How do you change, this polar form equation r(2 + cos theta) = 1 to Cartesian form?

Feb 15, 2016

$2 \sqrt{{x}^{2} + {y}^{2}} + x - 1$ I would leave it here optionally you can write (explanation)...
$5 {x}^{2} + 4 {y}^{2} + 4 x \sqrt{{x}^{2} + {y}^{2}} - 1 = 0$

Explanation:

1) GIVEN: r(2+costheta) = 1; r(theta) = 1/(2+cos(theta)
Transform $r \left(\theta\right) \implies f \left(x , y\right)$ Polar => Cartesian
2) RECIPE: r=x^2+y^2; y=rsintheta; x=rcostheta; theta = tan^-1(y/x)
3) Solution:
a) Multiply by r(theta) by r:
$2 r + r \cos \theta = 1$
b) Substitute from 2)
$2 \sqrt{{x}^{2} + {y}^{2}} + x = 1$ Now if you like you can square both sides or stop here...
a^2 + 2ax + x^2 = 1; " where " a = 2sqrt(x^2+y^2)
$4 \left({x}^{2} + {y}^{2}\right) + 4 x \sqrt{{x}^{2} + {y}^{2}} + {x}^{2} = 1$
$5 {x}^{2} + 4 {y}^{2} + 4 x \sqrt{{x}^{2} + {y}^{2}} - 1 = 0$