Chapter 8.5, Problem 24E

Solution

To find: The polar equation for the ellipse with a focus at the pole and the given polar coordinates as the endpoints of its major axis.

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

Given information:

The points are $\left(3,\frac{\pi }{2}\right)$ and $\left(0.75,-\frac{\pi }{2}\right)$ .

Formula used:

The focus at the pole is nearer to the point $\left(0.75,-\frac{\pi }{2}\right)$ and the directrix is horizontal, below the pole, so use the following polar equation.

  $r=\frac{ke}{1-e\sin \theta }$   ..... (1)

Calculation:

For illustration purposes, the locations of the endpoints of the major axis are on the line $\theta =\frac{\pi }{2}$ as shown below:

  

Substitute $3$ for $r$ and $\frac{\pi }{2}$ for $\theta$ in equation (1).

  $\begin{array}{c}3=\frac{ke}{1-e\sin \frac{\pi }{2}}\\ 3=\frac{ke}{1-e}\\ 3-3e=ke\end{array}$   ..... (2)

Substitute $0.75$ for $r$ and $-\frac{\pi }{2}$ for $\theta$ in equation (1).

  $\begin{array}{c}0.75=\frac{ke}{1-e\sin \left(-\frac{\pi }{2}\right)}\\ 0.75=\frac{ke}{1+e}\\ 0.75+0.75e=ke\end{array}$   ..... (3)

Subtract equation (2) and (3).

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

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

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

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