Chapter 8.5, Problem 27E

Solution

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

The required equation is $r=\frac{12}{2+3\sin \theta }$ .

Given information:

The points are $\left(2.4,\frac{\pi }{2}\right)$ and $\left(-12,\frac{3\pi }{2}\right)$ .

Formula used:

The focus at the pole is nearer to the point $\left(2.4,\frac{\pi }{2}\right)$ and the directrix is horizontal, above 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 transverse axis are on the polar axis as shown below:

  

Substitute $2.4$ for $r$ and $\frac{\pi }{2}$ for $\theta$ in equation (1).

  $\begin{array}{c}2.4=\frac{ke}{1+e\sin \frac{\pi }{2}}\\ 2.4=\frac{ke}{1+e}\\ 2.4+2.4e=ke\end{array}$   ..... (2)

Substitute $-12$ for $r$ and $\frac{3\pi }{2}$ for $\theta$ in equation (1).

  $\begin{array}{c}-12=\frac{ke}{1+e\sin \frac{3\pi }{2}}\\ -12=\frac{ke}{1-e}\\ -12+12e=ke\end{array}$   ..... (3)

Subtract equation (2) and (3).

  $\begin{array}{c}\left(2.4+2.4e\right)-\left(-12+12e\right)=ke-ke\\ 14.4-9.6e=0\\ 9.6e=14.4\\ e=\frac{3}{2}\end{array}$

Substitute $\frac{3}{2}$ for $e$ in equation (2) to find $k$ .

  $\begin{array}{c}2.4+2.4\left(\frac{3}{2}\right)=k\left(\frac{3}{2}\right)\\ k=\frac{2.4+2.4\left(1.5\right)}{1.5}\\ =4\end{array}$

Substitute $\frac{3}{2}$ for $e$ and $4$ for $k$ in equation (1).

  $\begin{array}{c}r=\frac{4\left(\frac{3}{2}\right)}{1+\frac{3}{2}\sin \theta }\\ r=\frac{12}{2+3\sin \theta }\end{array}$

Therefore, the required equation is $r=\frac{12}{2+3\sin \theta }$ .