The Alternate Polar Equation of Conic Sections


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. If 0 < e < 1, it is an ellipse. And If e > 1, it is a hyperbola. In this equation, e cannot be 0, although if e = 0, it is a circle.

Surprisingly, the polar equation of any conic section can also be written simply as \(r(\theta) = \frac{1}{1-e\sin\theta}\)! How is this possible? How can the numerator be replaced with 1 and we still have a conic section? Why is the distance d not necessary?

Figure 1

Figure 1 shows two hyperbolas with the same eccentricity of 1.5. The equation of the blue hyperbola is \(r(\theta) = \frac{1.5\cdot 2}{1-1.5\sin\theta}\) and the hyperbola in green is \(r(\theta) = \frac{1}{1-1.5\sin\theta}\).

As one notices, although the eccentricity is the same, the shape appears totally different. The green hyperbola seems to have pointed vertices and the blue hyperbola has much smoother vertices.

Could \(r(\theta) = \frac{1}{1-1.5\sin\theta}\) be a conic section but the value of e no longer holds as in the first equation? If the eccentricity is, in fact, the same for both, then what exactly is the eccentricity if it has no bearing on the shape of the hyperbola?

Actually, the eccentricity is still e in both equations! Moreover, this equation does allow e to be 0 and it reduces to \(r(\theta) = 1\), which is a circle with radius 1.

So what is going on?

Rectangular Conversion

Before we answer the question above, let us convert the equation \( r = \dfrac{ed}{1-e\sin\theta}\) to the rectangular form and show that the value of e still holds by proving that \(\frac{c}{a} = e\).

(i) \( r = \dfrac{ed}{1-e\sin\theta}\)

(ii) \( r(1-e\sin\theta) = ed \)

(iii) \( r = ed + (e\sin\theta)r \)

(iv) \( \sqrt{x^2 + y^2} = ed + e\cdot \dfrac{y}{r}\cdot r \)

(v) \( x^2 + y^2 = e^2d^2 + 2e^2dy + e^2y^2 \)

(vi) \( x^2 + (1-e^2)y^2 - 2e^2dy = e^2d^2 \)

(vii) \( x^2 + (1-e^2)\left(y - \dfrac{e^2d}{1-e^2}\right)^2 = e^2d^2 + \dfrac{e^4d^2}{1-e^2} \)

(viii) \( x^2 + (1-e^2)\left(y - \dfrac{e^2d}{1-e^2}\right)^2 = \dfrac{e^2d^2}{1-e^2} \)

(ix) \( \dfrac{x^2}{1-e^2} + \left(y - \dfrac{e^2d}{1-e^2}\right)^2 = \dfrac{e^2d^2}{(1-e^2)^2} \)

(x) \( \dfrac{x^2}{\frac{e^2d^2}{1-e^2}} + \dfrac{\left(y - \dfrac{e^2d}{1-e^2}\right)^2}{\frac{e^2d^2}{(1-e^2)^2}} = 1 \)

Equation (x) is the equation of the conic section in rectangular form with its center at \( \left(0, \, \frac{e^2d}{1-e^2}\right) \). Since we used sine for the conic section, its major axis is along the y-axis. Therefore, the value a is \( a = \frac{ed}{1-e^2} \) and the value b is \(b = \frac{ed}{\sqrt{1-e^2}} \).

To find the c value, we need to account for whether the conic is a hyperbola or an ellipse. If the conic is a hyperbola, then e > 1. Therefore, equation (x) looks like this instead: \( -\frac{x^2}{\frac{e^2d^2}{e^2-1}} + \frac{\left(y + \frac{e^2d}{e^2-1}\right)^2}{\frac{e^2d^2}{(e^2-1)^2}} = 1 \). The equation does not allow e = 1 for a parabola.

The equation of an ellipse with eccentricity of e and a distance of d from the focus to the directrix is \( \dfrac{x^2}{\frac{e^2d^2}{1-e^2}} + \dfrac{\left(y - \dfrac{e^2d}{1-e^2}\right)^2}{\frac{e^2d^2}{(1-e^2)^2}} = 1 \) and the equation of a hyperbola is \( -\dfrac{x^2}{\dfrac{e^2d^2}{e^2-1}} + \dfrac{\left(y + \dfrac{e^2d}{e^2-1}\right)^2}{\dfrac{e^2d^2}{(e^2-1)^2}} = 1 \).

The c Value

Now, we can calculate c for a hyperbola:

(i) \(c^2 = a^2 + b^2\)

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

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

(iv) \(c^2 = \dfrac{e^4d^2}{(e^2-1)^2} \)

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

Therefore, the eccentricity is \(\frac{c}{a} = \frac{\frac{e^2d}{e^2-1}}{\frac{ed}{e^2-1}} = e \).

In the same way, if 0 < e < 1 for an ellipse, we can show the same hold true. In the case of an ellipse, \(c = \dfrac{e^2d}{1-e^2} \).

The Answer

The answer to our original question is very simple. Since the numerator is 1, the quantity ed is equal to 1. Therefore, \( d = \frac{1}{e}\). This is an alternate way of simplifying the equation if the distance between the directrix to the focus does not matter.

Simplifying The Conic Equation

Now, we can substitute \(d = \frac{1}{e}\) in equation (x) and see the result.

(i) \( \dfrac{x^2}{\frac{e^2\cdot\frac{1}{e^2}}{1-e^2}} + \dfrac{\left(y - \dfrac{e^2\cdot \frac{1}{e}}{1-e^2}\right)^2}{\frac{e^2\cdot\frac{1}{e^2}}{(1-e^2)^2}} = 1 \)

(ii) \( \dfrac{x^2}{\frac{1}{1-e^2}} + \dfrac{\left(y - \dfrac{e}{1-e^2}\right)^2}{\frac{1}{(1-e^2)^2}} = 1 \) (for an ellipse)

(iii) \( -\dfrac{x^2}{\frac{1}{e^2-1}} + \dfrac{\left(y + \dfrac{e}{e^2-1}\right)^2}{\frac{1}{(e^2-1)^2}} = 1 \) (for a hyperbola)

Based on this equation, \(c = \dfrac{e}{e^2-1} \) for a hyperbola and \(c = \dfrac{e}{1-e^2} \) for an ellipse.

The directrix is defined by the equation \(y = -\frac{1}{e}\).

The Eccentricity

So, what exactly is the eccentricity if it does not determine the shape of the hyperbola or the ellipse? Clearly from Figure 1, for a hyperbola the shapes are different even though the eccentricity is 1.5 for both. Let us investigate if e determines the shape of an ellipse by comparing the a and b of 2 ellipses with the same eccentricity but choosing our own d for one of them. We want to know if one is a scalar of the other, meaning if a and b are proportionate in the two.

Let d = 2 and e = 0.8. Then our first equation is \( \frac{x^2}{\frac{(\frac{4}{5})^2(2)^2}{1-(\frac{4}{5})^2}} + \frac{\left(y - \frac{(\frac{4}{5})^2(2)}{1-(\frac{4}{5})^2}\right)^2}{\frac{(\frac{4}{5})^2(2)^2}{(1-(\frac{4}{5})^2)^2}} = 1 \) or \( \frac{x^2}{\frac{64}{9}} + \dfrac{(y - \frac{32}{9})^2}{\frac{1600}{81}} = 1 \).

This ellipse has an a value of \(a = \sqrt{\frac{1600}{81}} = \frac{40}{9} \approx 4.444\) and a b value of \( b = \sqrt{\frac{64}{9}} = \frac{8}{3} \approx 2.666 \). The ratio of a to b gives us the proportion of the ellipse: \(\frac{a}{b} = \frac{\frac{40}{9}}{\frac{8}{3}} = \frac{5}{3} \). The c value is \(c = \frac{32}{9}\).

Using the second and simple equation, the ellipse is \( \frac{x^2}{\frac{1}{1-(\frac{4}{5})^2}} + \frac{\left(y - \frac{\frac{4}{5}}{1-(\frac{4}{5})^2}\right)^2}{\frac{1}{(1-(\frac{4}{5})^2)^2}} = 1 \) or \( \frac{x^2}{\frac{25}{9}} + \frac{\left(y - \frac{20}{9}\right)^2}{\frac{625}{81}} = 1 \).

This ellipse has an a value of \(a = \sqrt{\frac{625}{81}} = \frac{25}{9} \approx 2.778\) and a b value of \( b = \sqrt{\frac{25}{9}} = \frac{5}{3} \approx 1.667 \). The ratio of a to b gives us the proportion of the ellipse: \(\frac{a}{b} = \frac{\frac{25}{9}}{\frac{5}{3}} = \frac{5}{3} \). The c value is \(c = \frac{20}{9}\).

We got the same ratio. That means the ellipses are, in fact, proportional in size as can be seen in Figure 2. However, we got two different c values. That means we can get the same eccentricity with different values of a and b.

Figure 2

So why is the shape different for hyperbolas? Or is the shape different? More precisly, is it the same scale?

Let’s try a hyperbola with e = 1.6 and d = 2. We will skips the steps, but the equation is \( -\frac{x^2}{\frac{256}{39}} + \frac{(y + \frac{128}{39})^2}{\frac{6400}{1521}} = 1 \).

In this equation, \(a = \sqrt{\frac{6400}{1521}} = \frac{80}{39} \approx 2.0512\) and \(b = \sqrt{\frac{256}{39}} = \frac{16}{\sqrt{39}} \approx 2.562\). The ratio of a to b is \(\frac{a}{b} \approx \frac{2.0512}{2.562} \approx = 0.8006 \).

Using the second and simple equation, the hyperbola equation is \( -\frac{x^2}{\frac{25}{39}} + \frac{(y + \frac{40}{39})^2}{\frac{625}{1521}} = 1 \).

In the second equation, \(a = \sqrt{\frac{625}{1521}} = \frac{25}{39} \approx 0.6410\) and \(b = \sqrt{\frac{25}{39}} = \frac{5}{\sqrt{39}} \approx 0.8006\). The ratio of a to b is \(\frac{a}{b} = \frac{5}{\sqrt{39}} \approx 0.8006 \).

The ratios are the same again. That means the asymptotes of both hyperbolas have the same slope, but not necessarily the same intersection point. If they have the same slope, that means they both have the same proportion when the eccentricity is equal.

The Concept of Scalability

What we can say about the eccentricity and the shape (or curvature) of a hyperbola is that the proportionality of hyperbolas does not correlate with the shape (or curvature). In the same way, the proportionality does not correlate with the shape in a parabola, although all parabolas can be considered proportional, regardless of a.

Circles and ellipses are closed curves. Therefore, we can see their scalability. However, hyperbolas and parabolas are not closed curves and we cannot see they are scaled in the same way.

The concept of scalability is different for shapes like squares and triangles and circles and ellipses vs parabolas and hyperbolas.

For example, how do we define scaled triangles and squares? The sides have the same ratio of a constant such as k and the angle measures are equal. That works for all shapes with straight sides.

But for circles and squares which do not have angles and straight sides, that no longer works. So scalability of these shapes is defined by scalars of the major and minor axes lengths.

So, for parabolas and hyperbolas, we have to rethink scalability because, if the eccentricity is the same, all hyperbolas are scalar; and all parabolas are scalar since their eccentricity is 1.

The Scalability of Conics

Let’s look at the two equations of conics again: \( r_1(\theta) = \frac{ed}{1-e\sin\theta} \) and \( r_2(\theta) = \frac{1}{1-e\sin\theta} \).

Notice the denominator is the same in both. Only the numerator differs. If we take the ratio of the radii at same angle for both, we get a constant: \( \frac{r_1(\theta)}{r_2(\theta)} = \frac{\frac{ed}{1-e\sin\theta}}{\frac{1}{1-e\sin\theta}} = ed \). For any ray from the pole crossing these two conics, the radii always have a constant ratio of ed! The focus is the focus for scalability!

We have to think of the scalability of these conics from the focus point of view.

Therefore, if the eccentricity is the same, the conics are scalar even if the shape looks different, as in the case of hyperbolas and parabolas.

Figure 3

Figure 3 shows two hyperbolas. The green is \( r(\theta) = \frac{1.5\cdot2}{1-1.5\sin\theta} \) and the red is \( r(\theta) = \frac{1}{1-1.5\sin\theta} \). Therefore, we expect the scalar to be \(ed = 1.5\cdot 2 = 3 \). The figure shows two lines drawn from the pole. One crosses both hyperbolas at two points and both sets of points give the ratio of 3. The second ray only crosses both hyperbolas once and it also gives a ratio of 3.

Figure 4 shows the intersection points on two parabolas from two rays extending from the pole. The green parabola distance is 3. One point switched from being father to closer so its ratio is the inverse. I added the last point that was missing from the hyperbola above.

Figure 4

Note that the green parabola is wider than the red parabola despite the scalability from the focus.

Therefore, given a fixed eccentricity, the conics are scalable from the perspective of the focus. And it is easy to see that if the focus is located at the pole.

Scalability from the Center

The ellipse and the hyperbola are also scalable from the perspective of the center if the eccentricity is the same. (Parabolas do not have a center.) The easiest way to see their scalability is to consider their polar equation with the center at the pole.

The equation of the ellipse is \( r(\theta) = \frac{ab}{\sqrt{(b\cos\theta)^2 + (a\sin\theta)^2}} \). When replacing a with ka and b with kb and then finding the ratio of the radii, the result is k. Same goes for the hyperbola.

Figure 5

Both ellipses are similar even though it does not seem that way. Perhaps our notion of similarity for the ellipse would be that the curves line up or both are equidistant at every point - perhaps from the tangent at every point. But such a curve would not be an ellipse.

This is best illustrated by a Geogebra activity below. Drag the points around the actual ellipse in orange. It has the parametric equation \( x = 3\cos(t); \, y = 4\sin(t) \). You will see the sum IK + KJ is constant and equal to 8. The curve in blue is a pseudo-ellipse. Its parametric equation, derived from all tangent points equidistant to the orange ellipse by 1 unit, is \( x = 3\cos(t) + \cos(\arctan(\frac{3}{4}\tan(t))); \) \( y=4\sin(t) + \sin(\arctan(\frac{3}{4}\tan(t))) \). Curiously, the right half is 1 unit more and the left half is 1 unit less.

Since only a true ellipse has foci, we pretend the foci of the blue pseudo-ellipse are positioned the same distance from the center as a real ellipse, by using the formula \( c = \sqrt{a^2 - b^2} \), where c is the distance to the foci from the center. An easy way to find the foci is to note that the distance FN is equal to half the major axis. As you move point A around the pseudo-ellipse, you will see the sum, which should be 2a = 10, is not constant, although very close to being constant.

The blue curve and the orange ellipse look similar in every way, yet they are not scalar like true ellipses in Figure 5. If we show the ellipse \( \frac{x^2}{16} + \frac{y^2}{25}=1 \) which would correpsond to the same major and minor axis as the pseudo-ellipse, you can see how closely they overlap. However, zooming in, we notice that they are actually different curves.

Figure 6: The right half of pseudo-ellipse and true ellipse seems to overlap perfectly
Figure 7: Zooming in reveals they do not overlap

Pleasing to the Eye

To get ellipses that are pleasing-to-the-eye “similar”, we can just add some constant to both a and b. These two ellipses will be very similar but not scalar.

Figure 8 shows 2 ellipses with the equation \( \frac{x^2}{16} + \frac{y^2}{9} \) and \( \frac{x^2}{25} + \frac{y^2}{16} \). As can be seen, the ratios of the radii are different but the similarity is more pleasing.

Figure 8

The distance of one ellipse to the other is not constant if we take the distance of a tangent to one to the other.

Infinite Ways

There are infinite ways we can choose a and b to arrive at the same eccentricity. The equation of the eccentricity can be written as \(e = \frac{\sqrt{a^2+b^2}}{a}\) for a hyperbola. We can manipulate that to arrive at the equation \(b^2 = (e^2 - 1)a^2\). That means for some fixed value of e, whatever we choose for a, there is some b that will satisfy the equation. And each time, the ratio of a to b will be the same. If we manipulate this equation, we can arrive at \(\frac{a}{b} = \sqrt{\frac{1}{e^2 - 1}}\). For a fixed eccentricity, the ratio is the same no matter what values of a and b we choose because they are all related.