Identify the eccentricity of the conic section whose equation is given. r = 3/5 - 3cosθ3/515/33

Answer:3/5Step-by-step explanation:A conic section with a focus at the origin, a directrix of x = ±p where p is a positive real number and positive eccentricity (e) has a polar equation:[tex]r=\frac{ep}{1 \pm e*cos(\theta)}[/tex]Given the conic equation   [tex]r=\frac{3}{5-3cos(\theta)}[/tex]We have to make the conic equation to be in the form [tex]r=\frac{ep}{1 \pm e*cos(\theta)}[/tex].[tex]r=\frac{3}{5-3cos(\theta)}\\\\Multiply\ the\ numerator\ and \ denominator\ by \ 1/5\\r=\frac{3*\frac{1}{5} }{(5-3cos(\theta))*\frac{1}{5}}\\\\r=\frac{3*\frac{1}{5} }{5*\frac{1}{5}-3cos(\theta)*\frac{1}{5}}\\\\r=\frac{\frac{3}{5} }{1-\frac{3}{5}cos(\theta)}[/tex]Comparing with  [tex]r=\frac{ep}{1 \pm e*cos(\theta)}[/tex]. gives:e = 3/5, p = 1The eccentricity is 3/5