How do you convert #r(2 - cosx) = 2# to rectangular form?

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Jan 29, 2017

Answer:

The explanation is written with the assumption that you meant #theta# for the argument of the cosine function.

Explanation:

Use the distributive property :

#2r - rcos(theta) = 2#

#2r = rcos(theta) + 2#

Substitute x for #rcos(theta)# and #sqrt(x^2 + y^2)# for r:

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

Square both sides:

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

#4x^2 + 4y^2 = x^2 + 4x + 4#

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

Add #3h^2# to both sides:

#3x^2 - 4x + 3h^2 + 4y^2 = 3h^2+ 4#

Remove a factor of the 3 from the first 3 terms:

#3(x^2 - 4/3x + h^2) + 4y^2 = 3h^2+ 4#

Use the middle in the right side of the pattern #(x - h)^2 = x^2 - 2hx + h^2# and middle term of the equation to find the value of h:

#-2hx = -4/3x#

#h = 2/3#

Substitute the left side of the pattern into the equation:

#3(x - h)^2 + 4y^2 = 3h^2+ 4#

Substitute #2/3# for h and insert a -0 into the y term:

#3(x- 2/3)^2 + 4(y-0)^2 = 3(2/3)^2+ 4#

#3(x- 2/3)^2 + 4(y-0)^2 = 16/3#

Divide both sides by #16/3#

#3(x- 2/3)^2/(16/3) + 4(y-0)^2/(16/3) = 1#

#(x- 2/3)^2/(16/9) + (y-0)^2/(16/12) = 1#

Write the denominators as squares:

#(x- 2/3)^2/(4/3)^2 + (y-0)^2/(4/sqrt(12))^2 = 1#

The is standard Cartesian form of the equation of an ellipse with a center at #(2/3,0)#; its semi-major axis is #4/3# units long and is parallel to the x axis and its semi-minor axis is #4/sqrt(12)# units long and is parallel to the y axis.

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