# Polar Forms

TonyC
I am working a problem:
Find the polar form of the equation 6x^2 + 3xy + 6y^2 = -3
I have run this 4 separate times and have come up with 4 different answers.

r^2= 6 over -6 - 3sin(theta)cos(theta)

r^2=-3 over 6 - 3sin(theta)cos(theta)

r^2=9 over 6 + 3sin(theta)cos(theta)

r^2=-3 over 6 + 3sin(theta)cos(theta)

Homework Helper
Don't make things hard.
recall (i an using t instead of theta)
x^2+y^2=r^2
2xy=r^2sin(2t)
we have
6x^2 + 3xy + 6y^2 = -3
get nice numbers by multiplying by 2/3
4x^2 + 2xy + 4y^2 = -2
group by know facts
4(x^2+y^2)+2xy=-2
use known fact
4r^2+r^2sin(2t)=-2
factor r^2
r^2(4+sin(2t))=-2
divide
r^2=-2/(4+sin(2t))
remember different equation can mean the same thing
I think some of your answers may be correct, but I have not checked them in detail.
r^2= 6 over -6 - 3sin(theta)cos(theta)
r^2=-3 over 6 + 3sin(theta)cos(theta)
keep in mind we have r^2<0 hence r is and imaginary number
the original equation can be written
3x^2+(x+y)^2 + 3y^2 = -2
clearly this has no solution when x and y are restricted to real numbers.

Dr.Brain
Rearrange your equation into the form $6(x^2+y^2) + 3xy + 3 = 0$ , now do the following substitutions:

x=r cosQ
y=r sinQ

and

$r^2 = x^2 + y^2$

BJ