The General Polar Equation of Conic Sections - Part 2

Introduction

We looked at the general polar equation of conics that are rotated at any angle in The General Polar Equation of Conic Sections - Part 1. The focus was at the origin and the equation turned out to be quite simple. Now, we will look at the polar equation of a conic where the directrix passes through the origin.

Deriving the General Equation

The steps to determine this equation in brief are to equate the distance from the focus to a point on the parabola to the ratio distance of the same point to the directrix. Then, we will isolate r to solve for r in terms of θ.

Figure 1

We will put the focus at the polar coordinate \(\left(a, \, \phi + \frac{\pi}{2} \right)\), where a is the distance of the focus from the pole. And the directrix will pass through the pole creating an angle φ with the polar axis.

First, notice that the distance PD can be given by the relationship \(d = r\sin(\theta - \phi)\) because of the right triangle PDO.

Because the focal segment OF is perpendicular to the directrix, the angle OF creates with the positive x-axis is \(\phi + \frac{\pi}{2}\). Therefore, the angle POF is equal to \( \phi + \frac{\pi}{2} - \theta\). Using the Law of Cosines, the distance FP is equal to \((ed)^2 = a^2 + r^2 - 2ar\cos(\phi + \frac{\pi}{2} - \theta) \). Now, we can substitute d from one to the other and follow the steps to isolate r.

(i) \( (er\sin(\theta - \phi))^2 = a^2 + r^2 - 2ar\cos(\phi + \frac{\pi}{2} - \theta) \)

Let’s simplify \(\cos(\phi + \frac{\pi}{2} - \theta)\) using the sum identity of cosine: \(\cos(\phi + \frac{\pi}{2} - \theta) = \) \( \cos(\phi - \theta)\cos(\frac{\pi}{2}) - \sin(\phi - \theta)\sin(\frac{\pi}{2}) = \) \( -\sin(\phi - \theta) = \sin(\theta - \phi) \).

(ii) \( (e^2\sin^{2}(\theta - \phi))r^2 = a^2 + r^2 - 2ar\sin(\theta - \phi) \)

(iii) \( (1 - e^2\sin^{2}(\theta - \phi))r^2 - 2a\sin(\theta - \phi)r + a^2 = 0 \)

Equation (iii) is a quadratic equation we can solve easily with the quadratic formula.

(iv) \( r = \dfrac{2a\sin(\theta - \phi) \pm \sqrt{4a^2\sin^{2}(\theta - \phi) - 4a^2(1-e^2\sin^{2}(\theta-\phi))}}{2(1-e^2\sin^{2}(\theta-\phi))} \)

(v) \( r = \dfrac{2a\sin(\theta - \phi) \pm 2a\sqrt{(1+e^2)\sin^{2}(\theta-\phi)-1}}{2(1-e^2\sin^{2}(\theta-\phi))} \)

(vi) \( r = \dfrac{a\sin(\theta - \phi) \pm a\sqrt{(1+e^2)\sin^{2}(\theta-\phi)-1}}{1-e^2\sin^{2}(\theta-\phi)} \)

The polar equation of a parabola that has the focus located at the polar coordinate \(\left(a, \, \phi + \frac{\pi}{2} \right) \) and the directrix creates an angle φ with the polar axis is given by:

\[ r(\theta) = \dfrac{a\sin(\theta - \phi) \pm a\sqrt{(1+e^2)\sin^{2}(\theta-\phi)-1}}{1 - e^2\sin^{2}(\theta - \phi)} \]

The Parabola

In equation (vi), if we let e = 1, then we get a simplified equation of a parabola.

The polar equation of a parabola that has the focus located at the polar coordinate \(\left(a, \, \phi + \frac{\pi}{2}\right)\) and the directrix creates the angle φ with the polar axis is given by:

\[ r(\theta) = \dfrac{a\sin(\theta - \phi) \pm a\sqrt{2\sin^{2}(\theta-\phi)-1}}{\cos^{2}(\theta - \phi)} \]

where a is the distance of the focus from the pole.

The rectangular position of the focus is \( \left( -a\sin\phi,\, a\cos\phi \right) \) and the equation of the directrix is \( y = (\tan\phi)x\).

We used the sine version of the cosine double-angle but we can also choose the cosine version. It would have been more concise to use \(-\cos(2\theta - 2\phi)\) but the negative sign seems out of place in the radical.

The equations are not as consice as the general equation for conics, but it does give a bit more leverage for choosing our focus point and the directrix.

From initial graphing tests, the plus/minus sign does not really change anything. However, there are gaps in the domains because of the radical, which probably lead to gaps in the range. When graphing \(f(\theta) = \sqrt{-\cos(2\theta - 2\phi)}\) as a rectangular function, there are gaps which are not filled by graphing both the plus and minus portions. However, Geogebra graphs the equation with a complete parabola without a hiccup. The only problem is a point on the parabola being tracked disappears at some intervals - probably at angles where the radical is undefined.

Other Conics

Figure 2: A hyperbola for φ = π/3, e = 3/2, and a = 1/2

The hyperbola in Figure 2 shows the value of e holds up. The equation graphed is \( r(\theta) = \frac{\frac{1}{2}\sin(\theta - \frac{\pi}{3}) + \frac{1}{2}\sqrt{1 + \frac{9}{4}\sin^{2}(\theta-\frac{\pi}{3})}}{1 - \frac{9}{4}\sin^{2}(\theta - \frac{\pi}{3})} \) for the focal distance a = 1/2, the eccentricity of e = 3/2, and the directrix equation \(y = (\tan\frac{\pi}{3})x \) or \(y = \sqrt{3}x\).

Figure 3: An ellipse for φ = 2π/3, e = 5/6, and a = 1/2

The ellipse in Figure 3 shows the value of e holds up for an ellipse also. The equation graphed is \( r(\theta) = \frac{\frac{1}{2}\sin(\theta - \frac{2\pi}{3}) + \frac{1}{2}\sqrt{1 + \frac{25}{36}\sin^{2}(\theta-\frac{2\pi}{3})}}{1 - \frac{25}{36}\sin^{2}(\theta - \frac{2\pi}{3})} \) for the focal distance a = 1/2, the eccentricity of e = 5/6, and the directrix equation \(y = (\tan\frac{2\pi}{3})x \) or \(y = -\sqrt{3}x\).