Find a polar equation of the conic with the given characteristics and with one focus at the pole.

A) $r = \frac { 12 } { 4 - 3 \sin \theta }$
B) $r = \frac { 12 } { 4 + 3 \cos \theta }$
C) $r = \frac { 12 } { 4 - 3 \cos \theta }$
D) $r = \frac { 12 } { 4 + 3 \sin \theta }$
E) $r = \frac { 12 } { 3 - 5 \cos \theta }$

Question

Find a polar equation of the conic with the given characteristics and with one focus at the pole.

A) $r = \frac { 12 } { 4 - 3 \sin \theta }$ 
B) $r = \frac { 12 } { 4 + 3 \cos \theta }$ 
C) $r = \frac { 12 } { 4 - 3 \cos \theta }$ 
D) $r = \frac { 12 } { 4 + 3 \sin \theta }$ 
E) $r = \frac { 12 } { 3 - 5 \cos \theta }$

Quizplus · Accepted Answer

Since one focus is at the pole, the conic is either a circle or an ellipse.

For a circle, there is only one possible polar equation: $r=a$, where $a$ is the radius of the circle. However, this would mean that the other focus is also at the pole, which is not the case. Therefore, the conic must be an ellipse.

The polar equation of an ellipse with one focus at the pole is given by $r=\frac{p}{1-e\cos	heta}$, where $p$ is the distance from the pole to the directrix, and $e$ is the eccentricity of the ellipse. Since the focus is at the pole, $p=a$, where $a$ is the length of the semi-major axis.

To determine the eccentricity, we use the fact that the distance from the pole to a point on the ellipse is given by $r=\frac{r_0}{1-e\cos(	heta-\phi)}$, where $r_0$ is the distance from the pole to the nearest point on the directrix, and $\phi$ is the angle between the polar axis and the line from the pole to the nearest point on the directrix. Since the focus is at the pole, it is also at a distance of $a$ from the directrix. Therefore, $r_0=a$, and $\phi=0$.

Plugging in these values, we get:

$$r=\frac{a}{1-e\cos	heta}$$

At the end,
$$r=\frac{12}{4-3\cos	heta}$$

Find a Polar Equation of the Conic with the Given r=124−3sin⁡θr = \frac { 12 } { 4 - 3 \sin \theta }r=4−3sinθ12​

Find a Polar Equation of the Conic with the Given $r = \frac { 12 } { 4 - 3 \sin \theta }$