Chapter 9.4, Problem 6CFU

To determine

To write: The given equation in polar form and round $\phi$ to the nearest degree

Answer to Problem 6CFU

The polar form of the equation is $\frac{9}{2\sqrt{5}}=r\cos \left(\theta -{117}^{\circ }\right)$ and the nearest degree to the value of $\phi \text{ is }{117}^{\circ }$

Explanation of Solution

Given information:

The equation is $-2x+4y=9$

Calculation:

The standard form of the equation is $-2x+4y-9=0$ . First, write the equation in normal form to find the values of $p\text{ and }\phi$ . To convert to normal form, it is necessary to find the value of $\pm \sqrt{A^2+B^2}$

  $\begin{array}{l}\pm \sqrt{A^2+B^2}\\ \pm \sqrt{{\left(-2\right)}^2+{\left(4\right)}^2}\\ \pm \sqrt{4+16}\\ \pm 2\sqrt{5}\end{array}$

Since $C$ is negative, use $+2\sqrt{5}$

The normal form of the equation is,

  $-\frac{2}{2\sqrt{5}}x+\frac{4}{2\sqrt{5}}y-\frac{9}{2\sqrt{5}}=0$

The normal form is $x\cos \phi -y\sin \phi -p=0$

So, it can be seen from the normal form that $\cos \phi =-\frac{2}{2\sqrt{5}},\sin \phi =\frac{4}{2\sqrt{5}}\text{ and }p=\frac{9}{2\sqrt{5}}$

Since $\cos \phi <0\text{ and }\sin \phi >0$ , so the normal lies in the second quadrant

  $\begin{array}{l}\tan \phi =\frac{\sin \phi }{\cos \phi }\\ \tan \phi =\frac{\frac{4}{2\sqrt{5}}}{-\frac{2}{2\sqrt{5}}}\\ \tan \phi =-\frac{4}{2}\\ \phi \approx {117}^{\circ }\left[\text{add 180 to the arctangent value}\right]\end{array}$

Substitute the values for $p\text{ and }\phi$ into the polar form

  $\begin{array}{l}p=r\cos \left(\theta -\phi \right)\\ \Rightarrow \frac{9}{2\sqrt{5}}=r\cos \left(\theta -{117}^{\circ }\right)\end{array}$

Therefore, the polar form of the equation is $\frac{9}{2\sqrt{5}}=r\cos \left(\theta -{117}^{\circ }\right)$ and the nearest degree to the value of $\phi \text{ is }{117}^{\circ }$