Question

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

3/5
1
5/3
3

Answer 1

3/5

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:

`r=(ep)/(1 \pm e*cos(\theta))`

Given the conic equation
`r=(3)/(5-3cos(\theta))`

We have to make the conic equation to be in the form
`r=(ep)/(1 \pm e*cos(\theta))`.

`r=(3)/(5-3cos(\theta))\\\\Multiply\ the\ numerator\ and \ denominator\ by \ 1/5\\r=(3*(1)/(5) )/((5-3cos(\theta))*(1)/(5))\\\\r=(3*(1)/(5) )/(5*(1)/(5)-3cos(\theta)*(1)/(5))\\\\r=((3)/(5) )/(1-(3)/(5)cos(\theta))`

Comparing with
`r=(ep)/(1 \pm e*cos(\theta))`. gives:

e = 3/5, p = 1

The eccentricity is 3/5