How do you graph the polar equation #2=rcos(theta+60^circ)#?

Question

2007 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-27T13:52:15

#x - sqrt 3 y = 4#..

Use #r = sqrt ( x^2 + y^2 ) >= 0#

# and r( cos theta, sin theta ) = ( x, y )#

#2 = r (cos theta cos (pi/3) - sin theta sin (pi/3 ))#

#= r/sqrt2 ( cos theta - sqrt3 sin theta#) converts to

#x - sqrt3 y = 4#.

Note that the perpendicular form of the polar equation of a straight

line is

#r cos (theta - alpha ) = p#, where

#( p, alpha )# is th foot of the perpendicular to the line, from the

pole #r = 0#. See graph of the given equation, with the foot of the

perpendicular, with #p = 2 and alpha = - pi/3#.

graph{(x-sqrt 3 y-4)((x-1)^2+(y+sqrt3)^2-0.001)(y+sqrt3 x)=0[0 4 -2 0]}