Graph the parabola or ellipse. Include the directrix that corresponds to the focus at the origin.
-$r = \frac { 8 } { 2 - \cos \theta }$

Question

Quizplus · Accepted Answer

The given equation is in polar form. To plot it, we can convert it to rectangular form using $r^2=x^2+y^2$ and $x=r\cos\theta$, $y=r\sin\theta$. We get:
\begin{align*}
r&=\frac{8}{2-\cos\theta}\
r(2-\cos\theta)&=8\
r&=\frac{8}{2-\cos\theta}\cdot\frac{2+\cos\theta}{2+\cos\theta}\
r&=\frac{16+8\cos\theta}{3+\cos\theta}\
x^2+y^2&=\frac{(16+8\cos\theta)^2}{(3+\cos\theta)^2}\
(x^2+y^2)(3+\cos\theta)^2&=(16+8\cos\theta)^2
\end{align*}
Simplifying this equation gives us:
$$9x^2+9y^2-32x-24y+64=0$$
To find the directrix, we need to first identify the type of conic section we have. We can do this by finding the eccentricity:
\begin{align*}
\frac{r}{e}&=1+\frac{\cos\theta}{2-\cos\theta}\
\frac{8}{e(2-\cos\theta)}&=1+\frac{\cos\theta}{2-\cos\theta}\
8&=e(2-\cos\theta)+\cos\theta e\
8&=e(2\cos\theta+2)\
e&=\frac{4}{2+\cos\theta}
\end{align*}
Since $e<1$, the conic section is an ellipse. The focus is at the origin, so the directrix is the line $x=-\frac{c^2}{a}$, where $c$ is the distance from the origin to the focus, and $a$ is the length of the semi-major axis. To find these values, we need to write the equation of the ellipse in standard form:
$$\frac{(x-2)^2}{\frac{64}{27}}+\frac{(y-1)^2}{\frac{64}{27}}=1$$
So $a=\frac{8}{\sqrt{27}}=\frac{8\sqrt{3}}{9}$, and $c=\sqrt{a^2-b^2}=\frac{5\sqrt{3}}{3}$. Thus, the directrix is $x=-\frac{25}{27}$.

In the given options, only option B accurately depicts the position of the directrix, thus the answer is B.

Graph the Parabola or Ellipse r=82−cos⁡θr = \frac { 8 } { 2 - \cos \theta }r=2−cosθ8​

Graph the Parabola or Ellipse $r = \frac { 8 } { 2 - \cos \theta }$