Question

Convert each polar equation into a rectangular equation.

$r=\frac{8}{2+4\cos \left(\theta \right)}$

Short answer

The rectangular equation is

$y^2-3x^2+16x=16$

Step-by-step solution

Step 1. Given information

The polar equation is

$r=\frac{8}{2+4\cos \left(\theta \right)}$

Step 2. Convert into the rectangular equation

The polar equation is

$r=\frac{8}{2+4\cos \left(\theta \right)}$

Multiply both sides by the denominator

$$\begin{gathered} r\times (2+4\cos (\theta \left)\right)=\frac{8}{2+4\cos \left(\theta \right)}\times 2+4\cos \left(\theta \right) \\ r(2+4\cos (\theta \left)\right)=8 \\ 2r(1+2\cos (\theta \left)\right)=8 \\ r(1+2\cos (\theta \left)\right)=4 \\ r+2r\cos \left(\theta \right)=4 \end{gathered}$$

Apply the conversion formula

$r=\sqrt{x^2+y^2},x=r\cos \left(\theta \right)$

$\sqrt{x^2+y^2}+2x=4$

Subtract both sides by $2x$

$$\begin{gathered} \sqrt{x^2+y^2}+2x-2x=4-2x \\ \sqrt{x^2+y^2}=4-2x \end{gathered}$$

Take square on both sides and simplify it

$$\begin{gathered} {\left(\sqrt{x^2+y^2}\right)}^2={(4-2x)}^2{x}^2+y^2=16+4x^2-16x \\ {x}^2+y^2-4x^2+16x=16 \\ {y}^2-3x^2+16x=16 \end{gathered}$$

So, the rectangular equation is $y^2-3x^2+16x=16$