Problem 26
Question
(a) Graph the conics $$r=\frac{e d}{1+e \sin \theta}$$ for \(e=1\) and various values of \(d\) . How does the value of \(d\) affect the shape of the conic? (b) Graph these conics for \(d=1\) and various values of \(e\) . How does the value of \(e\) affect the shape of the conic?
Step-by-Step Solution
Verified Answer
Varying d scales the conic's size for parabolas; varying e changes the conic type.
1Step 1: Understanding the Conic Equation
The given conic section equation is in the polar form \( r = \frac{ed}{1 + e \sin \theta} \), where \( e \) is the eccentricity and \( d \) is the semi-latus rectum. Changing these parameters will influence the shape of the conic.
2Step 2: Case (a): Varying d with e=1
For \( e = 1 \), the conic is a parabola. By varying \( d \), the distance from the focus to the directrix changes. Graph \( r = \frac{1 \cdot d}{1 + 1 \cdot \sin \theta} = \frac{d}{1 + \sin \theta} \) for different values of \( d \). The shape remains a parabola, but its size increases as \( d \) increases.
3Step 3: Influence of d in Case (a)
Since \( e = 1 \) results in a parabola, changing \( d \) only affects the scaling of the parabola. A larger \( d \) results in a larger parabola that is more spread out, but the overall shape remains unchanged.
4Step 4: Case (b): Varying e with d=1
For \( d = 1 \), graph \( r = \frac{e}{1 + e \sin \theta} \) using different values of \( e \). The value of \( e \) determines the type of conic: if \( e < 1 \), the conic is an ellipse; if \( e = 1 \), it is a parabola; and if \( e > 1 \), it is a hyperbola.
5Step 5: Influence of e in Case (b)
By varying \( e \), you actively change the type of conic. Lowering \( e \) below 1 creates an ellipse (almost circular as \( e \to 0 \)), while increasing \( e \) above 1 results in a hyperbola, indicating that \( e \) plays a crucial role in determining the conic type.
Key Concepts
EccentricityPolar CoordinatesParabolaEllipseHyperbola
Eccentricity
Eccentricity is a fundamental concept when it comes to conic sections. It is denoted by the letter \( e \) and is a measure of how much a conic section deviates from being circular. The value of \( e \) determines the shape of the conic:
- If \( e = 0 \), the conic is a circle.
- If \( 0 < e < 1 \), it is an ellipse.
- If \( e = 1 \), the conic is a parabola.
- If \( e > 1 \), it becomes a hyperbola.
Polar Coordinates
Polar coordinates offer a different way to represent points on a plane compared to Cartesian coordinates. Instead of using \((x, y)\), polar coordinates use \((r, \theta)\), where:
- \( r \) is the radial distance from the origin to the point.
- \( \theta \) is the angle between the positive x-axis and the line connecting the origin to the point.
Parabola
A parabola is one of the simpler conic sections and occurs when the eccentricity \( e = 1 \). Parabolas have several notable properties:
- They have a single focus point.
- A directrix line is equidistant from any point on the parabola.
- They are symmetrical with respect to their axis.
Ellipse
Ellipses occur when the eccentricity \( 0 < e < 1 \). These shapes resemble stretched circles and have several key characteristics:
- Two foci located on the major axis of the ellipse.
- The sum of distances from any point on the ellipse to both foci is constant.
- Ellipses maintain symmetry along both their major and minor axes.
Hyperbola
A hyperbola is another type of conic that occurs when the eccentricity \( e > 1 \). Hyperbolas have distinct features:
- They consist of two separate branches.
- Each branch is mirrored about two axes.
- The difference in distances from any point on a branch to its two foci is constant.
Other exercises in this chapter
Problem 26
Use a graphing device to graph the ellipse. $$ x^{2}+\frac{y^{2}}{12}=1 $$
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(a) Use the discriminant to determine whether the graph of the equation is a parabola, an ellipse, or a hyperbola. (b) Use a rotation of axes to eliminate the \
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Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find th
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Use a graphing device to graph the hyperbola. \(\frac{x^{2}}{100}-\frac{y^{2}}{64}=1\)
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