Chapter 10.3, Problem 69E

(a)

To determine

To plot: The graph of $r=\frac{2}{1+e\cos \theta }$ for $e=-0.1$ , $-0.3$ , $-0.5$ , $-0.7$ and $-0.9$ .

Answer to Problem 69E

The graph of $r=\frac{2}{1+e\cos \theta }$ for $e=-0.1$ , $-0.3$ , $-0.5$ , $-0.7$ and $-0.9$ is an ellipse shown in Figure (1).

Explanation of Solution

Given information: The given equation is $r=\frac{2}{1+e\cos \theta }$ . The given values are $e=-0.1$ , $-0.3$ , $-0.5$ , $-0.7$ and $-0.9$ .

Calculation:

Consider the given equation $r=\frac{2}{1+e\cos \theta }$ .

Draw the graph of $r\left(e\right)$ for $e=-0.1$ , $-0.3$ , $-0.5$ , $-0.7$ and $-0.9$ on the same plane as shown below.

  

Figure (1)

From the graph it can be observed that the graph of $r=\frac{2}{1+e\cos \theta }$ for $e=-0.1$ , $-0.3$ , $-0.5$ , $-0.7$ and $-0.9$ are ellipses.

Therefore, the graphs of $r=\frac{2}{1+e\cos \theta }$ for $e=-0.1$ , $-0.3$ , $-0.5$ , $-0.7$ and $-0.9$ is an ellipse shown in Figure (1).

(b)

To determine

To explain: The relationship between the graphs of conic section with eccentricity $e$ and the graph with eccentricity $-e$ .

Answer to Problem 69E

The graphs of $r=\frac{2}{1+e\cos \theta }$ with eccentricity $e$ and eccentricity $-e$ approaches the circle with radius $2$ and center at the origin and symmetric about the $x$ -axis.

Explanation of Solution

Given information: The given equation is $r=\frac{2}{1+e\cos \theta }$ .

From part (a), the graph of $r=\frac{2}{1+e\cos \theta }$ for $e=-0.1$ , $-0.3$ , $-0.5$ , $-0.7$ and $-0.9$ approaches the circle with radius $2$ and center at the origin from the opposite side as $e\to 0^{+}$ .

Also, the graph of $r=\frac{2}{1+e\cos \theta }$ for eccentricity $e$ approaches the circle with radius $2$ and center at the origin as $e\to 0^{+}$ .

Therefore, the graphs of $r=\frac{2}{1+e\cos \theta }$ with eccentricity $e$ and eccentricity $-e$ approaches the circle with radius $2$ and center at the origin and symmetric about the $x$ -axis.