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Conic Sections in Polar Coordinates

Identify a conic in polar form.

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 \(x=2+{y}^{2}\) shown in .

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 \(P(r,\theta )\) 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, \(x=\pm p,\) 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 \(\theta .\)

Condensed — the full section is in OpenStax Precalculus 2e.

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.

Example

Try it.

Graph \(r=\frac{5}{3+3\ \cos \ \theta }.\)

Solution

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

\[\begin{array}{l}\begin{array}{l} \\ \\ r=\frac{5}{3+3\ \cos \ \theta }=\frac{5(\frac{1}{3})}{3(\frac{1}{3})+3(\frac{1}{3})\cos \ \theta }\end{array} \\ r=\frac{\frac{5}{3}}{1+\cos \ \theta }\end{array}\]

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

\[\begin{array}{l}\frac{5}{3}=ep \\ \frac{5}{3}=(1)p \\ \frac{5}{3}=p\end{array}\]

The directrix is \(x=\frac{5}{3}.\)

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

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

Condensed — the full section is in OpenStax Precalculus 2e.

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.

Example

Try it.

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

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

\[r=\frac{ep}{1-e\ \sin \ \theta }\]

and \(e=3\) and \(|-2|=2=p.\)

Therefore,

\[\begin{array}{l} \\ \begin{array}{l}r=\frac{(3)(2)}{1-3\ \sin \ \theta } \\ r=\frac{6}{1-3\ \sin \ \theta }\end{array}\end{array}\]

Example

Try it.

Find the polar form of a conic given a focus at the origin, \(e=\frac{3}{5},\) and directrix \(x=4.\)

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

\[r=\frac{ep}{1+e\ \cos \ \theta }\]

and \(e=\frac{3}{5}\) and \(|4|=4=p.\)

Therefore,

\[\begin{array}{l}\begin{array}{l} \\ \\ r=\frac{(\frac{3}{5})(4)}{1+\frac{3}{5}\ \cos \ \theta }\end{array} \\ r=\frac{\frac{12}{5}}{1+\frac{3}{5}\ \cos \ \theta } \\ r=\frac{\frac{12}{5}}{1(\frac{5}{5})+\frac{3}{5}\ \cos \ \theta } \\ r=\frac{\frac{12}{5}}{\frac{5}{5}+\frac{3}{5}\ \cos \ \theta } \\ r=\frac{12}{5}⋅\frac{5}{5+3\ \cos \ \theta } \\ r=\frac{12}{5+3\ \cos \ \theta }\end{array}\]

Condensed — the full section is in OpenStax Precalculus 2e.

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 \(P(r,\theta )\) 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 .
The polar equations of conics can be graphed. See , , and .
Conics can be defined in terms of a focus, a directrix, and eccentricity. See and .
We can use the identities \(r=\sqrt{{x}^{2}+{y}^{2}},x=r\ \cos \ \theta ,\) and \(y=r\ \sin \ \theta\) to convert the equation for a conic from polar to rectangular form. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity.
1. \(r=\frac{6}{3+2\ \sin \ \theta }\)
2. \(r=\frac{12}{4+5\ \cos \ \theta }\)
3. \(r=\frac{7}{2-2\ \sin \ \theta }\)
Paljasta vastaus
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.
1. Multiply the numerator and denominator by \(\frac{1}{3}.\) \[r=\frac{6}{3+2\sin \ \theta }⋅\frac{(\frac{1}{3})}{(\frac{1}{3})}=\frac{6(\frac{1}{3})}{3(\frac{1}{3})+2(\frac{1}{3})\sin \ \theta }=\frac{2}{1+\frac{2}{3}\ \sin \ \theta }\]
  Because \(\sin \ \theta\) is in the denominator, the directrix is \(y=p.\) Comparing to standard form, note that \(e=\frac{2}{3}.\) Therefore, from the numerator,
  \[\begin{array}{l}2=ep \\ 2=\frac{2}{3}p \\ (\frac{3}{2})2=(\frac{3}{2})\frac{2}{3}p \\ 3=p\end{array}\]
  Since \(e<1,\) the conic is an ellipse. The eccentricity is \(e=\frac{2}{3}\) and the directrix is \(y=3.\)
2. Multiply the numerator and denominator by \(\frac{1}{4}.\) \[\begin{array}{l}\begin{array}{l} \\ \\ r=\frac{12}{4+5\ \cos \ \theta }⋅\frac{(\frac{1}{4})}{(\frac{1}{4})}\end{array} \\ r=\frac{12(\frac{1}{4})}{4(\frac{1}{4})+5(\frac{1}{4})\cos \ \theta } \\ r=\frac{3}{1+\frac{5}{4}\ \cos \ \theta }\end{array}\]
  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,
  \[\begin{array}{l}\ 3=ep \\ \ 3=\frac{5}{4}p \\ (\frac{4}{5})3=(\frac{4}{5})\frac{5}{4}p \\ \ \frac{12}{5}=p\end{array}\]
  Since \(e>1,\) the conic is a hyperbola. The eccentricity is \(e=\frac{5}{4}\) and the directrix is \(x=\frac{12}{5}=2.4.\)
3. Multiply the numerator and denominator by \(\frac{1}{2}.\) \[\begin{array}{l} \\ \\ \begin{array}{l}r=\frac{7}{2-2\ \sin \ \theta }⋅\frac{(\frac{1}{2})}{(\frac{1}{2})} \\ r=\frac{7(\frac{1}{2})}{2(\frac{1}{2})-2(\frac{1}{2})\ \sin \ \theta } \\ r=\frac{\frac{7}{2}}{1-\sin \ \theta }\end{array}\end{array}\]
  Because sine is in the denominator, the directrix is \(y=-p.\) Comparing to standard form, \(e=1.\) Therefore, from the numerator,
  \[\begin{array}{l}\frac{7}{2}=ep \\ \frac{7}{2}=(1)p \\ \frac{7}{2}=p\end{array}\]
  Because \(e=1,\) the conic is a parabola. The eccentricity is \(e=1\) and the directrix is \(y=-\frac{7}{2}=-3.5.\)
Identify the conic with focus at the origin, the directrix, and the eccentricity for \(r=\frac{2}{3-\cos \ \theta }.\)

Paljasta vastaus

ellipse; \(e=\frac{1}{3};\ x=-2\)

Graph \(r=\frac{5}{3+3\ \cos \ \theta }.\)

Paljasta vastaus

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

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

\[\begin{array}{l}\frac{5}{3}=ep \\ \frac{5}{3}=(1)p \\ \frac{5}{3}=p\end{array}\]

The directrix is \(x=\frac{5}{3}.\)

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

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

Graph \(r=\frac{8}{2-3\ \sin \ \theta }.\)

Paljasta vastaus

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

\[\begin{array}{l}\begin{array}{l} \\ \\ r=\frac{8}{2-3\sin \ \theta }=\frac{8(\frac{1}{2})}{2(\frac{1}{2})-3(\frac{1}{2})\sin \ \theta }\end{array} \\ r=\frac{4}{1-\frac{3}{2}\ \sin \ \theta }\end{array}\]

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

\[\begin{array}{l}4=ep \\ 4=(\frac{3}{2})p \\ 4(\frac{2}{3})=p \\ \frac{8}{3}=p\end{array}\]

The directrix is \(y=-\frac{8}{3}.\)

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

	A	B	C	D
\(\theta\)	\(0\)	\(\frac{\pi }{2}\)	\(\pi\)	\(\frac{3\pi }{2}\)
\[r=\frac{8}{2-3\sin \ \theta }\]	\(4\)	\(-8\)	\(4\)	\(\frac{8}{5}=1.6\)

Graph \(r=\frac{10}{5-4\ \cos \ \theta }.\)

Paljasta vastaus

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

\[\begin{array}{l} \\ \begin{array}{l}r=\frac{10}{5-4\cos \ \theta }=\frac{10(\frac{1}{5})}{5(\frac{1}{5})-4(\frac{1}{5})\cos \ \theta } \\ r=\frac{2}{1-\frac{4}{5}\ \cos \ \theta }\end{array}\end{array}\]

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

\[\begin{array}{l}2=ep \\ 2=(\frac{4}{5})p \\ 2(\frac{5}{4})=p \\ \frac{5}{2}=p\end{array}\]

The directrix is \(x=-\frac{5}{2}.\)

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

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

Graph \(r=\frac{2}{4-\cos \ \theta }.\)
Find the polar form of the conic given a focus at the origin, \(e=3\) and directrix \(y=-2.\)

Paljasta vastaus

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
\[r=\frac{ep}{1-e\ \sin \ \theta }\]
and \(e=3\) and \(|-2|=2=p.\)
Therefore,
\[\begin{array}{l} \\ \begin{array}{l}r=\frac{(3)(2)}{1-3\ \sin \ \theta } \\ r=\frac{6}{1-3\ \sin \ \theta }\end{array}\end{array}\]
Find the polar form of a conic given a focus at the origin, \(e=\frac{3}{5},\) and directrix \(x=4.\)

Paljasta vastaus

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
\[r=\frac{ep}{1+e\ \cos \ \theta }\]
and \(e=\frac{3}{5}\) and \(|4|=4=p.\)
Therefore,
\[\begin{array}{l}\begin{array}{l} \\ \\ r=\frac{(\frac{3}{5})(4)}{1+\frac{3}{5}\ \cos \ \theta }\end{array} \\ r=\frac{\frac{12}{5}}{1+\frac{3}{5}\ \cos \ \theta } \\ r=\frac{\frac{12}{5}}{1(\frac{5}{5})+\frac{3}{5}\ \cos \ \theta } \\ r=\frac{\frac{12}{5}}{\frac{5}{5}+\frac{3}{5}\ \cos \ \theta } \\ r=\frac{12}{5}⋅\frac{5}{5+3\ \cos \ \theta } \\ r=\frac{12}{5+3\ \cos \ \theta }\end{array}\]
Find the polar form of the conic given a focus at the origin, \(e=1,\) and directrix \(x=-1.\)

Paljasta vastaus

\(r=\frac{1}{1-\cos \theta }\)
Convert the conic \(r=\frac{1}{5-5\sin \ \theta }\) to rectangular form.

Paljasta vastaus

We will rearrange the formula to use the identities \(r=\sqrt{{x}^{2}+{y}^{2}},x=r\ \cos \ \theta ,\text{and }y=r\ \sin \ \theta .\)
\[\begin{array}{ll}\ r=\frac{1}{5-5\ \sin \ \theta } & \\ r⋅(5-5\ \sin \ \theta )=\frac{1}{5-5\ \sin \ \theta }⋅(5-5\ \sin \ \theta ) & \text{Eliminate the fraction}. \\ \ 5r-5r\ \sin \ \theta =1 & \text{Distribute}. \\ \ 5r=1+5r\ \sin \ \theta & \text{Isolate }5r. \\ \ 25{r}^{2}={(1+5r\ \sin \ \theta )}^{2} & \text{Square both sides}. \\ \ 25({x}^{2}+{y}^{2})={(1+5y)}^{2} & \text{Substitute }r=\sqrt{{x}^{2}+{y}^{2}}\ \text{and }y=r\ \sin \ \theta . \\ \ 25{x}^{2}+25{y}^{2}=1+10y+25{y}^{2} & \text{Distribute and use FOIL}. \\ \ 25{x}^{2}-10y=1 & \text{Rearrange terms and set equal to 1}.\end{array}\]
Convert the conic \(r=\frac{2}{1+2\ \cos \ \theta }\) to rectangular form.

Paljasta vastaus

\(4-8x+3{x}^{2}-{y}^{2}=0\)
Explain how eccentricity determines which conic section is given.

Paljasta vastaus

If eccentricity is less than 1, it is an ellipse. If eccentricity is equal to 1, it is a parabola. If eccentricity is greater than 1, it is a hyperbola.
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 \(\sin \ \theta ,\) what conclusion can be drawn about the directrix?

Paljasta vastaus

The directrix will be parallel to the polar axis.
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?

Paljasta vastaus

One of the foci will be located at the origin.
\(r=\frac{6}{1-2\ \cos \ \theta }\)
\(r=\frac{3}{4-4\ \sin \ \theta }\)

Paljasta vastaus

Parabola with \(e=1\) and directrix \(\frac{3}{4}\) units below the pole.
\(r=\frac{8}{4-3\ \cos \ \theta }\)
\(r=\frac{5}{1+2\ \sin \ \theta }\)

Paljasta vastaus

Hyperbola with \(e=2\) and directrix \(\frac{5}{2}\) units above the pole.
\(r=\frac{16}{4+3\ \cos \ \theta }\)
\(r=\frac{3}{10+10\ \cos \ \theta }\)

Paljasta vastaus

Parabola with \(e=1\) and directrix \(\frac{3}{10}\) units to the right of the pole.
\(r=\frac{2}{1-\cos \ \theta }\)
\(r=\frac{4}{7+2\ \cos \ \theta }\)

Paljasta vastaus

Ellipse with \(e=\frac{2}{7}\) and directrix \(2\) units to the right of the pole.
\(r(1-\cos \ \theta )=3\)
\(r(3+5\sin \ \theta )=11\)

Paljasta vastaus

Hyperbola with \(e=\frac{5}{3}\) and directrix \(\frac{11}{5}\) units above the pole.
\(r(4-5\sin \ \theta )=1\)
\(r(7+8\cos \ \theta )=7\)

Paljasta vastaus

Hyperbola with \(e=\frac{8}{7}\) and directrix \(\frac{7}{8}\) units to the right of the pole.
\(r=\frac{4}{1+3\ \sin \ \theta }\)
\(r=\frac{2}{5-3\ \sin \ \theta }\)

Paljasta vastaus

\(25{x}^{2}+16{y}^{2}-12y-4=0\)
\(r=\frac{8}{3-2\ \cos \ \theta }\)
\(r=\frac{3}{2+5\ \cos \ \theta }\)

Paljasta vastaus

\(21{x}^{2}-4{y}^{2}-30x+9=0\)
\(r=\frac{4}{2+2\ \sin \ \theta }\)
\(r=\frac{3}{8-8\ \cos \ \theta }\)

Paljasta vastaus

\(64{y}^{2}=48x+9\)
\(r=\frac{2}{6+7\ \cos \ \theta }\)
\(r=\frac{5}{5-11\ \sin \ \theta }\)

Paljasta vastaus

\(96{y}^{2}-25{x}^{2}+110y+25=0\)
\(r(5+2\ \cos \ \theta )=6\)
\(r(2-\cos \ \theta )=1\)

Paljasta vastaus

\(3{x}^{2}+4{y}^{2}-2x-1=0\)
\(r(2.5-2.5\ \sin \ \theta )=5\)
\(r=\frac{6\text{sec}\ \theta }{-2+3\ \text{sec}\ \theta }\)

Paljasta vastaus

\(5{x}^{2}+9{y}^{2}-24x-36=0\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\theta

theta
The usual name for an angle.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

\approx

approximately equal
Equal to the precision shown, not exactly.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

How to: Conic Sections in Polar Coordinates

Identify a conic in polar form.
Graph the polar equations of conics.
Deﬁne conics in terms of a focus and a directrix.
if
if
if
Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form.
Identify the eccentricity

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Kokeile omaasi

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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