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

1 Answer
May 3, 2016

In Catrersian form it is 3x^2+4y^2-4x-4=03x2+4y24x4=0 an ellipse.

Explanation:

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

x=rcosthetax=rcosθ, y=rsinthetay=rsinθ, r^2=x^2+y^2r2=x2+y2 and tantheta=y/xtanθ=yx

Or, costheta=x/rcosθ=xr, sintheta=y/rsinθ=yr, theta=tan^(-1)(y/x)θ=tan1(yx) and cottheta=x/ycotθ=xy.

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

2r-rcostheta=22rrcosθ=2

2(x^2+y^2)^(1/2)-x=22(x2+y2)12x=2 or

2(x^2+y^2)^(1/2)=2+x2(x2+y2)12=2+x or

4(x^2+y^2)=(2+x)^24(x2+y2)=(2+x)2 or

4x^2+4y^2=4+x^2+4x4x2+4y2=4+x2+4x or

3x^2+4y^2-4x-4=03x2+4y24x4=0

As coefficients of x^2x2 and y^2y2 are both positive but not equal, this is an ellipse.

The above can be written as

3(x^2-4/3x+4/9)+4y^2-4-12/9-03(x243x+49)+4y241290

or 3(x-2/3)^2+4(y-0)^2=48/9=16/33(x23)2+4(y0)2=489=163

or 9/16(x-2/3)^2+3/4(y-0)^2=1916(x23)2+34(y0)2=1

or (x-2/3)^2/(16/9)+(y-0)^2/(4/3)=1(x23)2169+(y0)243=1

Center of ellipse is (2/3,0)(23,0)

Major axis is 2xx4/3=8/32×43=83 and minor axis is 2xx2/sqrt3=4/sqrt32×23=43

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