Need a Pre Calculus problem checked?

Question

Find the rectangular form of the polar equation
r=1/(2+cosӨ)
i think the answer is 3x^2+4y^2+2x-1=0

Answer 1

r=1/(2+cosӨ)
r=sqrt(x^2+y^2)
and
x=rcosӨ so cosӨ = x/sqrt(x^+y^2)

sqrt(x^2+y^2) = 1/(2+ x/sqrt(x^+y^2) )
(2+ x/sqrt(x^+y^2) ) sqrt(x^2+ y^2 ) =1
2sqrt(x^2+ y^2 ) + x = 1
2sqrt(x^2+ y^2 ) = - x + 1
4(x^2 +y^2) = x^2 -2x + 1
3x^2 + 4y^2 + 2x -1 =0 is the equation .

Answer 2

r=sqrt(x^2+y^2) r*cos(Theta)=x so 2+cos(theta)= (2r+x)/r

so r = r/(2r+x) simplifyinf 2r+x= 1 ===> 2r= 1-x

squaring both sides 4(x^2+y^2)= x^2-2x+1
3x^2+4y^2+2x-1=0

you are right