Question

How do you convert `r(2 + cos theta) = 1` into cartesian form?

Answer 1

The equation is `3x^2+4y^2+2x-1=0`

Explanation:

To transform from polar coordinates `(r,theta)` into rectangular coordinates `(x,y)`, we use the folowing

`x=rcostheta`

`y=rsintheta`

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

Therefore,

`r(2+costheta)=1`

`2r+rcostheta=1`

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

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

Squaring both sides,

`4(x^2+y^2)=(1-x)^2=1-2x+x^2`

`3x^2+4y^2+2x-1=0`

Answer 2

It is the ellipsys `3x^2+4y^2+2x-1=0`

Explanation:

`rho=root2(x^2+y^2)` and `x=rho*sintheta` and `y=rho*costheta`
`root2(x^2+y^2)*(2+x/root2(x^2+y^2))=1`
that becomes
`2root2(x^2+y^2)+x=1`
or
`x^2+y^2=((1-x)/2)^2`
or
`x^2+y^2-(1-x)^2/4=0`