Problem 6 - Algebra and Trigonometry

Question

a. Identify the conic section that each polarequation represents. b. Describe the location of a directrix from the focus located at the pole. $$ r=\frac{8}{2-2 \sin \theta} $$

Step-by-Step Solution

Answer

The given polar equation represents a hyperbola, with the equation of the directrix being \( r \cos \theta = \frac{4}{15} \)

Step 1: Identify the Conic Section

First, rewrite the given polar equation in terms of \( r \) and \( \theta \) to resemble the standard equation for conic sections in polar form: \( r = \frac{e}{1 - e \sin \theta} \). Here \( e \) is the eccentricity. Comparing this with the standard equation, we see that \( e = 4 \), which makes our conic section a hyperbola since \( e > 1 \).

Step 2: Locate the Directrix

The location of the directrix for a hyperbola in polar form is given by the equation \( r \cos \theta = \frac{e}{e^2 - 1} \). With our value of \( e = 4 \), the equation of the directrix simplifies to \( r \cos \theta = \frac{4}{15} \)

Key Concepts

Hyperbola in Polar Form

A hyperbola can be elegantly expressed using polar coordinates, a form that can be particularly convenient for certain calculations or when assessing symmetry about the origin. Polar equations for conic sections are derived based on the focus-directrix property, where the distance to a focus divided by the distance to a directrix is constant, and this constant is the eccentricity, denoted by \( e \).

For a hyperbola in polar form, the equation is \( r = \frac{e}{1 - e \times \text{function of } \theta} \), where \( \text{function of } \theta \) depends on the orientation of the hyperbola. If the hyperbola's transverse axis is along the polar axis, the function is \( \text{function of } \theta = \text{cos} \theta \). Conversely, if the transverse axis is perpendicular to the polar axis, the function is \( \text{function of } \theta = \text{sin} \theta \). In the original exercise, the latter is the case.

A hyperbola will have an eccentricity \( e > 1 \). In the solved problem, with the given equation \( r = \frac{8}{2 - 2 \text{sin} \theta} \), by rearranging, we find that the eccentricity \( e = 4 \). The presence of \( \text{sin} \theta \) in the equation suggests that the hyperbola opens upwards or downwards with its transverse axis perpendicular to the initial line (the polar axis in this context).

Conic Section Directrix in Polar Form

In the realm of conic sections, the directrix serves as a fixed reference line, which, combined with a focus point, defines the set of points that form the conic in question. It's fundamentally involved in the focus-directrix definition of a conic section, where each point on the conic is at a distance from the focus that is a constant multiple (eccentricity) of the distance from the directrix.

For hyperbolas and ellipses, the equation for the directrix in polar coordinates relative to a focus at the pole generally takes the form \( r \text{cos} \theta = \frac{e}{e^2 - 1} \), where \( e \) is the eccentricity once again. The positive denominator indicates that this form will give the directrix of a hyperbola whose transverse axis aligns with the polar axis. If the transverse axis is perpendicular, a sine function would take the place of the cosine.

In the problem given, after identifying the hyperbola and its eccentricity as 4, the directrix is located via the equation \( r \text{cos} \theta = \frac{4}{15} \). It implies that the directrix is a straight line that does not pass through the pole since it's defined as a constant value for \( r \text{cos} \theta \), which is the rectangular coordinate \( x \) in polar form.

Polar Coordinates

Polar coordinates offer a two-dimensional coordinate system where each point on a plane is determined by an angle and a distance from a reference point known as the pole (similar to the origin in Cartesian coordinates). This system is particularly useful for dealing with problems involving circular or spherical symmetry, and is defined by the radius \( r \), which is the distance from the pole, and the angle \( \theta \), measured from the polar axis (analogous to the positive x-axis in Cartesian coordinates).

Converting between polar and rectangular (Cartesian) coordinates involves trigonometry. The relationships \( x = r \text{cos} \theta \) and \( y = r \text{sin} \theta \) are used to convert polar coordinates to rectangular coordinates. The inverse process can be done using \( r = \text{√}(x^2 + y^2) \) and \( \theta = \text{tan}^{-1}(y/x) \) (keeping in mind the correct quadrant for \( \theta \)).

Understanding polar coordinates is essential for solving problems like the one in the exercise, as it can simplify the representation of conic sections and make computations regarding their properties, like finding the location of a directrix, more intuitive.