Lesson 12.5: Conic Sections in Polar Coordinates

New Concept

Conic sections can be described with a single polar equation based on eccentricity, $e$. This value defines whether the conic is an ellipse ($0 \le e < 1$), a parabola ($e = 1$), or a hyperbola ($e > 1$).

What’s next

Next, you'll use this definition to identify, graph, and define conics from their polar equations through a series of interactive examples and practice problems.

Property

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. The set of all points $P$ such that $e = \frac{PF}{PD}$ is a conic.
For a conic with eccentricity $e$:

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

Examples

• If a conic has an eccentricity of $e = 0.5$, it's an ellipse because $0 \leq 0.5 < 1$.
• If a conic's eccentricity is $e = 1$, it is a parabola.
• A conic with an eccentricity of $e = 1.8$ is a hyperbola because $1.8 > 1$.

Explanation

Eccentricity is a single number that tells you the shape of your conic section. Think of it as a 'stretch' factor. A value less than 1 gives a closed loop (ellipse), 1 gives a U-shape (parabola), and greater than 1 gives two separate branches (hyperbola).

Property

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

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

Examples

• An equation like $r = \frac{10}{1 + 2 \cos \theta}$ describes a conic with a vertical directrix at $x=p$.
• An equation such as $r = \frac{5}{1 - 0.5 \sin \theta}$ describes a conic with a horizontal directrix at $y=-p$.
• The equation $r = \frac{3}{1 + \sin \theta}$ represents a conic with a horizontal directrix at $y=p$ and an eccentricity of $e=1$.

Explanation

These formulas neatly package all the key info about a conic. The focus is always at the origin. Use cosine for vertical directrices ($x=\pm p$) and sine for horizontal ones ($y=\pm p$). The sign in the denominator tells you if the directrix is on the positive or negative side.

Property

To identify the type of conic, the directrix, and the eccentricity from a polar equation:

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

Examples

• For $r = \frac{10}{5 + 4 \sin \theta}$, divide by 5 to get $r = \frac{2}{1 + 0.8 \sin \theta}$. Here, $e=0.8$ (ellipse) and $ep=2$, so the directrix is $y=2.5$.
• For $r = \frac{6}{2 - 3 \cos \theta}$, divide by 2 to get $r = \frac{3}{1 - 1.5 \cos \theta}$. Here, $e=1.5$ (hyperbola) and $ep=3$, so the directrix is $x=-2$.
• For $r = \frac{5}{3 - 3 \sin \theta}$, divide by 3 to get $r = \frac{5/3}{1 - \sin \theta}$. Here, $e=1$ (parabola) and $ep=5/3$, so the directrix is $y=-5/3$.

Explanation

To analyze a conic's polar equation, first make sure the denominator starts with '1'. The number multiplying sine or cosine is the eccentricity ($e$), which names the shape. The numerator, $ep$, helps you find the directrix line.

Property

To graph a conic in polar form, first rewrite the equation so that the denominator begins with 1. This helps determine the eccentricity $e$ and 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, allowing for a rough sketch of the graph.

Examples

• For the ellipse $r = \frac{3}{2 - \cos \theta}$, at $\theta=0$, the vertex is at $r = \frac{3}{2-1} = 3$.
• For the parabola $r = \frac{6}{3 + 3 \sin \theta}$, at $\theta=\frac{\pi}{2}$, the vertex is at $r = \frac{6}{3+3} = 1$.
• For the hyperbola $r = \frac{4}{1 - 2 \sin \theta}$, at $\theta=\frac{3\pi}{2}$, the vertex is at $r = \frac{4}{1 - 2(-1)} = \frac{4}{3}$.

Explanation

Graphing is a simple process: standardize the equation to identify the conic's shape. Then, find the four key points by plugging in angles $0, \pi/2, \pi,$ and $3\pi/2$. Plot these points and connect them to sketch your conic.

Property

Given the focus at the origin, eccentricity $e$, and directrix of a conic, determine the polar equation.

1. If the directrix is $y = \pm p$, use the sine form. If the directrix is $x = \pm p$, use the cosine form.
2. If the directrix is positive ($p>0$), use addition in the denominator. If it is negative ($p<0$), use subtraction.
3. The coefficient of the trigonometric function is the eccentricity $e$.
4. The numerator is the product of $e$ and the absolute value of $p$.

Examples

• Given $e = 2$ and directrix $y = 3$: The equation is $r = \frac{(2)(3)}{1 + 2 \sin \theta} = \frac{6}{1 + 2 \sin \theta}$.
• Given $e = 0.5$ and directrix $x = -4$: The equation is $r = \frac{(0.5)(4)}{1 - 0.5 \cos \theta} = \frac{2}{1 - 0.5 \cos \theta}$.
• Given $e = 1$ and directrix $x = 5$: The equation is $r = \frac{(1)(5)}{1 + \cos \theta} = \frac{5}{1 + \cos \theta}$.

Explanation

To build a conic's polar equation, pick sine for a horizontal directrix ($y=p$) and cosine for a vertical one ($x=p$). A positive directrix value means '+' in the denominator, while negative means '-'. The numerator is always $e$ times $p$.

Property

To convert a conic from polar to rectangular form, rearrange the formula to use the identities $r = \sqrt{x^2 + y^2}$, $x = r \operatorname{cos} \theta$, and $y = r \operatorname{sin} \theta$.

1. Multiply both sides by the denominator to clear the fraction.
2. Distribute $r$ inside the parentheses.
3. Isolate the term $cr$ on one side, where $c$ is a constant.
4. Substitute $r \operatorname{sin} \theta$ with $y$ or $r \operatorname{cos} \theta$ with $x$.
5. Square both sides of the equation.
6. Replace $r^2$ with $x^2 + y^2$ and simplify the equation.

Examples

• To convert $r = \frac{3}{1 - \cos \theta}$: $r(1 - \cos \theta) = 3 \Rightarrow r - r \cos \theta = 3 \Rightarrow \sqrt{x^2+y^2} - x = 3$. Squaring and simplifying gives $y^2 = 6x + 9$.
• To convert $r = \frac{8}{4 + 2 \sin \theta}$: $4r + 2r \sin \theta = 8 \Rightarrow 2r + y = 4 \Rightarrow 4r^2 = (4-y)^2$. This simplifies to $4x^2 + 3y^2 + 8y = 16$.
• To convert $r = \frac{2}{1 + \sin \theta}$: $r(1 + \sin \theta) = 2 \Rightarrow r + y = 2 \Rightarrow r^2 = (2-y)^2$. This simplifies to $x^2 + 4y = 4$.

Explanation

To switch from polar to rectangular form, your goal is to replace all $r$ and $\theta$ terms with $x$ and $y$. Use algebraic manipulation and the core conversion formulas: $x = r \cos \theta$, $y = r \sin \theta$, and $r^2 = x^2 + y^2$.