Question

How do you convert `r = 8 + 5cos(theta)` into cartesian form?

Answer 1

`x^4 - 10x^3 - 39x^2+2x^2y^2-10xy^2-64y^2+y^4 = 0`

Explanation:

Remember the conversions:

`x = rcostheta`

`y = rsintheta`

`x^2+y^2=r^2`

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

Note that `r=8+5costheta` is always positive, so we don't need the `+-` sign in front of our square root when making that substitution

`r = 8 + 5costheta`

`r^2 = 8r +5rcostheta`

`x^2+y^2 = 8sqrt(x^2+y^2) + 5x`

Since we have `x` and `y` mixed together, we shouldn't try to separate them out and solve for `y` by itself. Instead, let's try to get this equation into the form `f(x,y) = 0`.

`x^2-5x+y^2 = 8sqrt(x^2+y^2)`

`(x^2-5x+y^2)^2 = 64(x^2+y^2)`

`x^4-10x^3+2x^2y^2+25x^2-10xy^2+y^4 = 64x^2+64y^2`

`x^4 - 10x^3 - 39x^2+2x^2y^2-10xy^2-64y^2+y^4 = 0`

We could try to simplify this in other ways, but this is a valid form of this equation, so we'll leave it like this.

Final Answer