10.5 Conic sections in polar coordinates  (Page 2/8)

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 $\frac{1}{3}.$
   $$r=\frac{6}{3+2\sin \theta }\cdot \frac{\left(\frac{1}{3}\right)}{\left(\frac{1}{3}\right)}=\frac{6\left(\frac{1}{3}\right)}{3\left(\frac{1}{3}\right)+2\left(\frac{1}{3}\right)\sin \theta }=\frac{2}{1+\frac{2}{3}\sin \theta }$$

   Because $\sin \theta$ is in the denominator, the directrix is $y=p.$ Comparing to standard form, note that $e=\frac{2}{3}.$ Therefore, from the numerator,

   $$\begin{array}{l}2=ep\hfill \\ 2=\frac{2}{3}p\hfill \\ \left(\frac{3}{2}\right)2=\left(\frac{3}{2}\right)\frac{2}{3}p\hfill \\ 3=p\hfill \end{array}$$

   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 $\frac{1}{4}.$
   $$\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ r=\frac{12}{4+5\cos \theta }\cdot \frac{\left(\frac{1}{4}\right)}{\left(\frac{1}{4}\right)}\hfill \end{array}\hfill \\ r=\frac{12\left(\frac{1}{4}\right)}{4\left(\frac{1}{4}\right)+5\left(\frac{1}{4}\right)\cos \theta }\hfill \\ r=\frac{3}{1+\frac{5}{4}\cos \theta }\hfill \end{array}$$

   Because $\text{ cos}\theta$ is in the denominator, the directrix is $x=p.$ Comparing to standard form, $e=\frac{5}{4}.$ Therefore, from the numerator,

   $$\begin{array}{l}3=ep\hfill \\ 3=\frac{5}{4}p\hfill \\ \left(\frac{4}{5}\right)3=\left(\frac{4}{5}\right)\frac{5}{4}p\hfill \\ \frac{12}{5}=p\hfill \end{array}$$

   Since $e>1,$ the conic is a hyperbola    . The eccentricity is $e=\frac{5}{4}$ and the directrix is $x=\frac{12}{5}=\mathrm{2.4.}$

3. Multiply the numerator and denominator by $\frac{1}{2}.$
   $$\begin{array}{l}\hfill \\ \hfill \\ \begin{array}{l}r=\frac{7}{2-2\sin \theta }\cdot \frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)}\hfill \\ r=\frac{7\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right)-2\left(\frac{1}{2}\right)\sin \theta }\hfill \\ r=\frac{\frac{7}{2}}{1-\sin \theta }\hfill \end{array}\hfill \end{array}$$

   Because sine is in the denominator, the directrix is $y=-p.$ Comparing to standard form, $e=1.$ Therefore, from the numerator,

   $\begin{array}{l}\frac{7}{2}=ep\\ \frac{7}{2}=\left(1\right)p\\ \frac{7}{2}=p\end{array}$

   Because $e=1,$ the conic is a parabola    . The eccentricity is $e=1$ and the directrix is $y=-\frac{7}{2}=\mathrm{-3.5.}$

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is $\frac{1}{3}.$

Because $e=1,$ we will graph a parabola    with a focus at the origin. The function has a $\cos \theta ,$ and there is an addition sign in the denominator, so the directrix is $x=p.$

The directrix is $x=\frac{5}{3}.$

Plotting a few key points as in [link] will enable us to see the vertices. See [link] .