Problem 43
Question
The polar equation of an ellipse can be expressed in terms of its eccentricity \(e\) and the length \(a\) of its major axis. (a) Show that the polar equation of an ellipse with directrix \(x=-d\) can be written in the form $$r=\frac{a\left(1-e^{2}\right)}{1-e \cos \theta}$$ [Hint: Use the relation \(a^{2}=e^{2} d^{2} /\left(1-e^{2}\right)^{2}\) given in the proof on page 825 .] (b) Find an approximate polar equation for the elliptical orbit of the earth around the sun (at one focus) given that the eccentricity is about 0.017 and the length of the major axis is about \(2.99 \times 10^{8} \mathrm{km}\).
Step-by-Step Solution
Verified Answer
(a) Proved equation: \( r = \frac{a(1-e^2)}{1-e\cos\theta} \). (b) Earth's orbit: \( r \approx \frac{2.98848 \times 10^8}{1-0.017\cos\theta} \).
1Step 1: Understanding the Polar Equation of an Ellipse
An ellipse in polar coordinates with a focus at the origin is typically written as \( r = \frac{ed}{1 - e \cos \theta} \). Here \( r \) is the radial distance, \( e \) is the eccentricity, and \( d \) is the distance from the directrix to the ellipse.
2Step 2: Relate Parameters a and d
From the hint, we have the relationship \( a^2 = \frac{e^2 d^2}{(1-e^2)^2} \). Solving for \( d \), we get \( d = \frac{a (1-e^2)}{e} \). Substitute this directrix into the general polar equation \( r = \frac{ed}{1 - e \cos \theta} \).
3Step 3: Substitute Directrix in Polar Equation
Substitute \( d = \frac{a (1-e^2)}{e} \) into the ellipse equation: \( r = \frac{e \left(\frac{a (1-e^2)}{e}\right)}{1 - e \cos \theta} = \frac{a (1-e^2)}{1 - e \cos \theta} \). This confirms the equation given in part (a).
4Step 4: Substitute Known Values for Earth's Orbit
For part (b), substitute \( e = 0.017 \) and \( a = 2.99 \times 10^8 \) km into the equation: \( r = \frac{2.99 \times 10^8 (1 - (0.017)^2)}{1 - 0.017 \cos \theta} \).
5Step 5: Simplify Earth's Orbit Equation
First compute \( (0.017)^2 = 0.000289 \) and \( 1 - 0.000289 = 0.999711 \). Substitute these into the equation: \( r \approx \frac{2.99 \times 10^8 \times 0.999711}{1 - 0.017 \cos \theta} \).
6Step 6: Final Simplified Polar Equation
Calculate \( 2.99 \times 10^8 \times 0.999711 = 2.98848 \times 10^8 \). The approximate polar equation is \( r \approx \frac{2.98848 \times 10^8}{1 - 0.017 \cos \theta} \).
Key Concepts
EccentricityMajor AxisDirectrix
Eccentricity
Eccentricity is a crucial concept when discussing ellipses. It measures how much an ellipse deviates from being a perfect circle. You can think of it as how "stretched out" the ellipse is. The symbol used for eccentricity is \( e \). In mathematical terms:
This eccentricity value directly impacts the polar equation of the ellipse. Specifically, in the equation \( r = \frac{a(1-e^2)}{1 - e \cos \theta} \), it influences the numerator and can alter the shape described by the equation significantly.
- If \( e = 0 \), the ellipse is a perfect circle.
- If \( 0 < e < 1 \), the figure is an ellipse.
- The closer \( e \) is to 1, the more elongated the ellipse becomes.
This eccentricity value directly impacts the polar equation of the ellipse. Specifically, in the equation \( r = \frac{a(1-e^2)}{1 - e \cos \theta} \), it influences the numerator and can alter the shape described by the equation significantly.
Major Axis
The major axis of an ellipse is the longest diameter, stretching from one end of the ellipse to the other, passing through the center and both foci. It is denoted by the letter \( a \). This axis is pivotal because it determines the size of the ellipse and is used in forming the polar equation.
In polar coordinates, for an ellipse, the term involving \( a \), which is \( a(1-e^2) \), is used as part of the conversion between rectangular and polar forms to ensure the expression correctly represents the ellipse.
Understanding and calculating the major axis helps in applications such as astronomy, where celestial bodies often follow elliptical paths. In the problem given, the Earth's major axis is approximately \( 2.99 \times 10^8 \) km, showcasing its large orbital path size around the Sun.
In polar coordinates, for an ellipse, the term involving \( a \), which is \( a(1-e^2) \), is used as part of the conversion between rectangular and polar forms to ensure the expression correctly represents the ellipse.
Understanding and calculating the major axis helps in applications such as astronomy, where celestial bodies often follow elliptical paths. In the problem given, the Earth's major axis is approximately \( 2.99 \times 10^8 \) km, showcasing its large orbital path size around the Sun.
Directrix
A directrix in the context of conic sections, such as ellipses, is a fixed line used to derive the locus of the points forming the curve. The distance from this line helps define the shape and geometry of the ellipse.
The directrix plays a vital role when expressing the ellipse in polar form. For a given ellipse, an equation can be derived using the relationship between the eccentricity \( e \), the semi-major axis \( a \), and the directrix \( d \). This relationship is mathematically expressed as \( d = \frac{a(1-e^2)}{e} \), and it's substituted into the general polar form to account for the directrix's influence.
This conversion is especially important when changing coordinate systems, ensuring the ellipse's equation maintains its geometric properties no matter its representation. By using the directrix, the equation can adequately describe the position of the ellipse concerning its focus.
The directrix plays a vital role when expressing the ellipse in polar form. For a given ellipse, an equation can be derived using the relationship between the eccentricity \( e \), the semi-major axis \( a \), and the directrix \( d \). This relationship is mathematically expressed as \( d = \frac{a(1-e^2)}{e} \), and it's substituted into the general polar form to account for the directrix's influence.
This conversion is especially important when changing coordinate systems, ensuring the ellipse's equation maintains its geometric properties no matter its representation. By using the directrix, the equation can adequately describe the position of the ellipse concerning its focus.
Other exercises in this chapter
Problem 42
Find an equation for the hyperbola that satisfies the given conditions. Vertices: \((0, \pm 6),\) asymptotes: \(y=\pm \frac{1}{3} x\)
View solution Problem 42
Find an equation for the parabola that has its vertex at the origin and satisfies the given condition(s). Directrix: \(y=-5\)
View solution Problem 43
Finding the Equation of an Ellipse Find an equation for the ellipse that satisfies the given conditions. Foci: \(F(0, \pm \sqrt{10}),\) vertices: \((0, \pm 7)\)
View solution Problem 43
Find an equation for the hyperbola that satisfies the given conditions. Vertices: \((0, \pm 6),\) hyperbola passes through \((-5,9)\)
View solution