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

Question

Grace Ashcraft · Accepted Answer

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

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.