Example 1 from Algebra and Trigonometry, 12.5 Conic Sections in Polar Coordinates

Identifying a Conic Given the Polar Form

For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity.

$r = \frac{6}{3 + 2 \sin}$
$r = \frac{12}{4 + 5 \cos}$
$r = \frac{7}{2 - 2 \sin}$

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.

1
Solution
For each of the three conics, we will rewrite the equation in standard form. Standard form has a 1 as the constant in the denominator. Therefore, in all three parts, the first step will be to multiply the numerator and denominator by the reciprocal of the constant of the original equation, $\frac{1}{c},$ where $c$ is that constant.
1. Multiply the numerator and denominator by $\frac{1}{3} .$
  $r = \frac{6}{3 + 2 \sin θ} \cdot \frac{(\frac{1}{3})}{(\frac{1}{3})} = \frac{6 (\frac{1}{3})}{3 (\frac{1}{3}) + 2 (\frac{1}{3}) \sin θ} = \frac{2}{1 + \frac{2}{3} \sin}$
  
  Because $\sin θ$ is in the denominator, the directrix is $y = p .$ Comparing to standard form, note that $e = \frac{2}{3} .$ Therefore, from the numerator,
  
  $\begin{array}{l} 2 = e p \\ 2 = \frac{2}{3} p \\ (\frac{3}{2}) 2 = (\frac{3}{2}) \frac{2}{3} p \\ 3 = p \end{array}$
  Since $e < 1,$ the conic is an ellipse. The eccentricity is $e = \frac{2}{3}$ and the directrix is $y = 3.$
2. Multiply the numerator and denominator by $\frac{1}{4} .$
  $\begin{array}{l} \begin{array}{l} r = \frac{12}{4 + 5 \cos} \cdot \frac{(\frac{1}{4})}{(\frac{1}{4})} \end{array} \\ r = \frac{12 (\frac{1}{4})}{4 (\frac{1}{4}) + 5 (\frac{1}{4}) \cos θ} \\ r = \frac{3}{1 + \frac{5}{4} \cos} \end{array}$
  
  Because $cos θ$ is in the denominator, the directrix is $x = p .$ Comparing to standard form, $e = \frac{5}{4} .$ Therefore, from the numerator,
  
  $\begin{array}{l} 3 = e p \\ 3 = \frac{5}{4} p \\ (\frac{4}{5}) 3 = (\frac{4}{5}) \frac{5}{4} p \\ \frac{12}{5} = p \end{array}$
  
  Since $e > 1,$ the conic is a hyperbola. The eccentricity is $e = \frac{5}{4}$ and the directrix is $x = \frac{12}{5} = 2.4.$
3. Multiply the numerator and denominator by $\frac{1}{2} .$
  $\begin{array}{l} \begin{array}{l} r = \frac{7}{2 - 2 \sin} \cdot \frac{(\frac{1}{2})}{(\frac{1}{2})} \\ r = \frac{7 (\frac{1}{2})}{2 (\frac{1}{2}) - 2 (\frac{1}{2}) \sin} \\ r = \frac{\frac{7}{2}}{1 - \sin θ} \end{array} \end{array}$
  
  Because sine is in the denominator, the directrix is $y = - p .$ Comparing to standard form, $e = 1.$ Therefore, from the numerator,
  
  $\begin{array}{l} \frac{7}{2} = e p \\ \frac{7}{2} = (1) p \\ \frac{7}{2} = p \end{array}$
  
  Because $e = 1,$ the conic is a parabola. The eccentricity is $e = 1$ and the directrix is $y = - \frac{7}{2} = −3.5.$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Identify the conic with focus at the origin, the directrix, and the eccentricity for $r = \frac{2}{3 - \cos θ} .$

How did it go?

Solution

Your turn

Solution

Your turn