How do you change the polar equation #r(2+costheta)=1# into rectangular form?

Question

Caroline Adams · Accepted Answer

#(xsqrt(3)+1/sqrt(3))^2+4y^2-4/3=0#

supplying the pass formulas

#{(x=rcostheta),(y=rsintheta):}#

#r(2+costheta)=1 = r(2+x/r)=2r+x=1# then

#r = (1-x)/2->x^2+y^2=(1-2x+x^2)/4# and finally

#3x^2+4y^2+2x-1=0# or

#(xsqrt(3)+1/sqrt(3))^2+4y^2-4/3=0# which is an ellipse.

Kai Anderson · Answer

undefined To convert the polar equation $ r(2+\cos(\theta))=1 $ into rectangular form, you can use the following steps:

1. Substitute $ r = \sqrt{x^2 + y^2} $ and $ \cos(\theta) = \frac{x}{\sqrt{x^2 + y^2}} $ into the polar equation.
2. Simplify the equation.
3. Rearrange the equation to isolate $ y $, if necessary.

So, substituting into the equation:

$$ \sqrt{x^2 + y^2} (2 + \frac{x}{\sqrt{x^2 + y^2}}) = 1 $$

Simplify the equation:

$$ 2\sqrt{x^2 + y^2} + x = 1 $$

Rearrange to isolate $ y $:

$$ 2\sqrt{x^2 + y^2} = 1 - x $$

$$ \sqrt{x^2 + y^2} = \frac{1 - x}{2} $$

$$ x^2 + y^2 = \left(\frac{1 - x}{2}\right)^2 $$

$$ x^2 + y^2 = \frac{1 - 2x + x^2}{4} $$

$$ 4x^2 + 4y^2 = 1 - 2x + x^2 $$

$$ 3x^2 + 4y^2 + 2x - 1 = 0 $$

Therefore, the rectangular form of the polar equation $ r(2+\cos(\theta))=1 $ is $ 3x^2 + 4y^2 + 2x - 1 = 0 $.