Problem 11
Question
(a) Find the eccentricity, (b) identify the conic, (c) give an equation of the directrix, and (d) sketch the conic. $$I^{*}=\frac{12}{4-\sin \theta}$$
Step-by-Step Solution
Verified Answer
Eccentricity is \( \frac{3}{4} \), conic is an ellipse, directrix is \( x = 16 \).
1Step 1: Identify Standard Form
The given equation is \( I^{*} = \frac{12}{4 - \sin \theta} \). This needs to be compared to the general form of a conic section in polar coordinates \( r = \frac{ed}{1 - e \sin \theta} \) or \( r = \frac{ed}{1 + e \sin \theta} \).
2Step 2: Rearrange Equation
Rearrange \( I^{*} = \frac{12}{4 - \sin \theta} \) into one of the standard forms. Let \( r = I^{*}\), and compare it to \( r = \frac{ed}{1 - e \sin \theta} \). In this case, \( ed = 12 \) and \( 1 - e = \frac{1}{4} \).
3Step 3: Solve for Eccentricity (e)
From \( 1 - e = \frac{1}{4} \), calculate \( e \): \( e = 1 - \frac{1}{4} = \frac{3}{4} \).
4Step 4: Identify the Conic
Since \( e < 1 \), the conic is an ellipse. An ellipse has \( 0 < e < 1 \).
5Step 5: Find Directrix Equation
The equation for directrix involves \( ed = 12 \). Since \( e = \frac{3}{4} \), \( d = \frac{12}{\frac{3}{4}} = 16 \). For \( r = \frac{ed}{1 - e \sin \theta} \), the directrix (vertical) is \( x = \frac{ed}{e} = 16 \).
6Step 6: Sketch the Conic
Since the form is \( r = \frac{ed}{1 - e \sin \theta} \), the ellipse is oriented horizontally with focus at the origin. Sketch an ellipse with the major axis touching \( \theta = 0 \) and \( \theta = \pi \), center at the origin, and directrix is a vertical line at \( x = 16 \).
Key Concepts
EccentricityEllipseDirectrix
Eccentricity
Eccentricity is a measure of how much a conic section deviates from being a perfect circle. It is a number that helps us categorize conic sections into different types.
- If the eccentricity (\( e \)) is 0, the conic is a circle.
- If \( 0 < e < 1 \), it is an ellipse, which is our case here with \( e = \frac{3}{4} \).
- If \( e = 1 \), the conic section is a parabola.
- If \( e > 1 \), it is a hyperbola.
Ellipse
An ellipse is a type of conic section that resembles an elongated circle. It consists of all points where the sum of the distances to two fixed points, called foci, is constant. The fact that the eccentricity (\( e \)) of our conic section is \( \frac{3}{4} \) confirms that it is indeed an ellipse.Some key features of an ellipse include:
- Major Axis: The longest diameter of the ellipse, running through both foci.
- Minor Axis: The shortest diameter, perpendicular to the major axis at the center.
- Foci: Two points inside the ellipse, which relate to the ellipse's shape and orientation.
Directrix
The directrix of a conic section is a line outside the conic (in the case of an ellipse) used, along with the focus (another fixed point), to define the ellipses through a specific geometric property.For our ellipse defined by \( I^{*} = \frac{12}{4 - \sin \theta} \), the directrix is determined using the relationship \( ed = 12 \) and the previously found value of \( e \). Solving \( d = \frac{12}{\frac{3}{4}} = 16 \), we find the directrix at \( x = 16 \) in a vertical orientation.Understanding the role of the directrix is crucial:
- It helps in precisely defining the curve of the ellipse.
- The closer an ellipse is to its directrix, the more eccentric it becomes, or vice versa.
- The directrix supports the geometric definition of an ellipse where the ratio of the distance from any point on the ellipse to the focus and the distance to the directrix is constant.
Other exercises in this chapter
Problem 10
\(7-12\) Sketch the region in the plane consisting of points whose polar coordinates satisfy the given conditions. $$ 2
View solution Problem 10
(a) Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as \(t\) increases. (b)
View solution Problem 11
Find the vertices and foci of the ellipse and sketch its graph. $$\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$$
View solution Problem 11
\(9-14\) Sketch the curve and find the area that it encloses. $$ r^{2}=4 \cos 2 \theta $$
View solution