Chapter 12, Problem 7CLT

To determine

The polar coordinates of the rectangular coordinates ?

Answer to Problem 7CLT

The polar equation of the rectangular equation is $r=\frac{-5}{8\cos \theta +3\sin \theta }$ .

Explanation of Solution

Given:

The given rectangular coordinates are,

  $\begin{array}{l}-8x-3y-5=0\\ -8x-3y=5\end{array}$

Concept used:

The standard conversions of the coordinates are,

  $\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \\ r^2=x^2+y^2\\ \tan \theta =\frac{y}{x}\end{array}$

Calculation:

According to the question,

  $\begin{array}{l}-8x-3y-5=0\\ -8x-3y=5\end{array}$

Substitute the value

  $\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \\ -8x-3y-5=0\\ -8\left(r\cos \theta \right)-3\left(r\sin \theta \right)-5=0\\ -8r\cos \theta -3r\sin \theta =5\\ -r\left(8\cos \theta +3\sin \theta \right)=5\\ r=\frac{-5}{8\cos \theta +3\sin \theta }\end{array}$

Conclusion:

Hence, the polar equation of the rectangular equation is $r=\frac{-5}{8\cos \theta +3\sin \theta }$