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 , and the ratio of the distances of each to a point on the graph. Consider the $\,x=2+{y}^{2}\,$ shown in (Figure).

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 $\,P\left(r,\theta \right)\,$ at the pole, and a line, the directrix, which is perpendicular to the polar axis.

If $\,F\,$ is a fixed point, the focus, and $\,D\,$ is a fixed line, the directrix, then we can let $\,e\,$ 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 $\,P\,$ such that $\,e=\frac{PF}{PD}\,$ is a conic. In other words, we can define a conic as the set of all points $\,P\,$ with the property that the ratio of the distance from $\,P\,$ to $\,F\,$ to the distance from $\,P\,$ to $\,D\,$ is equal to the constant $\,e.$

For a conic with eccentricity $\,e,$

if $\,0\le e<1,$ the conic is an ellipse
if $\,e=1,$ the conic is a parabola
if $\,e>1,$ the conic is an hyperbola

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

The Polar Equation for a Conic

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

$r=\frac{ep}{1±e\text{ }\mathrm{cos}\text{ }\theta }$

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

$r=\frac{ep}{1±e\text{ }\mathrm{sin}\text{ }\theta }$

How To

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

Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form.
Identify the eccentricity $\,e\,$ as the coefficient of the trigonometric function in the denominator.
Compare $\,e\,$ with 1 to determine the shape of the conic.
Determine the directrix as $\,x=p\,$ if cosine is in the denominator and $\,y=p\,$ if sine is in the denominator. Set $\,ep\,$ equal to the numerator in standard form to solve for $\,x\,$ or $\,y.$

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.

$r=\frac{6}{3+2\text{ }\mathrm{sin}\text{ }\theta }$
$r=\frac{12}{4+5\text{ }\mathrm{cos}\text{ }\theta }$
$r=\frac{7}{2-2\text{ }\mathrm{sin}\text{ }\theta }$

Show 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, $\,\frac{1}{c},$ where $\,c\,$ is that constant.

Multiply the numerator and denominator by $\,\frac{1}{3}.$
$r=\frac{6}{3+2\mathrm{sin}\text{ }\theta }\cdot \frac{\left(\frac{1}{3}\right)}{\left(\frac{1}{3}\right)}=\frac{6\left(\frac{1}{3}\right)}{3\left(\frac{1}{3}\right)+2\left(\frac{1}{3}\right)\mathrm{sin}\text{ }\theta }=\frac{2}{1+\frac{2}{3}\text{ }\mathrm{sin}\text{ }\theta }$

Because $\mathrm{sin}\text{ }\theta$ is in the denominator, the directrix is $\,y=p.\,$ Comparing to standard form, note that $\,e=\frac{2}{3}.$ Therefore, from the numerator,

$\begin{array}{l}\text{ }2=ep\hfill \\ \text{ }2=\frac{2}{3}p\hfill \\ \left(\frac{3}{2}\right)2=\left(\frac{3}{2}\right)\frac{2}{3}p\hfill \\ \text{ }3=p\hfill \end{array}$

Since $\,e<1,$ the conic is an . The eccentricity is $\,e=\frac{2}{3}$ and the directrix is $\,y=3.$
Multiply the numerator and denominator by $\,\frac{1}{4}.$
$\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{12}{4+5\text{ }\mathrm{cos}\text{ }\theta }\cdot \frac{\left(\frac{1}{4}\right)}{\left(\frac{1}{4}\right)}\hfill \end{array}\hfill \\ r=\frac{12\left(\frac{1}{4}\right)}{4\left(\frac{1}{4}\right)+5\left(\frac{1}{4}\right)\mathrm{cos}\text{ }\theta }\hfill \\ r=\frac{3}{1+\frac{5}{4}\text{ }\mathrm{cos}\text{ }\theta }\hfill \end{array}$

Because $\text{ cos}\,\theta$ is in the denominator, the directrix is $\,x=p.\,$ Comparing to standard form, $\,e=\frac{5}{4}.\,$ Therefore, from the numerator,

$\begin{array}{l}\text{ }3=ep\hfill \\ \text{ }3=\frac{5}{4}p\hfill \\ \,\left(\frac{4}{5}\right)3=\left(\frac{4}{5}\right)\frac{5}{4}p\hfill \\ \text{ }\,\frac{12}{5}=p\hfill \end{array}$

Since $\,e>1,$ the conic is a . The eccentricity is $\,e=\frac{5}{4}\,$ and the directrix is $\,x=\frac{12}{5}=2.4.$
Multiply the numerator and denominator by $\,\frac{1}{2}.$
$\begin{array}{l}\hfill \\ \hfill \\ \begin{array}{l}r=\frac{7}{2-2\text{ }\mathrm{sin}\text{ }\theta }\cdot \frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)}\hfill \\ r=\frac{7\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)-2\left(\frac{1}{2}\right)\text{ }\mathrm{sin}\text{ }\theta }\hfill \\ r=\frac{\frac{7}{2}}{1-\mathrm{sin}\text{ }\theta }\hfill \end{array}\hfill \end{array}$

Because sine is in the denominator, the directrix is $\,y=-p.\,$ Comparing to standard form, $\,e=1.\,$ Therefore, from the numerator,

$\begin{array}{l}\frac{7}{2}=ep\\ \frac{7}{2}=\left(1\right)p\\ \frac{7}{2}=p\end{array}$

Because $\,e=1,$ the conic is a . The eccentricity is $\,e=1\,$ and the directrix is $\,y=-\frac{7}{2}=-3.5.$

Try It

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

Show Solution

ellipse; $\,e=\frac{1}{3};\,x=-2$

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 $\,e\,$ and, therefore, the shape of the curve. The next step is to substitute values for $\,\theta \,$ and solve for $\,r\,$ to plot a few key points. Setting $\,\theta \,$ equal to $\,0,\frac{\pi }{2},\pi ,$ and $\,\frac{3\pi }{2}\,$ provides the vertices so we can create a rough sketch of the graph.

Graphing a Parabola in Polar Form

Graph $\,r=\frac{5}{3+3\text{ }\mathrm{cos}\text{ }\theta }.$

Show Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is $\,\frac{1}{3}.$

$\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{5}{3+3\text{ }\mathrm{cos}\text{ }\theta }=\frac{5\left(\frac{1}{3}\right)}{3\left(\frac{1}{3}\right)+3\left(\frac{1}{3}\right)\mathrm{cos}\text{ }\theta }\hfill \end{array}\hfill \\ r=\frac{\frac{5}{3}}{1+\mathrm{cos}\text{ }\theta }\hfill \end{array}$

Because $\,e=1,$ we will graph a with a focus at the origin. The function has a $\mathrm{cos}\text{ }\theta ,$ and there is an addition sign in the denominator, so the directrix is $\,x=p.$

$\begin{array}{l}\frac{5}{3}=ep\\ \frac{5}{3}=\left(1\right)p\\ \frac{5}{3}=p\end{array}$

The directrix is $\,x=\frac{5}{3}.$

Plotting a few key points as in (Figure) will enable us to see the vertices. See (Figure).

	A	B	C	D
$\theta$	$0$	$\frac{\pi }{2}$	$\pi$	$\frac{3\pi }{2}$
$r=\frac{5}{3+3\text{ }\mathrm{cos}\text{ }\theta }$	$\frac{5}{6}\approx 0.83$	$\frac{5}{3}\approx 1.67$	undefined	$\frac{5}{3}\approx 1.67$

Figure 3.

Analysis

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

Figure 4.

Graphing a Hyperbola in Polar Form

Graph $\,r=\frac{8}{2-3\text{ }\mathrm{sin}\text{ }\theta }.$

Show Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is $\,\frac{1}{2}.$

$\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{8}{2-3\mathrm{sin}\text{ }\theta }=\frac{8\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)-3\left(\frac{1}{2}\right)\mathrm{sin}\text{ }\theta }\hfill \end{array}\hfill \\ r=\frac{4}{1-\frac{3}{2}\text{ }\mathrm{sin}\text{ }\theta }\hfill \end{array}$

Because $\,e=\frac{3}{2},e>1,$ so we will graph a with a focus at the origin. The function has a $\,\mathrm{sin}\text{ }\theta \,$ term and there is a subtraction sign in the denominator, so the directrix is $\,y=-p.$

$\begin{array}{l}\text{ }4=ep\hfill \\ \text{ }4=\left(\frac{3}{2}\right)p\hfill \\ 4\left(\frac{2}{3}\right)=p\hfill \\ \text{ }\frac{8}{3}=p\hfill \end{array}$

The directrix is $\,y=-\frac{8}{3}.$

Plotting a few key points as in (Figure) will enable us to see the vertices. See (Figure).

	A	B	C	D
$\theta$	$0$	$\frac{\pi }{2}$	$\pi$	$\frac{3\pi }{2}$
$r=\frac{8}{2-3\mathrm{sin}\,\theta }$	$4$	$-8$	$4$	$\frac{8}{5}=1.6$

Figure 5.

Graphing an Ellipse in Polar Form

Graph $\,r=\frac{10}{5-4\text{ }\mathrm{cos}\text{ }\theta }.$

Show Solution

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is $\,\frac{1}{5}.$

$\begin{array}{l}\hfill \\ \begin{array}{l}r=\frac{10}{5-4\mathrm{cos}\text{ }\theta }=\frac{10\left(\frac{1}{5}\right)}{5\left(\frac{1}{5}\right)-4\left(\frac{1}{5}\right)\mathrm{cos}\text{ }\theta }\hfill \\ r=\frac{2}{1-\frac{4}{5}\text{ }\mathrm{cos}\text{ }\theta }\hfill \end{array}\hfill \end{array}$

Because $\,e=\frac{4}{5},e<1,$ so we will graph an with a at the origin. The function has a $\,\text{cos}\,\theta ,$ and there is a subtraction sign in the denominator, so the is $\,x=-p.$

$\begin{array}{l}\text{ }2=ep\hfill \\ \text{ }2=\left(\frac{4}{5}\right)p\hfill \\ 2\left(\frac{5}{4}\right)=p\hfill \\ \text{ }\frac{5}{2}=p\hfill \end{array}$

The directrix is $\,x=-\frac{5}{2}.$

Plotting a few key points as in (Figure) will enable us to see the vertices. See (Figure).

	A	B	C	D
$\theta$	$0$	$\frac{\pi }{2}$	$\pi$	$\frac{3\pi }{2}$
$r=\frac{10}{5-4\text{ }\mathrm{cos}\text{ }\theta }$	$10$	$2$	$\frac{10}{9}\approx 1.1$	$2$

Figure 6.

Analysis

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

Figure 7. $r=\frac{10}{5-4\text{ }\mathrm{cos}\text{ }\theta }\,$ graphed on a viewing window of $\,\left[–3,12,1\right]\,$ by $\,\left[–4,4,1\right],\theta \,\text{min =}\,0$ and $\,\theta \,\text{max =}\,2\pi .$

Try It

Graph $\,r=\frac{2}{4-\mathrm{cos}\text{ }\theta }.$

Show 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.

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

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 given a at the origin, $\,e=3\,$ and $\,y=-2.$

Show Solution

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

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

$r=\frac{ep}{1-e\text{ }\mathrm{sin}\text{ }\theta }$

and $\,e=3\,$ and $\,|-2|=2=p.$

Therefore,

$\begin{array}{l}\hfill \\ \begin{array}{l}r=\frac{\left(3\right)\left(2\right)}{1-3\text{ }\mathrm{sin}\text{ }\theta }\hfill \\ r=\frac{6}{1-3\text{ }\mathrm{sin}\text{ }\theta }\hfill \end{array}\hfill \end{array}$

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

Find the given a at the origin, $\,e=\frac{3}{5},$ and $\,x=4.$

Show Solution

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

$r=\frac{ep}{1+e\text{ }\mathrm{cos}\text{ }\theta }$

and $\,e=\frac{3}{5}\,$ and $\,|4|=4=p.$

Therefore,

$\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{\left(\frac{3}{5}\right)\left(4\right)}{1+\frac{3}{5}\,\mathrm{cos}\,\theta }\hfill \end{array}\hfill \\ r=\frac{\frac{12}{5}}{1+\frac{3}{5}\,\mathrm{cos}\,\theta }\hfill \\ r=\frac{\frac{12}{5}}{1\left(\frac{5}{5}\right)+\frac{3}{5}\,\mathrm{cos}\,\theta }\hfill \\ r=\frac{\frac{12}{5}}{\frac{5}{5}+\frac{3}{5}\,\mathrm{cos}\,\theta }\hfill \\ r=\frac{12}{5}\cdot \frac{5}{5+3\,\mathrm{cos}\,\theta }\hfill \\ r=\frac{12}{5+3\,\mathrm{cos}\,\theta }\hfill \end{array}$

Try It

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

Show Solution

$r=\frac{1}{1-\mathrm{cos}\theta }$

Converting a Conic in Polar Form to Rectangular Form

Convert the conic $\,r=\frac{1}{5-5\mathrm{sin}\,\theta }$ to rectangular form.

Show Solution

We will rearrange the formula to use the identities $r=\sqrt{{x}^{2}+{y}^{2}},x=r\,\mathrm{cos}\,\theta ,\text{and }y=r\,\mathrm{sin}\,\theta .$

$\begin{array}{ll}\text{ }r=\frac{1}{5-5\,\mathrm{sin}\,\theta }\hfill & \hfill \\ r\cdot \left(5-5\,\mathrm{sin}\,\theta \right)=\frac{1}{5-5\,\mathrm{sin}\,\theta }\cdot \left(5-5\,\mathrm{sin}\,\theta \right)\hfill & \text{Eliminate the fraction}.\hfill \\ \text{ }\,5r-5r\,\mathrm{sin}\,\theta =1\hfill & \text{Distribute}.\hfill \\ \text{ }5r=1+5r\,\mathrm{sin}\,\theta \hfill & \text{Isolate }5r.\hfill \\ \text{ }25{r}^{2}={\left(1+5r\,\mathrm{sin}\,\theta \right)}^{2}\hfill & \text{Square both sides}.\hfill \\ \text{ }25\left({x}^{2}+{y}^{2}\right)={\left(1+5y\right)}^{2}\hfill & \text{Substitute }r=\sqrt{{x}^{2}+{y}^{2}}\text{ and }y=r\,\mathrm{sin}\,\theta .\hfill \\ \text{ }\,25{x}^{2}+25{y}^{2}=1+10y+25{y}^{2}\hfill & \text{Distribute and use FOIL}.\hfill \\ \text{ }\,25{x}^{2}-10y=1\hfill & \text{Rearrange terms and set equal to 1}.\hfill \end{array}$

Try It

Convert the conic $\,r=\frac{2}{1+2\text{ }\mathrm{cos}\text{ }\theta }\,$ to rectangular form.

Show Solution

$4-8x+3{x}^{2}-{y}^{2}=0$

Access these online resources for additional instruction and practice with conics in polar coordinates.

Visit this website for additional practice questions from Learningpod.

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 $\,P\left(r,\theta \right)\,$ at the pole, and a line, the directrix, which is perpendicular to the polar axis.
A conic is the set of all points $\,e=\frac{PF}{PD},$ where eccentricity $\,e\,$ is a positive real number. Each conic may be written in terms of its polar equation. See (Figure).
The polar equations of conics can be graphed. See (Figure), (Figure), and (Figure).
Conics can be defined in terms of a focus, a directrix, and eccentricity. See (Figure) and (Figure).
We can use the identities $\,r=\sqrt{{x}^{2}+{y}^{2}},x=r\text{ }\mathrm{cos}\text{ }\theta ,$ and $\,y=r\text{ }\mathrm{sin}\text{ }\theta \,$ to convert the equation for a conic from polar to rectangular form. See (Figure).

Section Exercises

Verbal

Explain how eccentricity determines which conic section is given.

Show Solution

If a conic section is written as a polar equation, what must be true of the denominator?

If a conic section is written as a polar equation, and the denominator involves $\,\mathrm{sin}\text{ }\theta ,$ what conclusion can be drawn about the directrix?

Show Solution

If the directrix of a conic section is perpendicular to the polar axis, what do we know about the equation of the graph?

What do we know about the focus/foci of a conic section if it is written as a polar equation?

Show Solution

Algebraic

For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity.

$r=\frac{6}{1-2\text{ }\mathrm{cos}\text{ }\theta }$

$r=\frac{3}{4-4\text{ }\mathrm{sin}\text{ }\theta }$

Show Solution

Parabola with $\,e=1\,$ and directrix $\,\frac{3}{4}\,$ units below the pole.

$r=\frac{8}{4-3\text{ }\mathrm{cos}\text{ }\theta }$

$r=\frac{5}{1+2\text{ }\mathrm{sin}\text{ }\theta }$

Show Solution

Hyperbola with $\,e=2\,$ and directrix $\,\frac{5}{2}\,$ units above the pole.

$r=\frac{16}{4+3\text{ }\mathrm{cos}\text{ }\theta }$

$r=\frac{3}{10+10\text{ }\mathrm{cos}\text{ }\theta }$

Show Solution

Parabola with $\,e=1\,$ and directrix $\,\frac{3}{10}\,$ units to the right of the pole.

$r=\frac{2}{1-\mathrm{cos}\text{ }\theta }$

$r=\frac{4}{7+2\text{ }\mathrm{cos}\text{ }\theta }$

Show Solution

Ellipse with $\,e=\frac{2}{7}\,$ and directrix $\,2\,$ units to the right of the pole.

$r\left(1-\mathrm{cos}\text{ }\theta \right)=3$

$r\left(3+5\mathrm{sin}\text{ }\theta \right)=11$

Show Solution

Hyperbola with $\,e=\frac{5}{3}\,$ and directrix $\,\frac{11}{5}\,$ units above the pole.

$r\left(4-5\mathrm{sin}\text{ }\theta \right)=1$

$r\left(7+8\mathrm{cos}\text{ }\theta \right)=7$

Show Solution

Hyperbola with $\,e=\frac{8}{7}\,$ and directrix $\,\frac{7}{8}\,$ units to the right of the pole.

For the following exercises, convert the polar equation of a conic section to a rectangular equation.

$r=\frac{4}{1+3\text{ }\mathrm{sin}\text{ }\theta }$

$r=\frac{2}{5-3\text{ }\mathrm{sin}\text{ }\theta }$

Show Solution

$25{x}^{2}+16{y}^{2}-12y-4=0$

$r=\frac{8}{3-2\text{ }\mathrm{cos}\text{ }\theta }$

$r=\frac{3}{2+5\text{ }\mathrm{cos}\text{ }\theta }$

Show Solution

$21{x}^{2}-4{y}^{2}-30x+9=0$

$r=\frac{4}{2+2\text{ }\mathrm{sin}\text{ }\theta }$

$r=\frac{3}{8-8\text{ }\mathrm{cos}\text{ }\theta }$

Show Solution

$64{y}^{2}=48x+9$

$r=\frac{2}{6+7\text{ }\mathrm{cos}\text{ }\theta }$

$r=\frac{5}{5-11\text{ }\mathrm{sin}\text{ }\theta }$

Show Solution

$96{y}^{2}-25{x}^{2}+110y+25=0$

$r\left(5+2\text{ }\mathrm{cos}\text{ }\theta \right)=6$

$r\left(2-\mathrm{cos}\text{ }\theta \right)=1$

Show Solution

$3{x}^{2}+4{y}^{2}-2x-1=0$

$r\left(2.5-2.5\text{ }\mathrm{sin}\text{ }\theta \right)=5$

$r=\frac{6\mathrm{sec}\text{ }\theta }{-2+3\text{ }\mathrm{sec}\text{ }\theta }$

Show Solution

$5{x}^{2}+9{y}^{2}-24x-36=0$

$r=\frac{6\mathrm{csc}\text{ }\theta }{3+2\text{ }\mathrm{csc}\text{ }\theta }$

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.

$r=\frac{5}{2+\mathrm{cos}\text{ }\theta }$

Show Solution

$r=\frac{2}{3+3\text{ }\mathrm{sin}\text{ }\theta }$

$r=\frac{10}{5-4\text{ }\mathrm{sin}\text{ }\theta }$

Show Solution

$r=\frac{3}{1+2\text{ }\mathrm{cos}\text{ }\theta }$

$r=\frac{8}{4-5\text{ }\mathrm{cos}\text{ }\theta }$

Show Solution

$r=\frac{3}{4-4\text{ }\mathrm{cos}\text{ }\theta }$

$r=\frac{2}{1-\mathrm{sin}\text{ }\theta }$

Show Solution

$r=\frac{6}{3+2\text{ }\mathrm{sin}\text{ }\theta }$

$r\left(1+\mathrm{cos}\text{ }\theta \right)=5$

Show Solution

$r\left(3-4\mathrm{sin}\text{ }\theta \right)=9$

$r\left(3-2\mathrm{sin}\text{ }\theta \right)=6$

Show Solution

$r\left(6-4\mathrm{cos}\text{ }\theta \right)=5$

For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix.

Directrix: $x=4;\,e=\frac{1}{5}$

Show Solution

$r=\frac{4}{5+\mathrm{cos}\theta }$

Directrix: $x=-4;\,e=5$

Directrix: $y=2;\,e=2$

Show Solution

$r=\frac{4}{1+2\mathrm{sin}\theta }$

Directrix: $y=-2;\,e=\frac{1}{2}$

Directrix: $x=1;\,e=1$

Show Solution

$r=\frac{1}{1+\mathrm{cos}\theta }$

Directrix: $x=-1;\,e=1$

Directrix: $x=-\frac{1}{4};\,e=\frac{7}{2}$

Show Solution

$r=\frac{7}{8-28\mathrm{cos}\theta }$

Directrix: $y=\frac{2}{5};\,e=\frac{7}{2}$

Directrix: $y=4;\,e=\frac{3}{2}$

Show Solution

$r=\frac{12}{2+3\mathrm{sin}\theta }$

Directrix: $x=-2;\,e=\frac{8}{3}$

Directrix: $x=-5;\,e=\frac{3}{4}$

Show Solution

$r=\frac{15}{4-3\mathrm{cos}\theta }$

Directrix: $y=2;\,e=2.5$

Directrix: $x=-3;\,e=\frac{1}{3}$

Show Solution

$r=\frac{3}{3-3\mathrm{cos}\theta }$

Extensions

Recall from Rotation of Axes that equations of conics with an $\,xy\,$ term have rotated graphs. For the following exercises, express each equation in polar form with $\,r\,$ as a function of $\,\theta .$

$xy=2$

${x}^{2}+xy+{y}^{2}=4$

Show Solution

$r=±\frac{2}{\sqrt{1+\mathrm{sin}\theta \mathrm{cos}\theta }}$

$2{x}^{2}+4xy+2{y}^{2}=9$

$16{x}^{2}+24xy+9{y}^{2}=4$

Show Solution

$r=±\frac{2}{4\mathrm{cos}\theta +3\mathrm{sin}\theta }$

$2xy+y=1$

Chapter Review Exercises

The Ellipse

For the following exercises, write the equation of the ellipse in standard form. Then identify the center, vertices, and foci.

$\frac{{x}^{2}}{25}+\frac{{y}^{2}}{64}=1$

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$\frac{{x}^{2}}{{5}^{2}}+\frac{{y}^{2}}{{8}^{2}}=1;\,$ center: $\,\left(0,0\right);\,$ vertices: $\,\left(5,0\right),\left(-5,0\right),\left(0,8\right),\left(0,-8\right);\,$ foci: $\,\left(0,\sqrt{39}\right),\left(0,-\sqrt{39}\right)$

$\frac{{\left(x-2\right)}^{2}}{100}+\frac{{\left(y+3\right)}^{2}}{36}=1$

$9{x}^{2}+{y}^{2}+54x-4y+76=0$

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$\frac{{\left(x+3\right)}^{2}}{{1}^{2}}+\frac{{\left(y-2\right)}^{2}}{{3}^{2}}=1\,\,\left(-3,2\right);\,\,\left(-2,2\right),\left(-4,2\right),\left(-3,5\right),\left(-3,-1\right);\,\,\left(-3,2+2\sqrt{2}\right),\left(-3,2-2\sqrt{2}\right)$

$9{x}^{2}+36{y}^{2}-36x+72y+36=0$

For the following exercises, graph the ellipse, noting center, vertices, and foci.

$\frac{{x}^{2}}{36}+\frac{{y}^{2}}{9}=1$

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center: $\,\left(0,0\right);\,$ vertices: $\,\left(6,0\right),\left(-6,0\right),\left(0,3\right),\left(0,-3\right);\,$ foci: $\,\left(3\sqrt{3},0\right),\left(-3\sqrt{3},0\right)$

$\frac{{\left(x-4\right)}^{2}}{25}+\frac{{\left(y+3\right)}^{2}}{49}=1$

$4{x}^{2}+{y}^{2}+16x+4y-44=0$

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center: $\,\left(-2,-2\right);\,$ vertices: $\,\left(2,-2\right),\left(-6,-2\right),\left(-2,6\right),\left(-2,-10\right);\,$ foci: $\,\left(-2,-2+4\sqrt{3},\right),\left(-2,-2-4\sqrt{3}\right)$

$\,2{x}^{2}+3{y}^{2}-20x+12y+38=0$

For the following exercises, use the given information to find the equation for the ellipse.

Center at $\,\left(0,0\right),$ focus at $\,\left(3,0\right),$ vertex at $\,\left(-5,0\right)$

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$\frac{{x}^{2}}{25}+\frac{{y}^{2}}{16}=1$

Center at $\,\left(2,-2\right),$ vertex at $\,\left(7,-2\right),$ focus at $\,\left(4,-2\right)$

A whispering gallery is to be constructed such that the foci are located 35 feet from the center. If the length of the gallery is to be 100 feet, what should the height of the ceiling be?

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The Hyperbola

For the following exercises, write the equation of the hyperbola in standard form. Then give the center, vertices, and foci.

$\frac{{x}^{2}}{81}-\frac{{y}^{2}}{9}=1$

$\frac{{\left(y+1\right)}^{2}}{16}-\frac{{\left(x-4\right)}^{2}}{36}=1$

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$\frac{{\left(y+1\right)}^{2}}{{4}^{2}}-\frac{{\left(x-4\right)}^{2}}{{6}^{2}}=1;\,$ center: $\,\left(4,-1\right);\,$ vertices: $\,\left(4,3\right),\left(4,-5\right);\,$ foci: $\,\left(4,-1+2\sqrt{13}\right),\left(4,-1-2\sqrt{13}\right)$

$9{y}^{2}-4{x}^{2}+54y-16x+29=0$

$3{x}^{2}-{y}^{2}-12x-6y-9=0$

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$\frac{{\left(x-2\right)}^{2}}{{2}^{2}}-\frac{{\left(y+3\right)}^{2}}{{\left(2\sqrt{3}\right)}^{2}}=1;\,$ center: $\,\left(2,-3\right);\,$ vertices: $\,\left(4,-3\right),\left(0,-3\right);\,$ foci: $\,\left(6,-3\right),\left(-2,-3\right)$

For the following exercises, graph the hyperbola, labeling vertices and foci.

$\frac{{x}^{2}}{9}-\frac{{y}^{2}}{16}=1$

$\frac{{\left(y-1\right)}^{2}}{49}-\frac{{\left(x+1\right)}^{2}}{4}=1$

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${x}^{2}-4{y}^{2}+6x+32y-91=0$

$2{y}^{2}-{x}^{2}-12y-6=0$

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For the following exercises, find the equation of the hyperbola.

Center at $\,\left(0,0\right),$ vertex at $\,\left(0,4\right),$ focus at $\,\left(0,-6\right)$

Foci at $\,\left(3,7\right)\,$ and $\,\left(7,7\right),$ vertex at $\,\left(6,7\right)$

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$\frac{{\left(x-5\right)}^{2}}{1}-\frac{{\left(y-7\right)}^{2}}{3}=1$

The Parabola

For the following exercises, write the equation of the parabola in standard form. Then give the vertex, focus, and directrix.

${y}^{2}=12x$

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

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${\left(x+2\right)}^{2}=\frac{1}{2}\left(y-1\right);\,$ vertex: $\,\left(-2,1\right);\,$ focus: $\,\left(-2,\frac{9}{8}\right);\,$ directrix: $\,y=\frac{7}{8}$

${y}^{2}-6y-6x-3=0$

${x}^{2}+10x-y+23=0$

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${\left(x+5\right)}^{2}=\left(y+2\right);\,$ vertex: $\,\left(-5,-2\right);\,$ focus: $\,\left(-5,-\frac{7}{4}\right);\,$ directrix: $\,y=-\frac{9}{4}$

For the following exercises, graph the parabola, labeling vertex, focus, and directrix.

${x}^{2}+4y=0$

${\left(y-1\right)}^{2}=\frac{1}{2}\left(x+3\right)$

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${x}^{2}-8x-10y+46=0$

$2{y}^{2}+12y+6x+15=0$

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For the following exercises, write the equation of the parabola using the given information.

Focus at $\,\left(-4,0\right);\,$ directrix is $\,x=4$

Focus at $\,\left(2,\frac{9}{8}\right);\,$ directrix is $\,y=\frac{7}{8}$

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${\left(x-2\right)}^{2}=\left(\frac{1}{2}\right)\left(y-1\right)$

A cable TV receiving dish is the shape of a paraboloid of revolution. Find the location of the receiver, which is placed at the focus, if the dish is 5 feet across at its opening and 1.5 feet deep.

Rotation of Axes

For the following exercises, determine which of the conic sections is represented.

$16{x}^{2}+24xy+9{y}^{2}+24x-60y-60=0$

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${B}^{2}-4AC=0,$
parabola

$4{x}^{2}+14xy+5{y}^{2}+18x-6y+30=0$

$4{x}^{2}+xy+2{y}^{2}+8x-26y+9=0$

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${B}^{2}-4AC=-31<0,$ ellipse

For the following exercises, determine the angle $\,\theta \,$ that will eliminate the $\,xy\,$ term, and write the corresponding equation without the $\,xy\,$ term.

${x}^{2}+4xy-2{y}^{2}-6=0$

${x}^{2}-xy+{y}^{2}-6=0$

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$\theta ={45}^{\circ },{{x}^{\prime }}^{2}+3{{y}^{\prime }}^{2}-12=0$

For the following exercises, graph the equation relative to the $\,{x}^{\prime }{y}^{\prime }\,$ system in which the equation has no $\,{x}^{\prime }{y}^{\prime }\,$ term.

$9{x}^{2}-24xy+16{y}^{2}-80x-60y+100=0$

${x}^{2}-xy+{y}^{2}-2=0$

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$\theta ={45}^{\circ }$

$6{x}^{2}+24xy-{y}^{2}-12x+26y+11=0$

For the following exercises, given the polar equation of the conic with focus at the origin, identify the eccentricity and directrix.

$r=\frac{10}{1-5\text{ }\mathrm{cos}\text{ }\theta }$

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Hyperbola with $\,e=5\,$ and directrix $\,2\,$ units to the left of the pole.

$r=\frac{6}{3+2\text{ }\mathrm{cos}\text{ }\theta }$

$r=\frac{1}{4+3\text{ }\mathrm{sin}\text{ }\theta }$

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Ellipse with $\,e=\frac{3}{4}\,$ and directrix $\,\frac{1}{3}\,$ unit above the pole.

$r=\frac{3}{5-5\text{ }\mathrm{sin}\text{ }\theta }$

For the following exercises, graph the conic given in polar form. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse or a hyperbola, label the vertices and foci.

$r=\frac{3}{1-\mathrm{sin}\text{ }\theta }$

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$r=\frac{8}{4+3\text{ }\mathrm{sin}\text{ }\theta }$

$r=\frac{10}{4+5\text{ }\mathrm{cos}\text{ }\theta }$

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$r=\frac{9}{3-6\text{ }\mathrm{cos}\text{ }\theta }$

For the following exercises, given information about the graph of a conic with focus at the origin, find the equation in polar form.

Directrix is $\,x=3\,$ and eccentricity $\,e=1$

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$r=\frac{3}{1+\mathrm{cos}\text{ } \theta }$

Directrix is $\,y=-2\,$ and eccentricity $\,e=4$

Practice Test

For the following exercises, write the equation in standard form and state the center, vertices, and foci.

$\frac{{x}^{2}}{9}+\frac{{y}^{2}}{4}=1$

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$\frac{{x}^{2}}{{3}^{2}}+\frac{{y}^{2}}{{2}^{2}}=1;\,$ center: $\,\left(0,0\right);\,$ vertices: $\,\left(3,0\right),\left(–3,0\right),\left(0,2\right),\left(0,-2\right);\,$ foci: $\left(\sqrt{5},0\right),\left(-\sqrt{5},0\right)$

$9{y}^{2}+16{x}^{2}-36y+32x-92=0$

For the following exercises, sketch the graph, identifying the center, vertices, and foci.

$\frac{{\left(x-3\right)}^{2}}{64}+\frac{{\left(y-2\right)}^{2}}{36}=1$

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center: $\,\left(3,2\right);\,$ vertices: $\,\left(11,2\right),\left(-5,2\right),\left(3,8\right),\left(3,-4\right);\,$ foci: $\,\left(3+2\sqrt{7},2\right),\left(3-2\sqrt{7},2\right)$

$2{x}^{2}+{y}^{2}+8x-6y-7=0$

Write the standard form equation of an ellipse with a center at $\,\left(1,2\right),$ vertex at $\,\left(7,2\right),$ and focus at $\,\left(4,2\right).$

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$\frac{{\left(x-1\right)}^{2}}{36}+\frac{{\left(y-2\right)}^{2}}{27}=1$

A whispering gallery is to be constructed with a length of 150 feet. If the foci are to be located 20 feet away from the wall, how high should the ceiling be?

For the following exercises, write the equation of the hyperbola in standard form, and give the center, vertices, foci, and asymptotes.

$\frac{{x}^{2}}{49}-\frac{{y}^{2}}{81}=1$

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$\frac{{x}^{2}}{{7}^{2}}-\frac{{y}^{2}}{{9}^{2}}=1;\,$ center: $\,\left(0,0\right);\,$ vertices $\,\left(7,0\right),\left(-7,0\right);\,$ foci: $\,\left(\sqrt{130},0\right),\left(-\sqrt{130},0\right);\,$ asymptotes: $\,y=±\frac{9}{7}x$

$16{y}^{2}-9{x}^{2}+128y+112=0$

For the following exercises, graph the hyperbola, noting its center, vertices, and foci. State the equations of the asymptotes.

$\frac{{\left(x-3\right)}^{2}}{25}-\frac{{\left(y+3\right)}^{2}}{1}=1$

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center: $\,\left(3,-3\right);\,$ vertices: $\,\left(8,-3\right),\left(-2,-3\right);$ foci: $\,\left(3+\sqrt{26},-3\right),\left(3-\sqrt{26},-3\right);\,$ asymptotes: $\,y=±\frac{1}{5}\left(x-3\right)-3$

${y}^{2}-{x}^{2}+4y-4x-18=0$

Write the standard form equation of a hyperbola with foci at $\,\left(1,0\right)\,$ and $\,\left(1,6\right),$ and a vertex at $\,\left(1,2\right).$

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$\frac{{\left(y-3\right)}^{2}}{1}-\frac{{\left(x-1\right)}^{2}}{8}=1$

For the following exercises, write the equation of the parabola in standard form, and give the vertex, focus, and equation of the directrix.

${y}^{2}+10x=0$

$3{x}^{2}-12x-y+11=0$

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${\left(x-2\right)}^{2}=\frac{1}{3}\left(y+1\right);\,$ vertex: $\,\left(2,-1\right);\,$ focus: $\,\left(2,-\frac{11}{12}\right);\,$ directrix: $\,y=-\frac{13}{12}$

For the following exercises, graph the parabola, labeling the vertex, focus, and directrix.

${\left(x-1\right)}^{2}=-4\left(y+3\right)$

${y}^{2}+8x-8y+40=0$

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Write the equation of a parabola with a focus at $\,\left(2,3\right)\,$ and directrix $\,y=-1.$

A searchlight is shaped like a paraboloid of revolution. If the light source is located 1.5 feet from the base along the axis of symmetry, and the depth of the searchlight is 3 feet, what should the width of the opening be?

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Approximately $\,8.49\,$ feet

For the following exercises, determine which conic section is represented by the given equation, and then determine the angle $\,\theta \,$ that will eliminate the $\,xy\,$ term.

$3{x}^{2}-2xy+3{y}^{2}=4$

${x}^{2}+4xy+4{y}^{2}+6x-8y=0$

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parabola; $\,\theta \approx {63.4}^{\circ }$

For the following exercises, rewrite in the $\,{x}^{\prime }{y}^{\prime }\,$ system without the $\,{x}^{\prime }{y}^{\prime }\,$ term, and graph the rotated graph.

$11{x}^{2}+10\sqrt{3}xy+{y}^{2}=4$

$16{x}^{2}+24xy+9{y}^{2}-125x=0$

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${{x}^{\prime }}^{2}-4{x}^{\prime }+3{y}^{\prime }=0$

For the following exercises, identify the conic with focus at the origin, and then give the directrix and eccentricity.

$r=\frac{3}{2-\mathrm{sin}\text{ }\theta }$

$r=\frac{5}{4+6\text{ }\mathrm{cos}\text{ }\theta }$

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Hyperbola with $\,e=\frac{3}{2},\,$ and directrix $\,\frac{5}{6}\,$ units to the right of the pole.

For the following exercises, graph the given conic section. If it is a parabola, label vertex, focus, and directrix. If it is an ellipse or a hyperbola, label vertices and foci.

$r=\frac{12}{4-8\text{ }\mathrm{sin}\text{ }\theta }$

$r=\frac{2}{4+4\text{ }\mathrm{sin}\text{ }\theta }$

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Find a polar equation of the conic with focus at the origin, eccentricity of $\,e=2,$ and directrix: $\,x=3.$

Glossary

eccentricity: the ratio of the distances from a point $\,P\,$ on the graph to the focus $\,F\,$ and to the directrix $\,D\,$ represented by $\,e=\frac{PF}{PD},$ where $\,e\,$ is a positive real number

polar equation: an equation of a curve in polar coordinates $\,r\,$ and $\,\theta$

Analytic Geometry