David Terr
Ph.D. Math, UC Berkeley


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9.6. Conic Sections in Polar Coordinates

Thus far in this chapter, we have looked at various equations for conic sections in rectangular coordinates. In this section, we will investigate conic sections in polar coordinates. As we will see, conic sections with one of their foci at the origin have a nice polar representation.

The general equation of a conic section in polar coordinates with one focus at the origin is as follows:

  • (9.6.1) r = 2p / [1 - e cos (θ - θ0)]

Here, p = a(1 - e2)/2 for an ellipse or a hyperbola and p is the parameter we defined in Section 9.3 for a parabola. The constant θ0 determines the orientation of the conic section. The standard orientation of a conic section is with θ0 = 0. We will look at this form for each of the three conic sections



An ellipse with standard orientation and one focus at the origin has the polar equation

  • (9.6.1) r = a(1 -e2) / (1 - e cos θ)

It is not too difficult to show that this is indeed the equation of an ellipse as claimed. Multiplying both sides of (9.6.1) by 1 - e cos θ, we find

r(1 - e cos θ) = r - ex = a(1 - e2).

Adding ex to each side and squaring both sides, we obtain

r2 = [a(1 - e2) + ex]2

x2 + y2 = a2(1 - e2)2 + 2ea(1 - e2)x + e2x2

Now subtracting e2x2 from both sides yields

(1 - e2)x2 + y2 = a2(1 - e2)2 + 2ea(1 - e2)x

Dividing both sides by 1 - e2 now yields

x2 + y2/(1 - e2) = a2(1 - e2) + 2eax

Subtracting 2eax from both sides, we find

x2 - 2eax + y2/(1 - e2) = a2(1 - e2)

To complete the square on the terms with x, we add e2a2 to both sides, obtaining

x2 - 2eax + e2a2 + y2/(1 - e2) = a2(1 - e2) + e2a2

(x - ea)2 + y2/(1 - e2) = a2

Finally, dividing both sides by a2 yields

(x - ea)2/a2 + y2/[a2(1 - e2)] = 1

(x - ea)2/a2 + y2/b2 = 1

where we used Equation (9.1.5) for b in the last step. Now we recognize the last equation as having the form of Equation (9.1.8) for an ellipse with semi-major axis a, semi-minor axis b = a√(1 - e2), and centered at the point (ea, 0), i.e. with the left focus at the origin.

From (9.6.1) it is easy to determine the points on the ellipse which lie closest to and farthest from the left focus, i.e. the origin. The maximum distance is achieved when θ = 0, for which we have r = a(1 - e2)/(1 - e) = a(1 + e), while the minimum distance is achieved when θ = π, for which we have r = a(1 - e2)/(1+ e) = a(1 - e). Since the orbit of a planet about the sun is an ellipse with the sun at one of the foci, we see that the maximum distance of the planet from the sun is a(1 + e) and the minimum distance is a(1 - e).



A hyperbola with standard orientation has the polar equation

  • (9.6.2) r = a(e2 - 1) / (e cos θ - 1)

This is really the same equation as (9.6.1), just written slightly differently. We have negated the numerator and denominator, since now we have e > 1. The derivation that Equation (9.6.2) represents a hyperbola is essentially the same as that of (9.6.1), the only difference being the sign change in front of the y term at the end. We leave the derivation as an exercise.

From (9.6.2), we can find the orientations of the asymptotes of the hyperbola. As e cos θ approaches 1, we see that the denominator becomes small, implying that r becomes large. When e cos θ = 1, the denominator is zero and r is undefined. Thus, we see that the orientations of the asymptotes are at the angles θ at which cos θ = 1/e, namely the angles θ = ±arccos(1/e). Note that by (9.6.2), r becomes negative when |θ| exceeds arccos(1/e). Negative values of r correspond to the second branch of the hyperbola. The first branch is farther from the origin than the second. As in the case of the ellipse, the center of the hyperbola lies at the point (ea, 0) and the second focus lies at (2ea, 0).



A parabola with standard orientation has the polar equation

  • (9.6.3) r = 2p / (1 - cos θ)

For this case, it is easy to convert the polar equation to rectangular coordinates and see that it really does correspond to a parabola. Multiplying both sides by 1 - cos θ, we find r(1 - cos θ) = r - x = 2p. Adding x to both sides, we get r = 2p + x. Squaring both sides, we find r2 = x2 + y2 = 4p2 + 4px + x2, implying y2 = 4px + 4p2 = 4p(x + p). Comparing with Equation (9.3.5), we see that this equation corresponds to a parabola opening up to the right with vertex at (0, -p). The directrix of this parabola is the line x = -2p.


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