Question

Convert each polar equation into a rectangular equation.

$r=\frac{6sec\left(\theta \right)}{2sec\left(\theta \right)-1}$

Short answer

The rectangular equation is

$3x^2+4y^2-12x=36$

Step-by-step solution

Step 1. Given information

The polar equation is

$r=\frac{6sec\left(\theta \right)}{2sec\left(\theta \right)-1}$

Step 2. Convert into the rectangular equation

$x$The polar equation is

$r=\frac{6sec\left(\theta \right)}{2sec\left(\theta \right)-1}$

Multiply top and bottom by $\cos \left(\theta \right)$

$$\begin{gathered} r=\frac{6sec\left(\theta \right)\times \cos \left(\theta \right)}{\left(2sec\right(\theta )-1)\times \cos \left(\theta \right)} \\ r=\frac{6\times {\displaystyle \frac{1}{\cos \left(\theta \right)}}\times \cos \left(\theta \right)}{(2\frac{1}{\cos \left(\theta \right)}\times \cos (\theta )-\cos (\theta \left)\right)} \\ r=\frac{6}{2-\cos \left(\theta \right)} \end{gathered}$$

Multiply denominator on both sides

$$\begin{gathered} r\times (2-\cos (\theta \left)\right)=\frac{6}{2-\cos \left(\theta \right)}\times 2-\cos \left(\theta \right) \\ r(2-\cos (\theta \left)\right)=6 \\ 2r-r\cos \left(\theta \right)=6 \end{gathered}$$

Apply the conversion formula

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

$2\sqrt{x^2+y^2}-x=6$

Add both sides by $x$

$$\begin{gathered} 2\sqrt{x^2+y^2}-x+x=6+x \\ 2\sqrt{x^2+y^2}=6+x \end{gathered}$$

Take square on both sides and then simplify

$$\begin{gathered} {\left(2\sqrt{x^2+y^2}\right)}^2={(6+x)}^{2}4(x^2+y^2)=36+x^2+12x \\ 4x^2+4y^2=36+x^2+12x \\ 4x^2+4y^2-x^2-12x=36 \\ 3x^2+4y^2-12x=36 \end{gathered}$$

So, the rectangular equation is

$3x^2+4y^2-12x=36$