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Study Guides > Precalculus II

Conic Sections in Polar Coordinates

Learning Objectives

By the end of this section, you will be able to:
  • Identify a conic in polar form.
  • Graph the polar equations of conics.
  • Define conics in terms of a focus and a directrix.
The planets and their orbits around the sun. (Pluto is included.) 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+{y}^{2} 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\left(r,\theta \right) 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=\frac{PF}{PD} 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 ee. For a conic with eccentricity ee,
  • if 0e<10\le e<1, the conic is an ellipse
  • if e=1e=1, the conic is a parabola
  • if e>1e>1, the conic is an hyperbola
With this definition, we may now define a conic in terms of the directrix, x=±px=\pm p, the eccentricity ee, and the angle θ\theta . Thus, each conic may be written as a polar equation, an equation written in terms of rr and θ\theta .

A General Note: The Polar Equation for a Conic

For a conic with a focus at the origin, if the directrix is x=±px=\pm p, where pp is a positive real number, and the eccentricity is a positive real number ee, the conic has a polar equation
r=ep1±e cos θr=\frac{ep}{1\pm e\text{ }\cos \text{ }\theta }
For a conic with a focus at the origin, if the directrix is y=±py=\pm p, where pp is a positive real number, and the eccentricity is a positive real number ee, the conic has a polar equation
r=ep1±e sin θr=\frac{ep}{1\pm e\text{ }\sin \text{ }\theta }

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 yy.

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+2 sin θr=\frac{6}{3+2\text{ }\sin \text{ }\theta }
  2. r=124+5 cos θr=\frac{12}{4+5\text{ }\cos \text{ }\theta }
  3. r=722 sin θr=\frac{7}{2 - 2\text{ }\sin \text{ }\theta }

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\frac{1}{c}, where cc is that constant.
  1. Multiply the numerator and denominator by 13\frac{1}{3}.
    r=63+2sin θ(13)(13)=6(13)3(13)+2(13)sin θ=21+23 sin θr=\frac{6}{3+2\sin \text{ }\theta }\cdot \frac{\left(\frac{1}{3}\right)}{\left(\frac{1}{3}\right)}=\frac{6\left(\frac{1}{3}\right)}{3\left(\frac{1}{3}\right)+2\left(\frac{1}{3}\right)\sin \text{ }\theta }=\frac{2}{1+\frac{2}{3}\text{ }\sin \text{ }\theta }
    Because sin θ\sin \text{ }\theta is in the denominator, the directrix is y=py=p. Comparing to standard form, note that e=23e=\frac{2}{3}. Therefore, from the numerator,
     2=ep 2=23p(32)2=(32)23p 3=p\begin{array}{l}\text{ }2=ep\hfill \\ \text{ }2=\frac{2}{3}p\hfill \\ \left(\frac{3}{2}\right)2=\left(\frac{3}{2}\right)\frac{2}{3}p\hfill \\ \text{ }3=p\hfill \end{array}
    Since e<1e<1, the conic is an ellipse. The eccentricity is e=23e=\frac{2}{3} and the directrix is y=3y=3.
  2. Multiply the numerator and denominator by 14\frac{1}{4}.
    r=124+5 cos θ(14)(14)r=12(14)4(14)+5(14)cos θr=31+54 cos θ\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{12}{4+5\text{ }\cos \text{ }\theta }\cdot \frac{\left(\frac{1}{4}\right)}{\left(\frac{1}{4}\right)}\hfill \end{array}\hfill \\ r=\frac{12\left(\frac{1}{4}\right)}{4\left(\frac{1}{4}\right)+5\left(\frac{1}{4}\right)\cos \text{ }\theta }\hfill \\ r=\frac{3}{1+\frac{5}{4}\text{ }\cos \text{ }\theta }\hfill \end{array}
    Because  cosθ\text{ cos}\theta is in the denominator, the directrix is x=px=p. Comparing to standard form, e=54e=\frac{5}{4}. Therefore, from the numerator,
     3=ep 3=54p(45)3=(45)54p 125=p\begin{array}{l}\text{ }3=ep\hfill \\ \text{ }3=\frac{5}{4}p\hfill \\ \left(\frac{4}{5}\right)3=\left(\frac{4}{5}\right)\frac{5}{4}p\hfill \\ \text{ }\frac{12}{5}=p\hfill \end{array}
    Since e>1e>1, the conic is a hyperbola. The eccentricity is e=54e=\frac{5}{4} and the directrix is x=125=2.4x=\frac{12}{5}=2.4.
  3. Multiply the numerator and denominator by 12\frac{1}{2}.
    r=722 sin θ(12)(12)r=7(12)2(12)2(12) sin θr=721sin θ\begin{array}{l}\hfill \\ \hfill \\ \begin{array}{l}r=\frac{7}{2 - 2\text{ }\sin \text{ }\theta }\cdot \frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)}\hfill \\ r=\frac{7\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)-2\left(\frac{1}{2}\right)\text{ }\sin \text{ }\theta }\hfill \\ r=\frac{\frac{7}{2}}{1-\sin \text{ }\theta }\hfill \end{array}\hfill \end{array}
    Because sine is in the denominator, the directrix is y=py=-p. Comparing to standard form, e=1e=1. Therefore, from the numerator,
    72=ep72=(1)p72=p\begin{array}{l}\frac{7}{2}=ep\\ \frac{7}{2}=\left(1\right)p\\ \frac{7}{2}=p\end{array}
    Because e=1e=1, the conic is a parabola. The eccentricity is e=1e=1 and the directrix is y=72=3.5y=-\frac{7}{2}=-3.5.

Try It 1

Identify the conic with focus at the origin, the directrix, and the eccentricity for r=23cos θr=\frac{2}{3-\cos \text{ }\theta }. Solution

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 θ\theta and solve for rr to plot a few key points. Setting θ\theta equal to 0,π2,π0,\frac{\pi }{2},\pi , and 3π2\frac{3\pi }{2} provides the vertices so we can create a rough sketch of the graph.

Example 2: Graphing a Parabola in Polar Form

Graph r=53+3 cos θr=\frac{5}{3+3\text{ }\cos \text{ }\theta }.

Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is 13\frac{1}{3}.
r=53+3 cos θ=5(13)3(13)+3(13)cos θr=531+cos θ\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{5}{3+3\text{ }\cos \text{ }\theta }=\frac{5\left(\frac{1}{3}\right)}{3\left(\frac{1}{3}\right)+3\left(\frac{1}{3}\right)\cos \text{ }\theta }\hfill \end{array}\hfill \\ r=\frac{\frac{5}{3}}{1+\cos \text{ }\theta }\hfill \end{array}
Because e=1e=1, we will graph a parabola with a focus at the origin. The function has a cos θ \cos \text{ }\theta , and there is an addition sign in the denominator, so the directrix is x=px=p.
53=ep53=(1)p53=p\begin{array}{l}\frac{5}{3}=ep\\ \frac{5}{3}=\left(1\right)p\\ \frac{5}{3}=p\end{array}
The directrix is x=53x=\frac{5}{3}. Plotting a few key points as in the table below will enable us to see the vertices.
A B C D
θ\theta 00 π2\frac{\pi }{2} π\pi 3π2\frac{3\pi }{2}
r=53+3 cos θr=\frac{5}{3+3\text{ }\cos \text{ }\theta } 560.83\frac{5}{6}\approx 0.83 531.67\frac{5}{3}\approx 1.67 undefined 531.67\frac{5}{3}\approx 1.67
Figure 3

Analysis of the Solution

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

Example 3: Graphing a Hyperbola in Polar Form

Graph r=823 sin θr=\frac{8}{2 - 3\text{ }\sin \text{ }\theta }.

Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is 12\frac{1}{2}.
r=823sin θ=8(12)2(12)3(12)sin θr=4132 sin θ\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{8}{2 - 3\sin \text{ }\theta }=\frac{8\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)-3\left(\frac{1}{2}\right)\sin \text{ }\theta }\hfill \end{array}\hfill \\ r=\frac{4}{1-\frac{3}{2}\text{ }\sin \text{ }\theta }\hfill \end{array}
Because e=32,e>1e=\frac{3}{2},e>1, so we will graph a hyperbola with a focus at the origin. The function has a sin θ\sin \text{ }\theta term and there is a subtraction sign in the denominator, so the directrix is y=py=-p.
 4=ep 4=(32)p4(23)=p 83=p\begin{array}{l}\text{ }4=ep\hfill \\ \text{ }4=\left(\frac{3}{2}\right)p\hfill \\ 4\left(\frac{2}{3}\right)=p\hfill \\ \text{ }\frac{8}{3}=p\hfill \end{array}
The directrix is y=83y=-\frac{8}{3}. Plotting a few key points as in the table below will enable us to see the vertices.
A B C D
θ\theta 00 π2\frac{\pi }{2} π\pi 3π2\frac{3\pi }{2}
r=823sinθr=\frac{8}{2 - 3\sin \theta } 44 8-8 44 85=1.6\frac{8}{5}=1.6
Figure 5

Example 4: Graphing an Ellipse in Polar Form

Graph r=1054 cos θr=\frac{10}{5 - 4\text{ }\cos \text{ }\theta }.

Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is 15\frac{1}{5}.
r=1054cos θ=10(15)5(15)4(15)cos θr=2145 cos θ\begin{array}{l}\hfill \\ \begin{array}{l}r=\frac{10}{5 - 4\cos \text{ }\theta }=\frac{10\left(\frac{1}{5}\right)}{5\left(\frac{1}{5}\right)-4\left(\frac{1}{5}\right)\cos \text{ }\theta }\hfill \\ r=\frac{2}{1-\frac{4}{5}\text{ }\cos \text{ }\theta }\hfill \end{array}\hfill \end{array}
Because e=45,e<1e=\frac{4}{5},e<1, so we will graph an ellipse with a focus at the origin. The function has a cosθ\text{cos}\theta , and there is a subtraction sign in the denominator, so the directrix is x=px=-p.
 2=ep 2=(45)p2(54)=p 52=p\begin{array}{l}\text{ }2=ep\hfill \\ \text{ }2=\left(\frac{4}{5}\right)p\hfill \\ 2\left(\frac{5}{4}\right)=p\hfill \\ \text{ }\frac{5}{2}=p\hfill \end{array}
The directrix is x=52x=-\frac{5}{2}. Plotting a few key points as in the table below will enable us to see the vertices.
A B C D
θ\theta 00 π2\frac{\pi }{2} π\pi 3π2\frac{3\pi }{2}
r=1054 cos θr=\frac{10}{5 - 4\text{ }\cos \text{ }\theta } 1010 22 1091.1\frac{10}{9}\approx 1.1 22
Figure 6

Analysis of the Solution

We can check our result using a graphing utility.
Figure 7. r=1054 cos θr=\frac{10}{5 - 4\text{ }\cos \text{ }\theta } graphed on a viewing window of [3,12,1]\left[-3,12,1\right] by [4,4,1],θmin =0\left[-4,4,1\right],\theta \text{min =}0 and θmax =2π\theta \text{max =}2\pi .

Try It 2

Graph r=24cos θr=\frac{2}{4-\cos \text{ }\theta }. Solution

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 yy, we use the general polar form in terms of sine. If the directrix is given in terms of xx, we use the general polar form in terms of cosine.
  2. Determine the sign in the denominator. If p<0p<0, use subtraction. If p>0p>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=2y=-2.

Solution

The directrix is y=py=-p, so we know the trigonometric function in the denominator is sine. Because y=2,2<0y=-2,-2<0, so we know there is a subtraction sign in the denominator. We use the standard form of
r=ep1e sin θr=\frac{ep}{1-e\text{ }\sin \text{ }\theta }
and e=3e=3 and 2=2=p|-2|=2=p. Therefore,
r=(3)(2)13 sin θr=613 sin θ\begin{array}{l}\hfill \\ \begin{array}{l}r=\frac{\left(3\right)\left(2\right)}{1 - 3\text{ }\sin \text{ }\theta }\hfill \\ r=\frac{6}{1 - 3\text{ }\sin \text{ }\theta }\hfill \end{array}\hfill \end{array}

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=35e=\frac{3}{5}, and directrix x=4x=4.

Solution

Because the directrix is x=px=p, we know the function in the denominator is cosine. Because x=4,4>0x=4,4>0, so we know there is an addition sign in the denominator. We use the standard form of
r=ep1+e cos θr=\frac{ep}{1+e\text{ }\cos \text{ }\theta }
and e=35e=\frac{3}{5} and 4=4=p|4|=4=p. Therefore,
r=(35)(4)1+35cosθr=1251+35cosθr=1251(55)+35cosθr=12555+35cosθr=12555+3cosθr=125+3cosθ\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{\left(\frac{3}{5}\right)\left(4\right)}{1+\frac{3}{5}\cos \theta }\hfill \end{array}\hfill \\ r=\frac{\frac{12}{5}}{1+\frac{3}{5}\cos \theta }\hfill \\ r=\frac{\frac{12}{5}}{1\left(\frac{5}{5}\right)+\frac{3}{5}\cos \theta }\hfill \\ r=\frac{\frac{12}{5}}{\frac{5}{5}+\frac{3}{5}\cos \theta }\hfill \\ r=\frac{12}{5}\cdot \frac{5}{5+3\cos \theta }\hfill \\ r=\frac{12}{5+3\cos \theta }\hfill \end{array}

Try It 3

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

Example 5: Converting a Conic in Polar Form to Rectangular Form

Convert the conic r=155sinθr=\frac{1}{5 - 5\sin \theta } to rectangular form.

Solution

We will rearrange the formula to use the identities r=x2+y2,x=rcosθ,and y=rsinθ r=\sqrt{{x}^{2}+{y}^{2}},x=r\cos \theta ,\text{and }y=r\sin \theta .
 r=155sinθr(55sinθ)=155sinθ(55sinθ)Eliminate the fraction. 5r5rsinθ=1Distribute. 5r=1+5rsinθIsolate 5r. 25r2=(1+5rsinθ)2Square both sides. 25(x2+y2)=(1+5y)2Substitute r=x2+y2 and y=rsinθ. 25x2+25y2=1+10y+25y2Distribute and use FOIL. 25x210y=1Rearrange terms and set equal to 1.\begin{array}{ll}\text{ }r=\frac{1}{5 - 5\sin \theta }\hfill & \hfill \\ r\cdot \left(5 - 5\sin \theta \right)=\frac{1}{5 - 5\sin \theta }\cdot \left(5 - 5\sin \theta \right)\hfill & \text{Eliminate the fraction}.\hfill \\ \text{ }5r - 5r\sin \theta =1\hfill & \text{Distribute}.\hfill \\ \text{ }5r=1+5r\sin \theta \hfill & \text{Isolate }5r.\hfill \\ \text{ }25{r}^{2}={\left(1+5r\sin \theta \right)}^{2}\hfill & \text{Square both sides}.\hfill \\ \text{ }25\left({x}^{2}+{y}^{2}\right)={\left(1+5y\right)}^{2}\hfill & \text{Substitute }r=\sqrt{{x}^{2}+{y}^{2}}\text{ and }y=r\sin \theta .\hfill \\ \text{ }25{x}^{2}+25{y}^{2}=1+10y+25{y}^{2}\hfill & \text{Distribute and use FOIL}.\hfill \\ \text{ }25{x}^{2}-10y=1\hfill & \text{Rearrange terms and set equal to 1}.\hfill \end{array}

Try It 4

Convert the conic r=21+2 cos θr=\frac{2}{1+2\text{ }\cos \text{ }\theta } to rectangular form. Solution

Key Concepts

  • Any conic may be determined by a single focus, the corresponding eccentricity, and the directrix. We can also define a conic in terms of a fixed point, the focus P(r,θ)P\left(r,\theta \right) at the pole, and a line, the directrix, which is perpendicular to the polar axis.
  • A conic is the set of all points e=PFPDe=\frac{PF}{PD}, where eccentricity ee is a positive real number. Each conic may be written in terms of its polar equation.
  • The polar equations of conics can be graphed.
  • Conics can be defined in terms of a focus, a directrix, and eccentricity.
  • We can use the identities r=x2+y2,x=r cos θr=\sqrt{{x}^{2}+{y}^{2}},x=r\text{ }\cos \text{ }\theta , and y=r sin θy=r\text{ }\sin \text{ }\theta to convert the equation for a conic from polar to rectangular form.

Glossary

eccentricity
the ratio of the distances from a point PP on the graph to the focus FF and to the directrix DD represented by e=PFPDe=\frac{PF}{PD}, where ee is a positive real number
polar equation
an equation of a curve in polar coordinates rr and θ\theta

Section Exercises

1. Explain how eccentricity determines which conic section is given. 2. If a conic section is written as a polar equation, what must be true of the denominator? 3. If a conic section is written as a polar equation, and the denominator involves sin θ\sin \text{ }\theta , what conclusion can be drawn about the directrix? 4. If the directrix of a conic section is perpendicular to the polar axis, what do we know about the equation of the graph? 5. What do we know about the focus/foci of a conic section if it is written as a polar equation? For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. 6. r=612 cos θr=\frac{6}{1 - 2\text{ }\cos \text{ }\theta } 7. r=344 sin θr=\frac{3}{4 - 4\text{ }\sin \text{ }\theta } 8. r=843 cos θr=\frac{8}{4 - 3\text{ }\cos \text{ }\theta } 9. r=51+2 sin θr=\frac{5}{1+2\text{ }\sin \text{ }\theta } 10. r=164+3 cos θr=\frac{16}{4+3\text{ }\cos \text{ }\theta } 11. r=310+10 cos θr=\frac{3}{10+10\text{ }\cos \text{ }\theta } 12. r=21cos θr=\frac{2}{1-\cos \text{ }\theta } 13. r=47+2 cos θr=\frac{4}{7+2\text{ }\cos \text{ }\theta } 14. r(1cos θ)=3r\left(1-\cos \text{ }\theta \right)=3 15. r(3+5sin θ)=11r\left(3+5\sin \text{ }\theta \right)=11 16. r(45sin θ)=1r\left(4 - 5\sin \text{ }\theta \right)=1 17. r(7+8cos θ)=7r\left(7+8\cos \text{ }\theta \right)=7 For the following exercises, convert the polar equation of a conic section to a rectangular equation. 18. r=41+3 sin θr=\frac{4}{1+3\text{ }\sin \text{ }\theta } 19. r=253 sin θr=\frac{2}{5 - 3\text{ }\sin \text{ }\theta } 20. r=832 cos θr=\frac{8}{3 - 2\text{ }\cos \text{ }\theta } 21. r=32+5 cos θr=\frac{3}{2+5\text{ }\cos \text{ }\theta } 22. r=42+2 sin θr=\frac{4}{2+2\text{ }\sin \text{ }\theta } 23. r=388 cos θr=\frac{3}{8 - 8\text{ }\cos \text{ }\theta } 24. r=26+7 cos θr=\frac{2}{6+7\text{ }\cos \text{ }\theta } 25. r=5511 sin θr=\frac{5}{5 - 11\text{ }\sin \text{ }\theta } 26. r(5+2 cos θ)=6r\left(5+2\text{ }\cos \text{ }\theta \right)=6 27. r(2cos θ)=1r\left(2-\cos \text{ }\theta \right)=1 28. r(2.52.5 sin θ)=5r\left(2.5 - 2.5\text{ }\sin \text{ }\theta \right)=5 29. r=6sec θ2+3 sec θr=\frac{6\sec \text{ }\theta }{-2+3\text{ }\sec \text{ }\theta } 30. r=6csc θ3+2 csc θr=\frac{6\csc \text{ }\theta }{3+2\text{ }\csc \text{ }\theta } For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. 31. r=52+cos θr=\frac{5}{2+\cos \text{ }\theta } 32. r=23+3 sin θr=\frac{2}{3+3\text{ }\sin \text{ }\theta } 33. r=1054 sin θr=\frac{10}{5 - 4\text{ }\sin \text{ }\theta } 34. r=31+2 cos θr=\frac{3}{1+2\text{ }\cos \text{ }\theta } 35. r=845 cos θr=\frac{8}{4 - 5\text{ }\cos \text{ }\theta } 36. r=344 cos θr=\frac{3}{4 - 4\text{ }\cos \text{ }\theta } 37. r=21sin θr=\frac{2}{1-\sin \text{ }\theta } 38. r=63+2 sin θr=\frac{6}{3+2\text{ }\sin \text{ }\theta } 39. r(1+cos θ)=5r\left(1+\cos \text{ }\theta \right)=5 40. r(34sin θ)=9r\left(3 - 4\sin \text{ }\theta \right)=9 41. r(32sin θ)=6r\left(3 - 2\sin \text{ }\theta \right)=6 42. r(64cos θ)=5r\left(6 - 4\cos \text{ }\theta \right)=5 For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. 43. Directrix: x=4;e=15x=4;e=\frac{1}{5} 44. Directrix: x=4;e=5x=-4;e=5 45. Directrix: y=2;e=2y=2;e=2 46. Directrix: y=2;e=12y=-2;e=\frac{1}{2} 47. Directrix: x=1;e=1x=1;e=1 48. Directrix: x=1;e=1x=-1;e=1 49. Directrix: x=14;e=72x=-\frac{1}{4};e=\frac{7}{2} 50. Directrix: y=25;e=72y=\frac{2}{5};e=\frac{7}{2} 51. Directrix: y=4;e=32y=4;e=\frac{3}{2} 52. Directrix: x=2;e=83x=-2;e=\frac{8}{3} 53. Directrix: x=5;e=34x=-5;e=\frac{3}{4} 54. Directrix: y=2;e=2.5y=2;e=2.5 55. Directrix: x=3;e=13x=-3;e=\frac{1}{3} Equations of conics with an xyxy term have rotated graphs. For the following exercises, express each equation in polar form with rr as a function of θ\theta . 56. xy=2xy=2 57. x2+xy+y2=4{x}^{2}+xy+{y}^{2}=4 58. 2x2+4xy+2y2=92{x}^{2}+4xy+2{y}^{2}=9 59. 16x2+24xy+9y2=416{x}^{2}+24xy+9{y}^{2}=4 60. 2xy+y=12xy+y=1