Answer:
To convert the polar equation to a rectangular equation .
Given polar equation is,
[tex]\theta=\frac{11\pi}{6}[/tex]
we know the convertion of polar coordinates (r,theta) to rectangular equation as,
[tex]\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \end{gathered}[/tex]
we get,
[tex]\theta=\frac{11\pi}{6}=(2\pi-\frac{\pi}{6})[/tex]
Substitute this in the above equation we get,
[tex]\begin{gathered} x=r\cos (2\pi-\frac{\pi}{6}) \\ \\ y=r\sin (2\pi-\frac{\pi}{6}) \end{gathered}[/tex]
Solving we get,
[tex]\begin{gathered} x=r\cos (\frac{\pi}{6}) \\ \\ y=-r\sin (\frac{\pi}{6}) \end{gathered}[/tex]
we get,
[tex]x=r(\frac{\sqrt[]{3}}{2})[/tex][tex]y=-r(\frac{1}{2})[/tex]
Substitute r=-2y in x we get,
[tex]x=-2y(\frac{\sqrt[]{3}}{2})[/tex][tex]y=-\frac{x}{\sqrt[]{3}}[/tex][tex]y=-\frac{\sqrt[]{3}x}{3}[/tex]
The required rectangular form of the given plar equation is,
[tex]y=-\frac{\sqrt[]{3}x}{3}[/tex]
