Chapter 8.5, Problem 57E

Solution

To prove: $r=\frac{ke}{\left(1-e\cos \theta \right)}$ is equivalent to $r=\frac{a\left(e^2-1\right)}{\left(1-e\cos \theta \right)}$ , where $a$ is the semi-transverse axis of the hyperbola.

It is proved that $r=\frac{a\left(e^2-1\right)}{\left(1-e\cos \theta \right)}$ .

Given information:

Polar equation: $r=\frac{ke}{\left(1-e\cos \theta \right)}$

Formula used:

  $k=ea-\frac{a}{e}$

Calculation:

Substitute $ea-\frac{a}{e}$ in $r$ .

  $\begin{array}{l}k=ea-\frac{a}{e}\\ k=a\left(e-\frac{1}{e}\right)\\ k=\frac{a}{e}\left(e^2-1\right)\\ ke=a\left(e^2-1\right)\end{array}$

Substitute $a\left(e^2-1\right)$ for $ke$ in $r$ .

  $r=\frac{a\left(e^2-1\right)}{\left(1-e\cos \theta \right)}$

Hence, it is proved that $r=\frac{a\left(e^2-1\right)}{\left(1-e\cos \theta \right)}$ .