Chapter 10, Problem 61RE

To determine

Tofind:the rectangular equation from the polar equation.

Answer to Problem 61RE

The rectangular equation of the given polar equation is $\text{-}\left(\text{3}{\left(\text{x-}\frac{\text{4}}{3}\right)}^2{\text{-y}}^2\text{-}\frac{\text{4}}{3}\right)=0$ .

Explanation of Solution

Given:

  $\text{r=}\frac{8}{\text{4+8cosθ}}$ .

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:

  $\text{r=}\frac{8}{\text{4+8cosθ}}$ .

Multiplying by $\left(\text{4+8cosθ}\right)$ both sides:

  $\text{r}\left(\text{4+8cosθ}\right)\text{=}\frac{8}{\left(\text{4+8cosθ}\right)}\left(\text{4+8cosθ}\right)$ .

  $\text{r}\left(\text{4+8cosθ}\right)\text{=8}\text{.}$

  $\text{4r+8rcosθ=8}\text{.}$

According to the complex formula:

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

  $\text{4r+8x=8}\text{.}$

  $\text{r=2-2x}\text{.}$

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

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

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

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

  $\text{-}\left(\text{3}{\left(\text{x-}\frac{\text{4}}{3}\right)}^2{\text{-y}}^2\text{-}\frac{\text{4}}{3}\right)=0$

Hence, the rectangular equation of the given polar equation is $\text{-}\left(\text{3}{\left(\text{x-}\frac{\text{4}}{3}\right)}^2{\text{-y}}^2\text{-}\frac{\text{4}}{3}\right)=0$ .