Chapter 10, Problem 62RE

To determine

Tofind:the rectangular equation from the polar equation.

Answer to Problem 62RE

The rectangular equation of the given polar equation is $\text{5}{\left(\text{x}+\frac{4}{5}\right)}^{\text{2}}{\text{+9y}}^{\text{2}}\text{-}\frac{36}{5}\text{=0}$ .

Explanation of Solution

Given:

  $\text{r=}\frac{2}{\text{3+2cosθ}}$ .

Concept used:

According to the complex formula:

  $\text{x=rcosθ}$ , $\text{y=rsinθ}$ and ${\text{r}}^{\text{2}}{\text{=x}}^{\text{2}}{\text{+y}}^{\text{2}}$ .

Calculation:

  $\text{r=}\frac{2}{\text{3+2cosθ}}$ .

Multiplying by $\text{3+2cosθ}$ both sides:

  $\text{r}\left(\text{3+2cosθ}\right)\text{=}\frac{2}{\left(\text{3+2cosθ}\right)}\left(\text{3+2cosθ}\right)$ .

  $\text{r}\left(\text{3+2cosθ}\right)\text{=2}$ .

  $\text{3r+2rcosθ=2}$ .

According to the complex formula:

  $\text{x=rcosθ}$ , $\text{y=rsinθ}$ and ${\text{r}}^{\text{2}}{\text{=x}}^{\text{2}}{\text{+y}}^{\text{2}}$ .

  $\text{3r+2x=2}$ .

  $\text{3r=2-2x}$ .

  ${\left(\text{3r}\right)}^2\text{=}{\left(\text{2-2x}\right)}^2$

  ${\text{9r}}^{\text{2}}{\text{=4-8x+4x}}^{\text{2}}$ .

  $\text{9}\left({\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}\right){\text{=4-8x+4x}}^{\text{2}}$ .

  ${\text{9x}}^{\text{2}}+{\text{9y}}^{\text{2}}{\text{-4x}}^{\text{2}}\text{+8x-4=0}$ .

  ${\text{5x}}^{\text{2}}\text{+8x-4}+{\text{9y}}^{\text{2}}\text{=0}$ .

  $\text{5}{\left(\text{x}+\frac{4}{5}\right)}^{\text{2}}{\text{+9y}}^{\text{2}}\text{-}\frac{36}{5}\text{=0}$

Hence, the rectangular equation of the given polar equation is $\text{5}{\left(\text{x}+\frac{4}{5}\right)}^{\text{2}}{\text{+9y}}^{\text{2}}\text{-}\frac{36}{5}\text{=0}$ .