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

2 Answers
Narad T.
Nov 26, 2016

The equation is 3x^2+4y^2+2x-1=03x2+4y2+2x−1=0

Explanation:

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

x=rcosthetax=rcosθ

y=rsinthetay=rsinθ

and x^2+y^2=r^2x2+y2=r2

Therefore,

r(2+costheta)=1r(2+cosθ)=1

2r+rcostheta=12r+rcosθ=1

2sqrt(x^2+y^2)+x=12√x2+y2+x=1

2sqrt(x^2+y^2)=1-x2√x2+y2=1−x

Squaring both sides,

4(x^2+y^2)=(1-x)^2=1-2x+x^24(x2+y2)=(1−x)2=1−2x+x2

3x^2+4y^2+2x-1=03x2+4y2+2x−1=0

Salvatore I.
Nov 26, 2016

It is the ellipsys 3x^2+4y^2+2x-1=03x2+4y2+2x−1=0

Explanation:

rho=root2(x^2+y^2)ρ=2√x2+y2 and x=rho*sinthetax=ρ⋅sinθ and y=rho*costhetay=ρ⋅cosθ
root2(x^2+y^2)*(2+x/root2(x^2+y^2))=12√x2+y2⋅(2+x2√x2+y2)=1
that becomes
2root2(x^2+y^2)+x=122√x2+y2+x=1
or
x^2+y^2=((1-x)/2)^2x2+y2=(1−x2)2
or
x^2+y^2-(1-x)^2/4=0x2+y2−(1−x)24=0

Related questions
See all questions in Converting Equations from Polar to Rectangular
Impact of this question
3653 views around the world
You can reuse this answer
Creative Commons License