Question

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

Answer 1

`(x+1/3)^2/(3/4)+y^2/(1/3)=1` See the markings at the center and the pole of the graph of this ellipse.

Explanation:

graph{(2sqrt(x^2+y^2)+x-1)(x^2+y^2-.0005)((x+1/3)^2+y^2-.0005)=0 [-1.25, 1.25, -0.625, 0.625]} The equation has the form

`(1/2)/r=1+1/2costheta`.

This represents the ellipse with focus S as pole, `theta=0` for S-side

major axis, eccentricity `e = 1/2` and semi major axis a = 2/3, from

`a(1-e^2)=l=1/2`.

The conversion formula is

`r(costheta, sin theta)=(x, y)`.

So, `2r+rcostheta=2sqrt(x^2+y^2)+x=1`, giving

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

In the standard form, this is

`(x+1/3)^2/(3/4)+y^2/(1/3)=1`

Answer 2

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

Explanation:

In converting from polar `(r,theta)` to Cartesian `(x,y)`
we have the relations:
`color(white)("XXX")color(red)(r=sqrt(x^2+y^2))`
and
`color(white)("XXX")color(blue)(cos(theta)=x/sqrt(x^2+y^2))`
(and others)

Polar form: `color(red)(r)(2+color(blue)(cos(theta)))=1`
becomes (in Cartesian form)
`color(white)("XXX")color(red)(sqrt(x^2+y^2)) * (2 +color(blue)(x/sqrt(x^2+y^2)))=1`

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

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

`color(white)("XXX")rarr x^2+y^2=(1-2x+x^2)/4`

`color(white)("XXX")rarr 4x^2+4y^2=1-2x+x^2`

`color(white)("XXX")rarr 3x^2+4y^2+2x=1`