10.5 Conic Sections in Polar Coordinates

Conic Sections in Polar Coordinates

Conic sections in polar coordinates offer a different way to describe circles, ellipses, parabolas, and hyperbolas. Instead of using $x$ and $y$ coordinates centered on the middle of the shape, polar form places one focus at the origin and describes every point using distance $(r)$ and angle $(\theta)$. This approach puts the focus-directrix relationship front and center, making eccentricity the single parameter that controls which type of conic you're looking at.

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Conic Sections in Polar Coordinates

Polar equations for conic sections

The general polar form of a conic section is:

$r = \frac{ep}{1 + e\cos\theta}$ or $r = \frac{ep}{1 - e\cos\theta}$

where $e$ is the eccentricity and $p$ is the distance from the focus (at the pole) to the directrix. You can also replace $\cos\theta$ with $\sin\theta$ to orient the conic vertically instead of horizontally.

The sign in the denominator controls which direction the conic opens. With $1 + e\cos\theta$, the directrix is to the right of the pole; with $1 - e\cos\theta$, it's to the left.

Eccentricity $(e)$ determines the type of conic:

• Circle: $e = 0$, giving a constant radius from the center
• Ellipse: $0 < e < 1$, a closed oval curve with two foci
• Parabola: $e = 1$, an open U-shaped curve with one focus
• Hyperbola: $e > 1$, an open curve with two separate branches

Polar equations for specific conics:

• A circle centered at the pole simplifies to $r = a$, where $a$ is the radius. If the center is offset from the pole, you get forms like $r = 2a\cos\theta$ or $r = 2a\sin\theta$.
• An ellipse can be written as $r = \frac{a(1 - e^2)}{1 + e\cos\theta}$, where $a$ is the semi-major axis. Notice that $a(1 - e^2)$ plays the role of $ep$ in the general form, so $p = \frac{a(1 - e^2)}{e}$.
• A parabola with $e = 1$ becomes $r = \frac{p}{1 + \cos\theta}$, where $p$ is the distance from the focus to the directrix. (Some texts write this as $r = \frac{2d}{1 + \cos\theta}$ where $d$ is the focal-to-directrix distance, so watch the conventions your course uses.)
• A hyperbola uses $r = \frac{a(e^2 - 1)}{1 + e\cos\theta}$, where $a$ is the semi-transverse axis and $e > 1$. The numerator is $e^2 - 1$ (not $1 - e^2$) to keep $r$ positive on the relevant branch.

Graphing conics in polar coordinates

To graph a conic from its polar equation, follow these steps:

1. Identify the conic type. Read off $e$ from the equation. Compare it to 0, 1, or greater than 1 to determine circle, ellipse, parabola, or hyperbola.
2. Find the vertices. Substitute $\theta = 0$ and $\theta = \pi$ into the equation. These give the points where the conic crosses the polar axis, and they're the easiest key points to calculate.
3. Locate the foci and directrix. The focus is already at the pole. The directrix sits at distance $p$ from the pole, perpendicular to the polar axis, on the side determined by the sign in the denominator.
4. Plot additional points. Substitute several values of $\theta$ (such as $\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}$, and their supplements) to get enough points for a smooth curve.
5. Connect the points smoothly, keeping the expected shape in mind.

Symmetry and key features:

• If the equation uses $\cos\theta$, the conic is symmetric about the horizontal polar axis ($\theta = 0$). If it uses $\sin\theta$, symmetry is about the vertical line $\theta = \frac{\pi}{2}$.
• Vertices are the closest and farthest points from the focus along the axis of symmetry.
• The directrix is a straight line perpendicular to the axis of symmetry, related to the focus by the eccentricity.

Transformations in polar form:

• Rotation: Replace $\theta$ with $\theta - \alpha$ to rotate the conic by angle $\alpha$. For example, $r = \frac{ep}{1 + e\cos(\theta - \frac{\pi}{4})}$ rotates the conic $45°$ counterclockwise.
• Translation of the pole is trickier in polar coordinates and generally produces more complex equations. Most problems at this level stick to rotation.

Focus-directrix in polar form

The focus-directrix definition says: a conic section is the set of all points where the ratio of the distance to a fixed point (the focus) to the distance to a fixed line (the directrix) equals a constant. That constant is the eccentricity $e$.

In symbols, for any point $P$ on the conic:

$\frac{\text{distance from } P \text{ to focus}}{\text{distance from } P \text{ to directrix}} = e$

This is exactly what the polar equation $r = \frac{ep}{1 + e\cos\theta}$ encodes. Here $r$ is the distance to the focus (at the pole), and the denominator accounts for the distance to the directrix.

Working with the focus-directrix form:

• To find $e$ and $p$ from a given equation, rewrite it so the denominator looks like $1 + e\cos\theta$ (or $1 - e\cos\theta$). You may need to divide numerator and denominator by the same constant. Then $e$ is the coefficient of $\cos\theta$ and the numerator equals $ep$.
• To build an equation from given information, identify $e$ (from the conic type or given directly) and $p$ (from the focus-directrix distance), then plug into the general form.

Applications of this relationship show up in optics and orbital mechanics. Parabolic mirrors use the reflective property that all rays parallel to the axis reflect through the focus. Planetary orbits are ellipses with the Sun at one focus, and their polar equations use exactly this form.

Polar Coordinate System and Conic Sections

Polar coordinates describe a point's position using a radial distance $(r)$ from a fixed point (the pole) and an angle $(\theta)$ measured from the polar axis. This system is a natural fit for conics because placing a focus at the pole lets the equation directly express how distance varies with angle.

The latus rectum is a chord perpendicular to the major axis that passes through a focus, with both endpoints on the curve. Its length tells you how "wide" the conic is at the focus. In polar form, you can find the semi-latus rectum by evaluating $r$ at $\theta = \frac{\pi}{2}$:

$r\Big|_{\theta = \frac{\pi}{2}} = \frac{ep}{1 + e\cos\frac{\pi}{2}} = \frac{ep}{1} = ep$

So the semi-latus rectum equals $ep$, and the full latus rectum has length $2ep$. This connects directly to the parameter $p$ in the general equation and gives you a quick way to check your work when graphing.