To write the polar equation of a conic with given focus, directrix, and eccentricity, follow these steps:
1. Understand the Problem:
- Focus at the origin.
- Directrix given as [tex]\( x = 3 \)[/tex].
- Eccentricity ([tex]\( e \)[/tex]) is [tex]\( \frac{2}{3} \)[/tex].
2. Concept of Conic Sections in Polar Coordinates:
- The general polar equation of a conic section with the focus at the origin and directrix [tex]\( x = d \)[/tex] is:
[tex]\[
r = \frac{ed}{1 + e \cos \theta}
\][/tex]
- Here [tex]\( d \)[/tex] represents the distance from the focus to the directrix (which is 3 in this case), and [tex]\( e \)[/tex] is the eccentricity.
3. Substitute the given values:
- Eccentricity ([tex]\( e \)[/tex]) is [tex]\( \frac{2}{3} \)[/tex].
- Directrix [tex]\( d \)[/tex] is 3.
Plugging these values into the general formula, we get:
[tex]\[
r = \frac{\left(\frac{2}{3}\right) \cdot 3}{1 + \left(\frac{2}{3}\right) \cos \theta}
\][/tex]
4. Simplify the Expression:
- Multiply the numerator:
[tex]\[
r = \frac{2}{1 + \left(\frac{2}{3}\right) \cos \theta}
\][/tex]
- Simplify the fraction in the denominator by obtaining a common denominator:
[tex]\[
r = \frac{2}{\frac{3 + 2 \cos \theta}{3}}
\][/tex]
- Multiply the numerator and the denominator by 3 to simplify the equation:
[tex]\[
r = \frac{6}{3 + 2 \cos \theta}
\][/tex]
Therefore, the polar equation of the conic is:
[tex]\[
r = \frac{6}{3 + 2 \cos \theta}
\][/tex]
Among the provided options, this matches the first choice:
\[
r = \frac{6}{3 + 2 \cos \theta}
\)
