Problem 42 - Contemporary Precalculus

Question

Find the polar equation of the conic section that has focus (0,0) and satisfies the given conditions. Eccentricity \(1 ;\) directrix: \(r=5 \sec \theta\)

Step-by-Step Solution

Answer

Answer: The polar equation of the conic section is \(r = \frac{5/2}{1 + \cos(\theta)}\).

Step 1: Identify the type of conic section

Using the given eccentricity value of 1, we can identify that the conic section is a parabola. In polar coordinates, parabolas have the general form \(r = \frac{p}{1 + e\cos(\theta)}\). Since the eccentricity is 1, we have \(r = \frac{p}{1 + \cos(\theta)}\). Our goal is to find the value of \(p\) to obtain the polar equation.

Step 2: Apply focus-directrix property

Recall that for parabolas, the distance from any point on the parabola to the focus is equal to the distance from that point to the directrix. In this case, the focus is at the origin (0, 0) and the directrix is \(r = 5 \sec(\theta)\). To find \(p\), we will use the focus-directrix property for parabolas. Let P be a point on the parabola with polar coordinates \((r, \theta)\). The distance from P to the focus (0, 0) is simply \(r\). The distance from P to the directrix, denoted as PD, can be calculated as the difference between the radii of P and the directrix given by the formula: \(PD = |r - 5 \sec(\theta)|\). Since the distance from P to the focus is equal to the distance from P to the directrix, we have \(r = |r - 5 \sec(\theta)|\).

Step 3: Solve for \(p\)

In order to find the value of \(p\), we'll solve the equation \(r = |r - 5 \sec(\theta)|\). We have two cases to consider due to the absolute value. Case 1: \(r = r - 5 \sec(\theta)\). This gives us \(5 \sec(\theta) = 0\), which is only satisfied for \(\theta \neq n\pi\) for any integer \(n\). In this case, there is no solution for \(p\). Case 2: \(r = -r + 5 \sec(\theta)\). This gives us \(2r = 5 \sec(\theta)\) or \(r = \frac{5}{2} \sec(\theta)\). Comparing this equation with the general form of a parabola \(r = \frac{p}{1 + \cos(\theta)}\), we can see that \(p = \frac{5}{2}\).

Step 4: Write the polar equation of the parabola

Now that we have the value of \(p\), we can write the polar equation of the parabola as follows: \(r = \frac{5/2}{1 + \cos(\theta)}\).

Key Concepts

Parabola in Polar Coordinates

A parabola represented in polar coordinates offers a unique view of this well-known shape. Unlike its Cartesian cousin, which is commonly given as either a vertical or horizontal equation like y=ax^2 or x=ay^2, a parabola in polar form has a central focus at the origin, typically designated as (0,0). The general expression for a parabola in polar coordinates takes the form

\[ r = \frac{p}{1 + e\cos(\theta)} \]
where \( r \) is the distance from the origin to a point on the parabola, \( \theta \) is the angle from the polar axis to the point, \( e \) is the eccentricity, and \( p \) is a positive constant that represents the semi-latus rectum of the parabola (the distance from the focus to the curve measured parallel to the directrix). The unique aspect of a polar parabola is that it does not open up or down or left or right, but instead opens away from or toward the pole, depending on whether \( e \) is positive or negative. The polar form is particularly useful when dealing with problems involving angles or rotations.

Eccentricity

Eccentricity plays a pivotal role in distinguishing between the types of conic sections: circle, ellipse, parabola, and hyperbola. It is defined as a nonnegative real number and symbolizes how much the conic section deviates from being circular.

A parabola is characterized by having an eccentricity of \( e = 1 \). This unique value sets parabolas apart from ellipses, which have eccentricities between 0 and 1, and hyperbolas, which have eccentricities greater than 1. The circle stands alone as the only conic with an eccentricity of 0, making it a special case of an ellipse. Eccentricity helps in formulating the equation of a conic in polar coordinates because it directly influences the denominator of the polar equation, where for a parabola, the equation simplifies to \[ r = \frac{p}{1 + \cos(\theta)} \] given that the eccentricity is 1. Understanding the concept of eccentricity allows students to quickly identify the type of conic section they are working with, enabling them to use the appropriate formula for the situation at hand.

Focus-Directrix Property

The focus-directrix property of a parabola provides a remarkable geometric relationship that helps to derive its polar equation. This property states that any point on the parabola is equidistant to a fixed point, called the focus, and a straight line, called the directrix.

To apply this principle in polar coordinates, consider a point P with coordinates \((r, \theta)\) on the parabola. The distance to the focus, which is at the origin, is simply \(r\). The distance from P to the directrix requires a bit more work, as you must take the absolute difference between \(r\) and the polar equation of the directrix.

For our example problem, the directrix was given by \[ r = 5 \sec(\theta) \] Substituting into the focus-directrix property, we have \( r = |r - 5 \sec(\theta)| \). Solving this equation with consideration of absolute values yields the value of \( p \), which is the key to drafting the full polar equation. The focus-directrix property is essential in defining parabolas and can also extend to other conic sections, making it a versatile tool in understanding and solving conic equations in polar form.