How do you graph #r(2 - costheta) = 2#?

1 Answer
May 3, 2016

In Catrersian form it is #3x^2+4y^2-4x-4=0# an ellipse.

Explanation:

If #(r,theta)# is in polar form and #(x,y)# in Cartesian form the relation between them is as follows:

#x=rcostheta#, #y=rsintheta#, #r^2=x^2+y^2# and #tantheta=y/x#

Or, #costheta=x/r#, #sintheta=y/r#, #theta=tan^(-1)(y/x)# and #cottheta=x/y#.

Hence, #r(2-costheta)=2# can be written as

#2r-rcostheta=2#

#2(x^2+y^2)^(1/2)-x=2# or

#2(x^2+y^2)^(1/2)=2+x# or

#4(x^2+y^2)=(2+x)^2# or

#4x^2+4y^2=4+x^2+4x# or

#3x^2+4y^2-4x-4=0#

As coefficients of #x^2# and #y^2# are both positive but not equal, this is an ellipse.

The above can be written as

#3(x^2-4/3x+4/9)+4y^2-4-12/9-0#

or #3(x-2/3)^2+4(y-0)^2=48/9=16/3#

or #9/16(x-2/3)^2+3/4(y-0)^2=1#

or #(x-2/3)^2/(16/9)+(y-0)^2/(4/3)=1#

Center of ellipse is #(2/3,0)#

Major axis is #2xx4/3=8/3# and minor axis is #2xx2/sqrt3=4/sqrt3#

graph{3x^2+4y^2-4x-4=0 [-3, 3, -1.5, 1.5]}