The eccentricity of the conic $$\frac {3}{r}=4-2 cos \theta$$ is​

Correct option is A

​​$$\textbf{Solution: Direct Eccentricity from Polar Equation of Conic} \\[10pt]\text{Recall the general polar form of a conic section is:} \\[4pt]r = \frac{ed}{1 + e \cos \theta} \quad \text{or} \quad r = \frac{ed}{1 - e \cos \theta} \\[12pt]\text{Let's match:} \\[6pt]\text{Given:} \quad r = \frac{4 - 2 \cos \theta}{3} = \frac{2(2 - \cos \theta)}{3} \\[10pt]\text{So,} \\[6pt]r = \frac{2}{3} \cdot (2 - \cos \theta) = \frac{2}{3} \cdot \left( 2 \left(1 - \frac{1}{2} \cos \theta \right) \right) \Rightarrow e = \frac{1}{2} \\[12pt]\boxed{\text{Final Answer:} \quad \boxed{\frac{1}{2}}}$$

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