Question

How do you convert `r = 1/(8 - cos(theta))` into rectangular form?

Answer 1

`8sqrt(x^2+y^2)-x=1`

Explanation:

In converting between polar and rectangular forms:
`color(white)("XXX")r=sqrt(x^2+y^2)`
and
`color(white)("XXX")cos(theta)=x/r = x/(sqrt(x^2+y^2))`

Therefore, given
`color(white)("XXX")r = 1/(8-cos(theta))`

We have
`color(white)("XXX")r = 1/(8-x/r)`

`color(white)("XXX")r= 1/((8r-x)/r)`

`color(white)("XXX")r= r/(8r-x)`

`color(white)("XXX")8r-x=1`

`color(white)("XXX")8(sqrt(x^2+y^2))-x = 1`