Question

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

Answer 1

`x - sqrt 3 y = 4`..

Explanation:

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]}