How do you convert #r = 3 / (3 - cos(theta))# to rectangular form?

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Answer 1 · 2016-09-24T13:13:51

#8x^2+y^2-2x-1=0#

#r = sqrt(x^2+y^2)#
#costheta = x/sqrt(x^2+y^2)#

substituting

#sqrt(x^2+y^2) = 3/(3- x/sqrt(x^2+y^2)) = sqrt(x^2+y^2)/(3sqrt(x^2+y^2)-x)#

simplifying

#3sqrt(x^2+y^2)=x+1#

squaring

#9(x^2+y^2)=x^2+2x+1# and finally

#8x^2+y^2-2x-1=0# which corresponds to an ellipse.

Answer 2 · 2016-09-24T14:44:39

#(x+3/8)^2/(9/8)^2+y^2/(3/(2sqrt 2))^2=1# that represents an ellipse.

The given equation can be written in the form

#1/r =1-(1/3)cos theta# that represents an ellipse of eccentricity 1/3.

Use the conversion equation #r(cos theta, sin theta)=(x, y)# that

gives # r = sqrt(x^2+y^2) and cos theta =x/r#..

Here, #r = 3/(3 -x/r)=(3r)/(3r-x). |r|3/|3-cos theta|>=3/4#. So, canceling r and

rearranging,

#3r=3sqrt(x^2+y^2)= x+3#

Rationalizing,

#9(x^2+y^2)=x^2-6x+9.#. Simplifying,
# 8x^2+9y^2+6x-9=0#.

This can further reduced to the standard form

#(x+3/8)^2/(9/8)^2+y^2/(3/(2sqrt 2))^2=1# that represents an ellipse..