Chapter 9.4, Problem 12E

To determine

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

Answer to Problem 12E

The polar form of $7x-24y+100=0$ is $4=r\cos \left[\theta -106.26°\right]$ and nearest of angle $\phi \approx 106.26°$ in degree

Explanation of Solution

Given information

The given equation is $7x-24y+100=0$

Formula used: $p=r\cos \left[\theta -\phi \right]$ .

Calculation:

The standard form of the equation is $7x-24y+100=0$

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 $\pm \sqrt{A^2+B^2}$ .

  $\begin{array}{l}\pm \sqrt{A^2+B^2}= \pm \sqrt{{\left(7\right)}^2+{\left(-24\right)}^2}\left[A=7,B=-24\right] \\ =\pm \sqrt{49+576}=\pm \sqrt{625}=\pm 25\end{array}$

Since $C$ is positive, use $-25$ .The normal form of the equation is $\frac{7}{\left(-25\right)}x+\frac{24}{\left(25\right)}y-4=0$ .

  $x\cos \phi +y\sin \phi -p=0$ We can see from the normal form that $p=4,\cos \phi =\frac{7}{\left(-25\right)}\text{ and}\sin \phi =-\frac{24}{\left(-25\right)}$ .Since $\cos \phi$ is negative and $\sin \phi$ is positive , the normal lies in the second quadrant.

  $\begin{array}{l}\tan \phi =\frac{\sin \phi }{\cos \phi }=\frac{-\frac{24}{\left(-25\right)}}{\frac{7}{\left(-25\right)}}=-\frac{24}{7}\\ \phi ={\tan }^{\left(-1\right)}-\frac{24}{7}\text{ [add }180°\text{ to the arctangent value]}\\ \phi \approx -73.74°+180°\\ \phi \approx 106.26°\end{array}$ [using the arctangent function].

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

  $p=r\cos \left[\theta -\phi \right]\to 4=r\cos \left[\theta -106.26°\right]$

The polar form of $7x-24y+100=0$ is $4=r\cos \left[\theta -106.26°\right]$