Chapter 8.5, Problem 56E

Solution

To derive: formulas for the focal width of an ellipse and the focal width of a hyperbola.

The focal width is $2ke$ .

Formula used:

Polar equation: $r=\frac{ke}{1\pm \cos \theta }$

Polar equation: $r=\frac{ke}{1\pm \sin \theta }$

Calculation:

The focal width lies in the line $\theta =\frac{\pi }{2}$ . So, the two endpoints of the focal width when $\theta =\frac{\pi }{2}$ and $\theta =\frac{3\pi }{2}$ can be found by substituting them in $r$ .

For $\theta =\frac{\pi }{2}$ ,

  $\begin{array}{c}r=\frac{ke}{1\pm \cos \frac{\pi }{2}}\\ =\frac{ke}{1\pm 0}\\ =ke\end{array}$

For $\theta =\frac{3\pi }{2}$ ,

  $\begin{array}{c}r=\frac{ke}{1\pm \cos \frac{3\pi }{2}}\\ =\frac{ke}{1\pm 0}\\ =ke\end{array}$

The endpoints are $\left(ke,\frac{\pi }{2}\right)$ and $\left(ke,\frac{3\pi }{2}\right)$ .

The focal width is the sum of the two $x$ co-ordinates.

  $ke+ke=2ke$

The polar equation: $r=\frac{ke}{1\pm \sin \theta }$

The focus is at the pole with the focal axis. The focal width lies on the $x$ -axis. Find the two endpoints when $\theta =0$ and $\theta =\pi$ .

Substitute $0$ for $\theta$ in the polar equation.

  $\begin{array}{l}r=\frac{ke}{1\pm \sin \theta }\\ =\frac{ke}{1\pm \sin 0}\\ =\frac{ke}{1\pm 0}\\ =ke\end{array}$

Substitute $\pi$ for $\theta$ in the polar equation.

  $\begin{array}{l}r=\frac{ke}{1\pm \sin \pi }\\ =\frac{ke}{1\pm 0}\\ =ke\end{array}$

The endpoints are $\left(ke,0\right)$ and $\left(ke,0\right)$ .

The focal width is the sum of the $x$ co-ordinates.

  $ke+ke=2ke$

Hence, the focal width is $2ke$ .