Chapter 9.5, Problem 76E

To determine

To calculate: To convert the polar equation to rectangular form

Answer to Problem 76E

The rectangular equation is ${\left(x^2+y^2\right)}^3=9{\left(x^2-y^2\right)}^2$

Explanation of Solution

Given information: Polar equation is $r=-3\cos 2\theta$

Formula Used:

Polar coordinate of a point $\left(r,\theta \right)$ in rectangular form is given as

  $\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \end{array}$

Where,

  $\begin{array}{l}{r}^2=x^2+y^2\\ \theta ={\tan }^{-1}\frac{y}{x}\end{array}$

Calculation:

Polarequation is given as

  $r=-3\cos 2\theta$

Simplifying the equation as

  $\begin{array}{l}r=-3\left({\cos }^2\theta -{\sin }^2\theta \right)\\ r^3=r^2\times -3\left({\cos }^2\theta -{\sin }^2\theta \right)\\ r^3=-3\left(r^2{\cos }^2\theta -r^2{\sin }^2\theta \right)\\ {\left(r^2\right)}^{\frac{3}{2}}=-3\left({\left(r\cos \theta \right)}^2-{\left(r\sin \theta \right)}^2\right)\end{array}$

Substituting $r^2=x^2+y^2$ and $r\cos \theta =x;r\sin \theta =y$ in above equation:

  $\begin{array}{l}{\left(x^2+y^2\right)}^{\frac{3}{2}}=-3\left(x^2-y^2\right)\\ {\left(x^2+y^2\right)}^3=9{\left(x^2-y^2\right)}^2\end{array}$

Thus, rectangularequation is ${\left(x^2+y^2\right)}^3=9{\left(x^2-y^2\right)}^2$

Conclusion:

Hence, rectangular equation is ${\left(x^2+y^2\right)}^3=9{\left(x^2-y^2\right)}^2$