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.
-$$e = \frac { 1 } { 3 } , y = 5$$
A) $r = \frac { 5 } { 3 - \sin \theta }$
B) $r = \frac { 5 } { 3 + \sin \theta }$
C) $r = \frac { 5 } { 1 + 3 \cos \theta }$
D) $r = \frac { 5 } { 3 + \cos \theta }$

Question

Quizplus · Accepted Answer

The directrix is horizontal (parallel to the x-axis), so we use the formula for a conic section in polar coordinates $r = \frac {ed}{1 + e \sin \theta}$ or $r = \frac {ed}{1 - e \cos \theta}$, where $d$ is the perpendicular distance from the directrix to the pole (origin), and $e$ is the eccentricity. Since the directrix is $y = 5$, $d = 5$, and $e = \frac{1}{3}$, the correct formula is $r = \frac {5}{1 + \frac{1}{3} \sin \theta}$, simplifying to $r = \frac {5}{3 + \sin \theta}$.

The Eccentricity Is Given of a Conic Section with One e=13,y=5e = \frac { 1 } { 3 } , y = 5e=31​,y=5

The Eccentricity Is Given of a Conic Section with One $e = \frac { 1 } { 3 } , y = 5$