Chapter 8.5, Problem 54E

Solution

a)

To draw: an appropriate diagram for the equation and derive it.

The equation $r=\frac{ke}{1-e\cos \theta }$ is derived.

Given information:

  

Formula used:

  $e=\frac{PF}{PD}$

Calculation:

The angle is formed counterclockwise from the $x$-axis. The directrix is a vertical line, which is on left side of the pole. The graph would be as follows:

  

The eccentricity is the ratio of the distance from focus to directrix. Therefore,

  $e=\frac{PF}{PD}$

Substitute

  $r$ for $PF$ and $k+r\cos \theta$ for $PD$ in $e$ .

  $e=\frac{r}{k+r\cos \theta }$

Solve for $r$ , by multiplying both sides of the equation by $k+r\cos \theta$ .

  $\begin{array}{l}e\left(k+r\cos \theta \right)=r\\ ke+er\cos \theta =r\\ r-er\cos \theta =ke\\ r\left(1-e\cos \theta \right)=ke\\ r=\frac{ke}{1-e\cos \theta }\end{array}$

Hence, the equation $r=\frac{ke}{1-e\cos \theta }$ is derived.

b)

To draw: an appropriate diagram for the equation and derive it.

The equation $r=\frac{ke}{1+e\sin \theta }$ is derived.

Given information:

  

Formula used:

  $e=\frac{PF}{PD}$

Calculation:

The angle is formed counterclockwise from the $x$-axis. The directrix is a horizontal line, which is on left side of the pole. The graph would be as follows:

  

The eccentricity is the ratio of the distance from focus to directrix. Therefore,

  $e=\frac{PF}{PD}$

Substitute $r$ for $PF$ and $k-r\sin \theta$ for $PD$ in $e$ .

  $e=\frac{r}{k-r\sin \theta }$

Solve for $r$ , by multiplying both sides of the equation by $k+r\cos \theta$ .

  $\begin{array}{l}e\left(k-r\sin \theta \right)=r\\ ke+er\sin \theta =r\\ r+er\sin \theta =ke\\ r\left(1+e\sin \theta \right)=ke\\ r=\frac{ke}{1+e\sin \theta }\end{array}$

Hence, the equation $r=\frac{ke}{1+e\sin \theta }$ is derived.

c)

To draw: an appropriate diagram for the equation and derive it.

The equation $r=\frac{ke}{1-e\sin \theta }$ is derived.

Given information:

  

Formula used:

Calculation:

The angle is formed counterclockwise from the $x$-axis. The directrix is a horizontal line, which is on below the pole. The graph would be as follows:

  

The eccentricity is the ratio of the distance from focus to directrix. Therefore,

  $e=\frac{PF}{PD}$

Substitute $r$ for $PF$ and $k+r\sin \theta$ for $PD$ in $e$ .

  $e=\frac{r}{k+r\sin \theta }$

Solve for $r$ , by multiplying both sides of the equation by $k+r\cos \theta$ .

  $\begin{array}{l}e\left(k+r\sin \theta \right)=r\\ ke+er\sin \theta =r\\ r-er\sin \theta =ke\\ r\left(1-e\sin \theta \right)=ke\\ r=\frac{ke}{1-e\sin \theta }\end{array}$

Hence, the equation $r=\frac{ke}{1-e\sin \theta }$ is derived.