Find the polar coordinates of the point(s)of intersection of the given curves for 0 $$0

Question

Find the polar coordinates of the point(s)of intersection of the given curves for 0 $$0 < \theta < 2 \pi .$$
-$$\mathrm { r } = 6 \cos \theta , \mathrm { r } = - 1 + 4 \cos \theta$$
A) $\left( 3 , \frac { 2 \pi } { 3 } \right) , \left( - 3 , \frac { 4 \pi } { 3 } \right)$
B) $\left( 3 , \frac { 5 \pi } { 6 } \right) , \left( 3 , \frac { \pi } { 6 } \right)$
C) $\left( - 3 , \frac { 5 \pi } { 3 } \right) , \left( - 3 , \frac { \pi } { 3 } \right)$
D) $\left( 3 , \frac { 2 \pi } { 3 } \right) , \left( 3 , \frac { 4 \pi } { 3 } \right)$

Quizplus · Accepted Answer

Explanation:Solving the given equations, we get$$6\cos \theta = -1+4\cos \theta$$$$2\cos \theta = -1$$$$\cos \theta = -\frac{1}{2}$$$$\theta = \frac{2\pi}{3}, \frac{4\pi}{3}$$Substituting these values of $\theta$ into the first equation, we get$$r = 6\cos \frac{2\pi}{3} = -3$$$$r = 6\cos \frac{4\pi}{3} = 3$$Therefore, the points of intersection are $(3, \frac{2\pi}{3})$ and $(-3, \frac{4\pi}{3})$.

Find the Polar Coordinates of the Point(s)of Intersection of the Given