Chapter 8.5, Problem 29E

Solution

To find: The polar equation for the conic with a focus at the pole and the given polar coordinates for the intercepts shown.

The required equation is $r=\frac{6}{5+3\cos \theta }$ .

Given information:

Consider the given graph.

Formula used:

The focus at the pole is nearer to the point $\left(0.75,0\right)$ and the directrix is vertical, to the right of the pole, so use the following polar equation.

  $r=\frac{ke}{1+e\cos \theta }$   ..... (1)

Calculation:

Substitute $0.75$ for $r$ and $0$ for $\theta$ in equation (1).

  $\begin{array}{c}0.75=\frac{ke}{1+e\cos 0}\\ 0.75=\frac{ke}{1+e}\\ 0.75+0.75e=ke\end{array}$   ..... (2)

Substitute $3$ for $r$ and $\pi$ for $\theta$ in equation (1).

  $\begin{array}{c}3=\frac{ke}{1+e\cos \pi }\\ 3=\frac{ke}{1-e}\\ 3-3e=ke\end{array}$   ..... (3)

Subtract equation (2) and (3).

  $\begin{array}{c}\left(0.75+0.75e\right)-\left(3-3e\right)=ke-ke\\ -2.25+3.75e=0\\ 3.75e=2.25\\ e=\frac{3}{5}\end{array}$

Substitute $\frac{3}{5}$ for $e$ in equation (3) to find $k$ .

  $\begin{array}{c}3-3\left(\frac{3}{5}\right)=k\left(\frac{3}{5}\right)\\ k=\frac{3-3\left(0.6\right)}{0.6}\\ =2\end{array}$

Substitute $\frac{3}{5}$ for $e$ and $2$ for $k$ in equation (1).

  $\begin{array}{c}r=\frac{\left(2\right)\left(\frac{3}{5}\right)}{1+\frac{3}{5}\cos \theta }\\ r=\frac{6}{5+3\cos \theta }\end{array}$

Therefore, the required equation is $r=\frac{6}{5+3\cos \theta }$ .