Question

Convert each polar equation into a rectangular equation.

$r=\frac{12}{4+8\sin \left(\theta \right)}$

Short answer

The rectangular equation is

$x^2-3y^2+12y=9$

Step-by-step solution

Step 1. Given information

The polar equation is

$r=\frac{12}{4+8\sin \left(\theta \right)}$

Step 2. Convert into the polar equation.

The polar equation is

$r=\frac{12}{4+8\sin \left(\theta \right)}$

Multiply denominator on both sides

$$\begin{gathered} r\times (4+8\sin (\theta \left)\right)=\frac{12}{4+8\sin \left(\theta \right)}\times (4+8\sin (\theta \left)\right) \\ r(4+8\sin (\theta \left)\right)=12 \\ 4r+8r\sin \left(\theta \right)=12 \end{gathered}$$

Divide both sides by $4$

role="math" localid="1647259342478" $r+2r\sin \left(\theta \right)=3$

Apply conversion equation formula

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

role="math" localid="1647259639691" $\sqrt{x^2+y^2}+2y=3$

Subtract both sides by y

role="math" localid="1647259661067" $$\begin{gathered} \sqrt{x^2+y^2}+2y-2y=3-2y \\ \sqrt{x^2+y^2}=3-2y \end{gathered}$$

Take square on both sides

$$\begin{gathered} {\left(\sqrt{x^2+y^2}\right)}^2={(3-2y)}^2{x}^2+y^2=9+4y^2-12y \\ {x}^2+y^2-4y^2+12y=9 \\ {x}^2-3y^2+12y=9 \end{gathered}$$

So, the rectangular equation is$x^2-3y^2+12y=9$