# Thread: Polar to Rectangular and Vice Versa

1. ## Polar to Rectangular and Vice Versa

Well I've figured out how to do most of my questions I need to get done, but there are still a few that I cannot quite figure out.

r = 1 + 18cosѲ
r^2 = r + 8rcosѲ
x^2 + y^1 = r + 8y
This is where I get stumped.

8x^2 + 8y^2 - 88y = 0
x^2 + Y^2 - 11y = 0
r^2 - 11cosѲ = 0
r(r - 11cosѲ) = 0
r=11cosѲ

Not sure if I did this one correctly.

2. Hello, Crell!

$r \:=\: 1 + 18\cos\theta$
$r^2 \:=\: r + 8r\cos\theta$
$x^2 + y^2 \:=\: r + 8y$
This is where I get stumped.
We must eliminate that last $r.$

We have: . $x^2+y^2-18x \:=\:r$

Square: . $(x^2+y^2 - 18x)^2 \:=\:r^2$

Therefore: . $(x^2+y^2 - 18x)^2 \:=\:x^2+y^2$

. . I'd leave it like that.

$8x^2 + 8y^2 - 88y \:=\: 0$
$x^2 + y^2 - 11y \:=\: 0$
$r^2 - 11r{\color{red}\cos}\theta \:=\: 0$ ?
No . . . $y \:=\:r\sin\theta$

3. Originally Posted by Soroban
Hello, Crell!

We must eliminate that last $r.$

We have: . $x^2+y^2-18x \:=\:r$

Square: . $(x^2+y^2 - 18x)^2 \:=\:r^2$

Therefore: . $(x^2+y^2 - 18x)^2 \:=\:x^2+y^2$

. . I'd leave it like that.

No . . . $y \:=\:r\sin\theta$

Thanks a lot, guess I was pretty much there. Lets hope this online assignments takes that for an answer.