Conic Sections in Polar Coordinates

Learning Objectives

In this section, you will:

• Identify a conic in polar form.
• Graph the polar equations of conics.
• Define conics in terms of a focus and a directrix.

Figure 1 Planets orbiting the sun follow elliptical paths. (credit: NASA Blueshift, Flickr)

Most of us are familiar with orbital motion, such as the motion of a planet around the sun or an electron around an atomic nucleus. Within the planetary system, orbits of planets, asteroids, and comets around a larger celestial body are often elliptical. Comets, however, may take on a parabolic or hyperbolic orbit instead. And, in reality, the characteristics of the planets’ orbits may vary over time. Each orbit is tied to the location of the celestial body being orbited and the distance and direction of the planet or other object from that body. As a result, we tend to use polar coordinates to represent these orbits.

In an elliptical orbit, the periapsis is the point at which the two objects are closest, and the apoapsis is the point at which they are farthest apart. Generally, the velocity of the orbiting body tends to increase as it approaches the periapsis and decrease as it approaches the apoapsis. Some objects reach an escape velocity, which results in an infinite orbit. These bodies exhibit either a parabolic or a hyperbolic orbit about a body; the orbiting body breaks free of the celestial body’s gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system.

Identifying a Conic in Polar Form

Any conic may be determined by three characteristics: a single focus, a fixed line called the directrix, and the ratio of the distances of each to a point on the graph. Consider the parabola x=2+y2𝑥=2+𝑦2 shown in Figure 2.

Figure 2

In The Parabola, we learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any conic in the polar coordinate system in terms of a fixed point, the focus P(r,θ)𝑃(𝑟,𝜃) at the pole, and a line, the directrix, which is perpendicular to the polar axis.

If F𝐹 is a fixed point, the focus, and D𝐷 is a fixed line, the directrix, then we can let e𝑒 be a fixed positive number, called the eccentricity, which we can define as the ratio of the distances from a point on the graph to the focus and the point on the graph to the directrix. Then the set of all points P𝑃 such that e=PFPD𝑒=𝑃𝐹𝑃𝐷 is a conic. In other words, we can define a conic as the set of all points P𝑃 with the property that the ratio of the distance from P𝑃 to F𝐹 to the distance from P𝑃 to D𝐷 is equal to the constant e.𝑒.

For a conic with eccentricity e,𝑒,

• if 0≤e<1,0≤𝑒<1, the conic is an ellipse
• if e=1,𝑒=1, the conic is a parabola
• if e>1,𝑒>1, the conic is an hyperbola

With this definition, we may now define a conic in terms of the directrix, x=±p,𝑥=±𝑝, the eccentricity e,𝑒, and the angle θ.𝜃. Thus, each conic may be written as a polar equation, an equation written in terms of r𝑟 and θ.𝜃.

THE POLAR EQUATION FOR A CONIC

For a conic with a focus at the origin, if the directrix is x=±p,𝑥=±𝑝, where p𝑝 is a positive real number, and the eccentricity is a positive real number e,𝑒, the conic has a polar equation

r=ep1±ecosθ𝑟=𝑒𝑝1±𝑒cos𝜃

For a conic with a focus at the origin, if the directrix is y=±p,𝑦=±𝑝, where p𝑝 is a positive real number, and the eccentricity is a positive real number e,𝑒, the conic has a polar equation

r=ep1±esinθ𝑟=𝑒𝑝1±𝑒sin𝜃

HOW TO

Given the polar equation for a conic, identify the type of conic, the directrix, and the eccentricity.

1. Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form.
2. Identify the eccentricity e𝑒 as the coefficient of the trigonometric function in the denominator.
3. Compare e𝑒 with 1 to determine the shape of the conic.
4. Determine the directrix as x=p𝑥=𝑝 if cosine is in the denominator and y=p𝑦=𝑝 if sine is in the denominator. Set ep𝑒𝑝 equal to the numerator in standard form to solve for x𝑥 or y.𝑦.

EXAMPLE 1

Identifying a Conic Given the Polar Form

For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity.

1. r=63+2sinθ𝑟=63+2sin𝜃
2. r=124+5cosθ𝑟=124+5cos𝜃
3. r=72−2sinθ𝑟=72−2sin𝜃

Solution

For each of the three conics, we will rewrite the equation in standard form. Standard form has a 1 as the constant in the denominator. Therefore, in all three parts, the first step will be to multiply the numerator and denominator by the reciprocal of the constant of the original equation, 1c,1𝑐, where c𝑐 is that constant.

1. Multiply the numerator and denominator by 13.13.r=63+2sinθ⋅(13)(13)=6(13)3(13)+2(13)sinθ=21+23sinθ𝑟=63+2sin𝜃⋅(13)(13)=6(13)3(13)+2(13)sin𝜃=21+23sin𝜃Because sinθsin𝜃 is in the denominator, the directrix is y=p.𝑦=𝑝. Comparing to standard form, note that e=23.𝑒=23. Therefore, from the numerator,2=ep2=23p(32)2=(32)23p3=p2=𝑒𝑝2=23𝑝(32)2=(32)23𝑝3=𝑝Since e<1,𝑒<1, the conic is an ellipse. The eccentricity is e=23𝑒=23 and the directrix is y=3.𝑦=3.
2. Multiply the numerator and denominator by 14.14.r=124+5cosθ⋅(14)(14)r=12(14)4(14)+5(14)cosθr=31+54cosθ𝑟=124+5cos𝜃⋅(14)(14)𝑟=12(14)4(14)+5(14)cos𝜃𝑟=31+54cos𝜃Because cosθ cos𝜃  is in the denominator, the directrix is x=p.𝑥=𝑝. Comparing to standard form, e=54.𝑒=54. Therefore, from the numerator,     3=ep     3=54p(45)3=(45)54p  125=p     3=𝑒𝑝     3=54𝑝(45)3=(45)54𝑝  125=𝑝Since e>1,𝑒>1, the conic is a hyperbola. The eccentricity is e=54𝑒=54 and the directrix is x=125=2.4.𝑥=125=2.4.
3. Multiply the numerator and denominator by 12.12.r=72−2sinθ⋅(12)(12)r=7(12)2(12)−2(12)sinθr=721−sinθ𝑟=72−2sin𝜃⋅(12)(12)𝑟=7(12)2(12)−2(12)sin𝜃𝑟=721−sin𝜃Because sine is in the denominator, the directrix is y=−p.𝑦=−𝑝. Comparing to standard form, e=1.𝑒=1. Therefore, from the numerator,72=ep72=(1)p72=p72=𝑒𝑝72=(1)𝑝72=𝑝Because e=1,𝑒=1, the conic is a parabola. The eccentricity is e=1𝑒=1 and the directrix is y=−72=−3.5.𝑦=−72=−3.5.

TRY IT #1

Identify the conic with focus at the origin, the directrix, and the eccentricity for r=23−cosθ.𝑟=23−cos𝜃.

Graphing the Polar Equations of Conics

When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with 1. This enables us to determine e𝑒 and, therefore, the shape of the curve. The next step is to substitute values for θ𝜃 and solve for r𝑟 to plot a few key points. Setting θ𝜃 equal to 0,π2,π,0,𝜋2,𝜋, and 3π23𝜋2 provides the vertices so we can create a rough sketch of the graph.

EXAMPLE 2

Graphing a Parabola in Polar Form

Graph r=53+3cosθ.𝑟=53+3cos𝜃.

Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is 13.13.

r=53+3cosθ=5(13)3(13)+3(13)cosθr=531+cosθ𝑟=53+3cos𝜃=5(13)3(13)+3(13)cos𝜃𝑟=531+cos𝜃

Because e=1,𝑒=1, we will graph a parabola with a focus at the origin. The function has a  cosθ, cos𝜃, and there is an addition sign in the denominator, so the directrix is x=p.𝑥=𝑝.

53=ep53=(1)p53=p53=𝑒𝑝53=(1)𝑝53=𝑝

The directrix is x=53.𝑥=53.

Plotting a few key points as in Table 1 will enable us to see the vertices. See Figure 3.

	A	B	C	D
θ𝜃	00	π2𝜋2	π𝜋	3π23𝜋2
r=53+3cosθ𝑟=53+3cos𝜃	56≈0.8356≈0.83	53≈1.6753≈1.67	undefined	53≈1.6753≈1.67

Figure 3

Analysis

We can check our result with a graphing utility. See Figure 4.

Figure 4

EXAMPLE 3

Graphing a Hyperbola in Polar Form

Graph r=82−3sinθ.𝑟=82−3sin𝜃.

Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is 12.12.

r=82−3sinθ=8(12)2(12)−3(12)sinθr=41−32sinθ𝑟=82−3sin𝜃=8(12)2(12)−3(12)sin𝜃𝑟=41−32sin𝜃

Because e=32,e>1,𝑒=32,𝑒>1, so we will graph a hyperbola with a focus at the origin. The function has a sinθsin𝜃 term and there is a subtraction sign in the denominator, so the directrix is y=−p.𝑦=−𝑝.

4=ep4=(32)p4(23)=p83=p4=𝑒𝑝4=(32)𝑝4(23)=𝑝83=𝑝

The directrix is y=−83.𝑦=−83.

Plotting a few key points as in Table 2 will enable us to see the vertices. See Figure 5.

	A	B	C	D
θ𝜃	00	π2𝜋2	π𝜋	3π23𝜋2
r=82−3sinθ𝑟=82−3sin𝜃	44	−8−8	44	85=1.685=1.6

Figure 5

EXAMPLE 4

Graphing an Ellipse in Polar Form

Graph r=105−4cosθ.𝑟=105−4cos𝜃.

Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is 15.15.

r=105−4cosθ=10(15)5(15)−4(15)cosθr=21−45cosθ𝑟=105−4cos𝜃=10(15)5(15)−4(15)cos𝜃𝑟=21−45cos𝜃

Because e=45,e<1,𝑒=45,𝑒<1, so we will graph an ellipse with a focus at the origin. The function has a cosθ,cos𝜃, and there is a subtraction sign in the denominator, so the directrix is x=−p.𝑥=−𝑝.

2=ep2=(45)p2(54)=p52=p2=𝑒𝑝2=(45)𝑝2(54)=𝑝52=𝑝

The directrix is x=−52.𝑥=−52.

Plotting a few key points as in Table 3 will enable us to see the vertices. See Figure 6.

	A	B	C	D
θ𝜃	00	π2𝜋2	π𝜋	3π23𝜋2
r=105−4cosθ𝑟=105−4cos𝜃	1010	22	109≈1.1109≈1.1	22

Figure 6

Analysis

We can check our result using a graphing utility. See Figure 7.

Figure 7 r=105−4cosθ𝑟=105−4cos𝜃 graphed on a viewing window of [–3,12,1][–3,12,1] by [–4,4,1],θmin =0[–4,4,1],𝜃min =0 and θmax =2π.𝜃max =2𝜋.

TRY IT #2

Graph r=24−cosθ.𝑟=24−cos𝜃.

Defining Conics in Terms of a Focus and a Directrix

So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation.

HOW TO

Given the focus, eccentricity, and directrix of a conic, determine the polar equation.

1. Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of y,𝑦, we use the general polar form in terms of sine. If the directrix is given in terms of x,𝑥, we use the general polar form in terms of cosine.
2. Determine the sign in the denominator. If p<0,𝑝<0, use subtraction. If p>0,𝑝>0, use addition.
3. Write the coefficient of the trigonometric function as the given eccentricity.
4. Write the absolute value of p𝑝 in the numerator, and simplify the equation.

EXAMPLE 5

Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix

Find the polar form of the conic given a focus at the origin, e=3𝑒=3 and directrix y=−2.𝑦=−2.

Solution

The directrix is y=−p,𝑦=−𝑝, so we know the trigonometric function in the denominator is sine.

Because y=−2,–2<0,𝑦=−2,–2<0, so we know there is a subtraction sign in the denominator. We use the standard form of

r=ep1−esinθ𝑟=𝑒𝑝1−𝑒sin𝜃

and e=3𝑒=3 and |−2|=2=p.|−2|=2=𝑝.

Therefore,

r=(3)(2)1−3sinθr=61−3sinθ𝑟=(3)(2)1−3sin𝜃𝑟=61−3sin𝜃

EXAMPLE 6

Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix

Find the polar form of a conic given a focus at the origin, e=35,𝑒=35, and directrix x=4.𝑥=4.

Solution

Because the directrix is x=p,𝑥=𝑝, we know the function in the denominator is cosine. Because x=4,4>0,𝑥=4,4>0, so we know there is an addition sign in the denominator. We use the standard form of

r=ep1+ecosθ𝑟=𝑒𝑝1+𝑒cos𝜃

and e=35𝑒=35 and |4|=4=p.|4|=4=𝑝.

Therefore,

r=(35)(4)1+35cosθr=1251+35cosθr=1251(55)+35cosθr=12555+35cosθr=125⋅55+3cosθr=125+3cosθ𝑟=(35)(4)1+35cos𝜃𝑟=1251+35cos𝜃𝑟=1251(55)+35cos𝜃𝑟=12555+35cos𝜃𝑟=125⋅55+3cos𝜃𝑟=125+3cos𝜃

TRY IT #3

Find the polar form of the conic given a focus at the origin, e=1,𝑒=1, and directrix x=−1.𝑥=−1.

EXAMPLE 7

Converting a Conic in Polar Form to Rectangular Form

Convert the conic r=15−5sinθ𝑟=15−5sin𝜃 to rectangular form.

Solution

We will rearrange the formula to use the identities  r=x2+y2−−−−−−√,x=rcosθ,and y=rsinθ. 𝑟=𝑥2+𝑦2,𝑥=𝑟cos𝜃,and 𝑦=𝑟sin𝜃.

                         r=15−5sinθ r⋅(5−5sinθ)=15−5sinθ⋅(5−5sinθ)      5r−5rsinθ=1                       5r=1+5rsinθ                   25r2=(1+5rsinθ)2        25(x2+y2)=(1+5y)2      25×2+25y2=1+10y+25y2        25×2−10y=1Eliminate the fraction.Distribute.Isolate 5r.Square both sides.Substitute r=x2+y2−−−−−−√and y=rsinθ.Distribute and use FOIL.Rearrange terms and set equal to 1.                         𝑟=15−5sin𝜃 𝑟⋅(5−5sin𝜃)=15−5sin𝜃⋅(5−5sin𝜃)Eliminate the fraction.      5𝑟−5𝑟sin𝜃=1Distribute.                       5𝑟=1+5𝑟sin𝜃Isolate 5𝑟.                   25𝑟2=(1+5𝑟sin𝜃)2Square both sides.        25(𝑥2+𝑦2)=(1+5𝑦)2Substitute 𝑟=𝑥2+𝑦2and 𝑦=𝑟sin𝜃.      25𝑥2+25𝑦2=1+10𝑦+25𝑦2Distribute and use FOIL.        25𝑥2−10𝑦=1Rearrange terms and set equal to 1.

TRY IT #4

Convert the conic r=21+2cosθ𝑟=21+2cos𝜃 to rectangular form.