The eccentricity is given of a conic section with one focus at the origin, along with the directrix corresponding to that focus. Find a polar equation for the conic section.
-e = 2, x = 9
A) $r = \frac { 18 } { 1 + 2 \sin \theta }$
B) $r = \frac { 18 } { 1 + 9 \cos \theta }$
C) $r = \frac { 18 } { 1 + 2 \cos \theta }$
D) $r = \frac { 18 } { 1 - 2 \cos \theta }$

Question

The eccentricity is given of a conic section with one focus at the origin, along with the directrix corresponding to that focus. Find a polar equation for the conic section.
-e = 2, x = 9 
A) $r = \frac { 18 } { 1 + 2 \sin \theta }$
B) $r = \frac { 18 } { 1 + 9 \cos \theta }$
C) $r = \frac { 18 } { 1 + 2 \cos \theta }$
D) $r = \frac { 18 } { 1 - 2 \cos \theta }$

Quizplus · Accepted Answer

Since the focus is at the origin, we know that the conic section is either an ellipse or a hyperbola. Since the eccentricity is greater than 1, we know it is a hyperbola.

The polar equation for a hyperbola with the focus at the origin and the directrix parallel to the $x$-axis is given by:
$$r=\frac{ed}{1\pm e\cos	heta}$$
where $e$ is the eccentricity, $d$ is the distance from the origin to the directrix, and the $\pm$ sign corresponds to the orientation of the hyperbola.

In this case, $e=2$ and the directrix is the vertical line $x=9$. The distance from the origin to the directrix is 9, since the directrix is perpendicular to the axis of symmetry of the hyperbola, which is the $y$-axis.

Thus, plugging in these values, we get:
$$r=\frac{(2)(9)}{1+2\cos	heta}=\frac{18}{1+2\cos	heta}$$

Therefore, the correct choice is $\boxed{	extbf{(C)}}$.

The Eccentricity Is Given of a Conic Section with One r=181+2sin⁡θr = \frac { 18 } { 1 + 2 \sin \theta }r=1+2sinθ18​

The Eccentricity Is Given of a Conic Section with One $r = \frac { 18 } { 1 + 2 \sin \theta }$