Graphing a Hyperbola in Polar Form · Physics Playground

Example 3 from Algebra and Trigonometry, 12.5 Conic Sections in Polar Coordinates

Graphing a Hyperbola in Polar Form

Graph $r = \frac{8}{2 - 3 \sin} .$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is $\frac{1}{2} .$

$\begin{array}{l} \begin{array}{l} r = \frac{8}{2 - 3 \sin θ} = \frac{8 (\frac{1}{2})}{2 (\frac{1}{2}) - 3 (\frac{1}{2}) \sin θ} \end{array} \\ r = \frac{4}{1 - \frac{3}{2} \sin} \end{array}$

Because $e = \frac{3}{2}, e > 1,$ so we will graph a hyperbola with a focus at the origin. The function has a $\sin θ$ term and there is a subtraction sign in the denominator, so the directrix is $y = - p .$

$\begin{array}{l} 4 = e p \\ 4 = (\frac{3}{2}) p \\ 4 (\frac{2}{3}) = p \\ \frac{8}{3} = p \end{array}$
The directrix is $y = - \frac{8}{3} .$

Plotting a few key points as in Table 2 will enable us to see the vertices. See Figure 5.

A B C D

$θ$ $0$ $\frac{π}{2}$ $π$ $\frac{3 π}{2}$

$r = \frac{8}{2 - 3 \sin θ}$ $4$ $- 8$ $4$ $\frac{8}{5} = 1.6$

Table 2

Figure 5

How did it go?