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