Chapter 9.4, Problem 10MCQ

To determine

To write each equation in polar form and nearest angle in degree

Answer to Problem 10MCQ

The polar form of $-2x-6y=2$ is $\sqrt{10}=r\cos \left[\theta -251.56°\right]$ and nearest of angle $\phi \approx 251.56°$ in degree

Explanation of Solution

Given information

The form of equation is $-2x-6y=2$

Calculation:

The standard form of the equation is $-2x-6y-2=0$

Firstly. write the equation in normal form to find the values of $p$ and $\phi$ .To convert to normal form, we need to find the value of $\pm \sqrt{A^2+B^2}$ .

  $\begin{array}{l}\pm \sqrt{A^2+B^2}= \pm \sqrt{{\left(-2\right)}^2+{\left(-6\right)}^2}\left[A=-2,B=-6\right] \\ =\pm \sqrt{4+36}=\pm \sqrt{40}\end{array}$

Since $C$ is negative, use $+\sqrt{40}$ .The normal form of the equation is $-\frac{2}{\left(\sqrt{40}\right)}x-\frac{6}{\left(\sqrt{40}\right)}y-\frac{2}{\sqrt{40}}=0$ . $x\cos \phi +y\sin \phi -p=0$ We can see from the normal form that $p=\frac{2}{\sqrt{40}}=\sqrt{10},\cos \phi =-\frac{2}{\sqrt{40}}\text{ and}\sin \phi =-\frac{6}{\sqrt{40}}$ .Since $\cos \phi \text{and}\sin \phi$ are both negative,the normal lies in the third quadrant.

  $\begin{array}{l}\tan \phi =\frac{\sin \phi }{\cos \phi }=\frac{-\frac{6}{\sqrt{40}}}{-\frac{2}{\sqrt{40}}}=3\\ \phi ={\tan }^{\left(-1\right)}3\left[add{180}^{°}\text{arctangent}\right]\\ \phi \approx 71.56°+180°\\ \phi \approx 251.56°\end{array}$ [using the arctangent function].

Substitute the values for $p$ and $\phi$ into the polar form.

  $p=r\cos \left[\theta -\phi \right]\to \sqrt{10}=r\cos \left[\theta -251.56°\right]$

The polar form of $-2x-6y=2$ is $\sqrt{10}=r\cos \left[\theta -251.56°\right]$