• Algebra > Parabola Polar Equation

The General Polar Equation of a Parabola

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 \(e=1\), the equation is a parabola. Therefore, the equation of a parabola is \(r(\theta)=\frac{d}{1-\sin\theta}\), where d is the distance of the focal point to the directrix. This parabola opens up or down depending on the sign of d.

The equation of a parabola that opens left or right is given by \(r(\theta) = \frac{d}{1+\cos\theta}\). The beauty of this equation of conics is the simplicity and being able to define all conics (except the circle) with one (or two depending on its orientation) equations.

The drawback, however, is that, for a parabola, we can only control its wideness or narrowness by increasing or decreasing d. We cannot rotate the parabola.

To fill that void, we will come up with an equation for a parabola rotated at any angle. For simplicity, the directrix of the parabola will pass 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 distance of the same point to the directrix. Then, we will isolate r to solve for r in terms of θ.

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 \(d^2 = a^2 + r^2 - 2ar\cos(\phi + \frac{\pi}{2} - \theta) \). Now, we can substitute d from one to the other and follow the steps to isolate r.

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

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

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

(iv) \( (\cos^{2}(\theta - \phi))r^2 - 2a\sin(\theta - \phi)r + a^2 = 0 \)

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

(v) \( r = \dfrac{2a\sin(\theta - \phi) \pm \sqrt{4a^2\sin^{2}(\theta - \phi) - 4a^2\cos^{2}(\theta - \phi)}}{2\cos^{2}(\theta - \phi)} \)

(vi) \( r = \dfrac{a\sin(\theta - \phi) \pm a\sqrt{\sin^{2}(\theta - \phi) - \cos^{2}(\theta - \phi)}}{\cos^{2}(\theta - \phi)} \)   (Which cosine double-angle identity shall we use?)

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

Equation (vii) represents the requation of the parabola we seek. 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 equation is 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

Epiphany! We can actually write the equation of the other conics with rotated axes in the same manner if we let the distance be in a ratio of e.

The equation is: \( r(\theta) = \dfrac{a\sin(\theta - \phi) \pm a\sqrt{(1+e^2)\sin^{2}(\theta-\phi)-1}}{1 - e^2\sin^{2}(\theta - \phi)} \).

If we let e = 1 in this equation, we get the equation we got above for the rotated parabola.

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 \(y = \sqrt{3}x\).

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 \(y = -\sqrt{3}x\).

The Focus on the Focus

The equation presented above has the directrix passing through the pole. This makes the equation complicated.

If we place the focus at the pole, then the equation is quite simple. Figure 4 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 a. This is also equal to BD where we have dranw 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) \( e = \frac{r}{PD} \)

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

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

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

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

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

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

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

If θ = 0, we essentially get the familiar equation of conics: \( r(\theta) = \frac{ea}{1+e\sin\theta} \).

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

The distance of the second vertex from the pole is:

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

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

The second vertex can also be written as: \( \left( -\frac{ea}{1-e}, \, \phi + \frac{\pi}{2} \right) \) 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 are 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 \( \left(\frac{2e^2a}{e^2-1}, \, \phi + \frac{\pi}{2} \right)\).

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