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Chapter 10: Problem 12

Find a polar equation for the ellipse that has its focus at the pole and satisfies the stated conditions. (a) Directrix to the left of the pole; \(b=4 ; e=\frac{3}{5}\). (b) Directrix above the pole; \(c=5 ; e=\frac{1}{5}\).

Short Answer

Expert verified
(a) \( r = \frac{4}{1 - \frac{3}{5}\cos\theta} \); (b) \( r = \frac{5}{1 - \frac{1}{5}\sin\theta} \).

Step by step solution

01

Understanding Parameters

The problem asks for the polar equations of ellipses. In the polar form, an ellipse with a focus at the pole is given by the equation \( r = \frac{ed}{1 - e\cos\theta} \) when the directrix is on the left, and by \( r = \frac{ed}{1 - e\sin\theta} \) when it's above the pole. The parameters \( e \) represents the eccentricity, \( d \) the distance to the directrix, and \( b \), \( c \) are related to the ellipse dimensions.For (a), given \( b = 4 \) and \( e = \frac{3}{5} \), we use the formula for \( d = \frac{b}{e} \).For (b), we are given \( c = 5 \) and \( e = \frac{1}{5} \), and similarly use \( d = \frac{c}{e}\).
02

Calculate Directrix Distance for Part (a)

The ellipse's directrix distance \( d \) for part (a) is calculated using \( d = \frac{b}{e} \).Substitute \( b = 4 \) and \( e = \frac{3}{5} \):\[ d = \frac{4}{\frac{3}{5}} = \frac{4 \times 5}{3} = \frac{20}{3}. \]
03

Write the Polar Equation for Part (a)

With the directrix on the left and using the distance calculated \( d = \frac{20}{3} \), substitute in the polar form formula:\[ r = \frac{\frac{3}{5} \cdot \frac{20}{3}}{1 - \frac{3}{5}\cos\theta} = \frac{4}{1 - \frac{3}{5}\cos\theta}. \]
04

Calculate Directrix Distance for Part (b)

The ellipse's directrix distance \( d \) for part (b) is calculated using \( d = \frac{c}{e} \).Substitute \( c = 5 \) and \( e = \frac{1}{5} \):\[ d = \frac{5}{\frac{1}{5}} = 5 \times 5 = 25. \]
05

Write the Polar Equation for Part (b)

With the directrix above the pole and using the distance calculated \( d = 25 \), substitute in the polar form:\[ r = \frac{\frac{1}{5} \cdot 25}{1 - \frac{1}{5}\sin\theta} = \frac{5}{1 - \frac{1}{5}\sin\theta}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Eccentricity
Eccentricity is a measure that helps us understand how much an ellipse deviates from being a circle. It is denoted by the symbol \( e \). Here's what you need to know about eccentricity in simple terms:
  • For a circle, \( e = 0 \). This means the circle is perfectly round.
  • For an ellipse, the value of \( e \) is between 0 and 1. This indicates how "stretched" the ellipse is. The closer \( e \) is to 0, the more it resembles a circle. The closer to 1, the more elongated it is.
  • In our example, the eccentricities \( e= \frac{3}{5} \) for part (a) and \( e= \frac{1}{5} \) for part (b) signify ellipses that are moderately elongated and almost circular respectively.
By understanding eccentricity, you can predict the shape of the ellipse even before seeing it. It plays a crucial role in defining the relationship between the distance from the focus to a point on the ellipse and the distance from this point to the directrix.
Exploring the Directrix
The directrix is a fixed line used to define a conic section like an ellipse. In the polar coordinates system, it helps determine the shape and orientation of the ellipse.
  • The directrix is involved in the equation: \( r = \frac{ed}{1 - e\cos\theta} \) or \( r = \frac{ed}{1 - e\sin\theta} \). It is critical for knowing where the ellipse will extend in the coordinate system.
  • In part (a), the directrix is located on the left of the pole, requiring the use of \( e\cos\theta \).
  • For part (b), the directrix is above the pole, so \( e\sin\theta \) comes into play.
  • The distance from the pole to the directrix is represented by \( d \), calculated easily through \( d = \frac{b}{e} \) or \( d = \frac{c}{e} \) depending on the initial conditions provided.
By understanding the role of the directrix, you can comprehend how ellipses are positioned in space and how they will look when plotted.
Polar Coordinates and Their Significance
Polar coordinates offer a way to represent points in a plane, using distance from a reference point and angle from a reference direction. This system is especially beneficial when dealing with circles or ellipses, as it simplifies the mathematical equations. Here鈥檚 how it works:
  • The position of a point is given by \( (r, \theta) \), where \( r \) is the distance from the pole (the origin in polar coordinates) and \( \theta \) is the angle from the positive x-axis.
  • This system is critical in formulating polar equations of ellipses like \( r = \frac{ed}{1 - e \cos \theta} \) or \( r = \frac{ed}{1 - e \sin \theta} \).
  • Using polar coordinates allows easy calculation of points on an ellipse that has a focus at the pole, as it incorporates both direction and distance.
Polar coordinates are key in understanding how ellipses are constructed and visualized when their focus is at the pole, providing a clear and straightforward way to describe their position.

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Most popular questions from this chapter

Prove that a hyperbola is an equilateral hyperbola if and only if \(e=\sqrt{2}\).

Consider the family of curves described by the parametric equations $$ x=a \cos t+h, \quad y=b \sin t+k \quad(0 \leq t<2 \pi) $$ where \(a \neq 0\) and \(b \neq 0\). Describe the curves in this family if (a) \(h\) and \(k\) are fixed but \(a\) and \(b\) can vary (b) \(a\) and \(b\) are fixed but \(h\) and \(k\) can vary (c) \(a=1\) and \(b=1\), but \(h\) and \(k\) vary so that \(h=k+1\).

Consider the second-degree equation $$A x^{2}+C y^{2}+D x+E y+F=0$$ where \(A\) and \(C\) are not both \(0 .\) Show by completing the square: (a) If \(A C>0\), then the equation represents an ellipse, a circle, a point, or has no graph. (b) If \(A C<0\), then the equation represents a hyperbola or a pair of intersecting lines. (c) If \(A C=0\), then the equation represents a parabola, a pair of parallel lines, or has no graph.

Find the highest point on the cardioid \(r=1+\cos \theta\).

If \(f^{\prime}(t)\) and \(g^{\prime}(t)\) are continuous functions, and if no segment of the curve $$ x=f(t), \quad y=g(t) \quad(a \leq t \leq b) $$ is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the \(x\) -axis is $$ S=\int_{a}^{b} 2 \pi y \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ and the area of the surface generated by revolving the curve about the \(y\) -axis is $$ S=\int_{a}^{b} 2 \pi x \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ [The derivations are similar to those used to obtain Formulas (4) and (5) in Section 6.5.] Use the formulas above in these exercises. $$ \begin{aligned} &\text { Find the area of the surface generated by revolving the curve }\\\ &x=e^{t} \cos t, y=e^{t} \sin t(0 \leq t \leq \pi / 2) \text { about the } x \text { -axis. } \end{aligned} $$
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