How do you convert #r(2 + cos theta) = 1# into cartesian form?

2 Answers
Nov 26, 2016

The equation is #3x^2+4y^2+2x-1=0#

Explanation:

To transform from polar coordinates #(r,theta)# into rectangular coordinates #(x,y)#, we use the folowing

#x=rcostheta#

#y=rsintheta#

and #x^2+y^2=r^2#

Therefore,

#r(2+costheta)=1#

#2r+rcostheta=1#

#2sqrt(x^2+y^2)+x=1#

#2sqrt(x^2+y^2)=1-x#

Squaring both sides,

#4(x^2+y^2)=(1-x)^2=1-2x+x^2#

#3x^2+4y^2+2x-1=0#

Nov 26, 2016

It is the ellipsys #3x^2+4y^2+2x-1=0#

Explanation:

#rho=root2(x^2+y^2)# and #x=rho*sintheta# and #y=rho*costheta#
#root2(x^2+y^2)*(2+x/root2(x^2+y^2))=1#
that becomes
#2root2(x^2+y^2)+x=1#
or
#x^2+y^2=((1-x)/2)^2#
or
#x^2+y^2-(1-x)^2/4=0#