1, \, 201^{\circ} \right) \). This vertex is in the 3rd quadrant also. This is the vertex closest to the pole.

The second vertex is at \( \left( \frac{ed}{1-e}, \, \phi + 270^{\circ} \right) \) = \( \left( \frac{2\cdot 1.5}{1-2}, \, 381^{\circ} \right) \) = \( \left( -3, \, 381^{\circ} \right) \). The second vertex angle makes a full 360° circle. We can subtract 180° from that value and take the positive radius to simplify the coordinate: \( \left( 3, \, 201^{\circ} \right) \).

The second vertex is also in the third quadrant. We expected the two vertices and the foci to be collinear.

The Directrix: The equation of the first directrix is \(y = (\tan\phi)x + d\sec\phi \) or \(y = \tan(111^{\circ})x + \frac{3}{2}\sec(111^{\circ}) \). The approximate equation is \( y = -2.605x - 4.186 \).

The equation of the second directrix is: \( y = (\tan\phi)x - \frac{d(1+e^2)}{1-e^2}\sec\phi \) or \( y = \tan(111^{\circ})x - \frac{1.5(1+2^2)}{1-2^2}\sec(111^{\circ}) \) or \( y = \tan(111^{\circ})x + \frac{5}{2}\sec(111^{\circ}) \). The approximate equation is \( y = -2.605x - 6.976 \).

Asymptotes: The rectangular equation of the first asymptote is:

(i) \( y = \cot\left( \cos^{-1}(\frac{1}{e}) - \phi \right)x \, + \) \( \frac{e^2d}{e^2-1}\sin\phi\cot\left( \cos^{-1}(\frac{1}{e}) - \phi \right) + \frac{e^2d}{e^2-1}\cos\phi \)

(ii) \( y = \cot\left( \cos^{-1}(0.5) - 111^{\circ} \right)x \, + \) \( \frac{2^2\cdot 1.5}{2^2-1}\sin(111^{\circ})\cot\left( \cos^{-1}(0.5) - 111^{\circ} \right) + \frac{2^2\cdot 1.5}{2^2-1}\cos(111^{\circ}) \)

The value of \(\cot\left( \cos^{-1}(0.5) - 111^{\circ} \right)\) is equal to approximately –0.809, \(\sin(111^{\circ})\) is approximately 0.934, and \(\cos(111^{\circ})\) is approximately –0.358.

(iii) \( y = -0.809x + (2)(0.934)(-0.809) + (2)(-0.358) \).

(iv) \( y = -0.809x - 2.227 \).

We can use some of the values we found above for the second asymptote. The rectangular equation of the second asymptote is:

(i) \( y = \cot\left( -\cos^{-1}(0.5) - 111^{\circ} \right)x \, + \) \( (2)(0.934)\cot\left( -\cos^{-1}(0.5) - 111^{\circ} \right) + (2)(-0.358) \).

The value of \(\cot\left( -\cos^{-1}(0.5) - 111^{\circ} \right)\) is equal to approximately 6.313.

(ii) \( y = 6.313x + (2)(0.934)(6.313) + (2)(-0.358) \).

(iii) \( y = 6.313x + 11.077 \)

Figure 5 shows the hyperbola with all the critical points and lines. Each critical point seems to be spaced out 1 unit apart. The eccentricity ratio with both directrixes has also been shown.

Figure 6 shows the zoomed out image with the asymptotes to make sure we have the asymptotes equations correct because the chance of making an error in the equation is high. Note that they both cross the center of the hyperbola.

Last Word

We looked at the general polar equation of conic sections and found all critical points and lines.

In Part 2, we look at another way of writing the equation of conic sections where the directrix passes through the pole.

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