Problem 29
Question
For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. $$ r=\frac{6 \sec \theta}{-2+3 \sec \theta} $$
Step-by-Step Solution
Verified Answer
The conic is an ellipse with eccentricity \( e = \frac{2}{3} \) and directrix \( d = 3 \).
1Step 1: Rewrite the Given Equation
The given equation is \( r = \frac{6 \sec \theta}{-2 + 3 \sec \theta} \). We know that \( \sec \theta = \frac{1}{\cos \theta} \), so substituting this into the equation, we get \( r = \frac{6}{-2 \cos \theta + 3} \).
2Step 2: Identify the Type of Conic Section
A conic section in polar form \( r = \frac{ed}{1 + e \cos \theta} \) has a focus at the origin and a directrix perpendicular to the polar axis. By comparing \( r = \frac{6}{-2 \cos \theta + 3} \) to the standard form \( r = \frac{ed}{1 + e \cos \theta} \), we can rewrite the expression as \( r = \frac{6/3}{-2/3 + \sec \theta} \). Here, we see \( ed = 6/3 = 2 \) and \( -e = -2/3 \). Solving for \( e \), we find \( e = \frac{2}{3} \).
3Step 3: Determine Eccentricity and Directrix
From the comparison, eccentricity \( e = \frac{2}{3} \) and is less than 1, so the conic section is an ellipse. The given polar equation after rearranging is \( r = \frac{2}{1+ \frac{2}{3}\cos \theta} \), which corresponds to the conic in standard form with \( ed = 2 \). Now, use \( e = \frac{2}{3} \) to find the directrix, \( d = \frac{ed}{e} = \frac{2}{2/3} = 3 \).
4Step 4: Finalize the Expression
The given conic is an ellipse as the eccentricity \( e \) is \( \frac{2}{3} \). Therefore, our final answer consists of the eccentricity \( e = \frac{2}{3} \) and the directrix \( d = 3 \). The conic section is an ellipse with these properties.
Key Concepts
Understanding EllipsesEccentricity ExplainedThe Role of the Directrix
Understanding Ellipses
An ellipse is a fascinating conic section and can be visualized as an elongated circle. Unlike a circle, which has one center point, an ellipse has two focal points known as foci. The shape of an ellipse is determined by these two foci and the distance between them. This characteristic gives an ellipse its unique appearance: it's slightly stretched if the foci are apart and more circular if they are closer together.
Mathematically, the standard form of an ellipse in polar coordinates is given by:
The condition that defines an ellipse is that the eccentricity \( e \) must be greater than 0 and less than 1 \((0 < e < 1)\). This means that the ellipse is always bounded, never stretching out to infinity, offering some symmetry and equilibrium in its dimensions.
Mathematically, the standard form of an ellipse in polar coordinates is given by:
- \( r = \frac{ed}{1 + e\cos\theta} \)
The condition that defines an ellipse is that the eccentricity \( e \) must be greater than 0 and less than 1 \((0 < e < 1)\). This means that the ellipse is always bounded, never stretching out to infinity, offering some symmetry and equilibrium in its dimensions.
Eccentricity Explained
Eccentricity, denoted as \( e \), is a crucial concept in understanding conic sections, particularly ellipses. It serves as a measure of how much an ellipse deviates from being circular. When you look at the formula for a conic section, eccentricity determines the shape and type of conic section:
- If \( e = 0 \), the conic is a circle.
- If \( 0 < e < 1 \), the conic is an ellipse.
- If \( e = 1 \), the conic is a parabola.
- If \( e > 1 \), the conic is a hyperbola.
The Role of the Directrix
The directrix is a significant concept in the study of conic sections. It is an imaginary line used to define and understand the properties of a conic. For an ellipse, the directrix complements the focus in describing the shape.
Certainly, every ellipse has a directrix just as it has a focus. The distance to the directrix helps us in conjunction with the eccentricity to describe the set of points that make up the ellipse; this relationship is described as:
Certainly, every ellipse has a directrix just as it has a focus. The distance to the directrix helps us in conjunction with the eccentricity to describe the set of points that make up the ellipse; this relationship is described as:
- The distance from any point on the ellipse to the focus divided by the perpendicular distance from the directrix gives the eccentricity \( e \).
Other exercises in this chapter
Problem 28
For the following exercises, determine the angle ? that will eliminate the xy term and write the corresponding equation without the \(xy\) term. $$x^{2}+4 x y+4
View solution Problem 28
Find the equations of the asymptotes for each hyperbola. \(\frac{(y-3)^{2}}{3^{2}}-\frac{(x+5)^{2}}{6^{2}}=1\)
View solution Problem 29
For the following exercises, convert the polar equation of a conic section to a rectangular equation. $$ r=\frac{6 \sec \theta}{-2+3 \sec \theta} $$
View solution Problem 29
Convert the polar equation of a conic section to a rectangular equation. $$ r=\frac{6 \sec \theta}{-2+3 \sec \theta} $$
View solution