Question

How do you convert `r = 3 / (3 - cos(theta))` to rectangular form?

Answer 1

`8x^2+y^2-2x-1=0`

Explanation:

`r = sqrt(x^2+y^2)`
`costheta = x/sqrt(x^2+y^2)`

substituting

`sqrt(x^2+y^2) = 3/(3- x/sqrt(x^2+y^2)) = sqrt(x^2+y^2)/(3sqrt(x^2+y^2)-x)`

simplifying

`3sqrt(x^2+y^2)=x+1`

squaring

`9(x^2+y^2)=x^2+2x+1` and finally

`8x^2+y^2-2x-1=0` which corresponds to an ellipse.

Answer 2

`(x+3/8)^2/(9/8)^2+y^2/(3/(2sqrt 2))^2=1` that represents an ellipse.

Explanation:

The given equation can be written in the form

`1/r =1-(1/3)cos theta` that represents an ellipse of eccentricity 1/3.

Use the conversion equation `r(cos theta, sin theta)=(x, y)` that

gives `r = sqrt(x^2+y^2) and cos theta =x/r`..

Here, `r = 3/(3 -x/r)=(3r)/(3r-x). |r|3/|3-cos theta|>=3/4`. So, canceling r and

rearranging,

`3r=3sqrt(x^2+y^2)= x+3`

Rationalizing,

`9(x^2+y^2)=x^2-6x+9.`. Simplifying,
`8x^2+9y^2+6x-9=0`.

This can further reduced to the standard form

`(x+3/8)^2/(9/8)^2+y^2/(3/(2sqrt 2))^2=1` that represents an ellipse..