How do you graph #r=(12)/(3+4 cos theta)#?

1 Answer
Nov 24, 2016

Please see the explanation.

Explanation:

Multiply both side by #3 + 4cos(theta)#:

#3r + 4rcos(theta) = 12#

Substitute x for #rcos(theta)#:

#3r + 4x = 12#

Subtract 4x from both sides:

#3r = 12 - 4x#

Square both sides:

#9r^2 = 144 - 96x + 16x^2#

Substitute #9x^2 + 9y^2# for #9r^2#:

#9x^2 + 9y^2 = 144 - 96x + 16x^2#

Move everything but the constant to the left and subtract #7h^2# from both sides:

#-7x^2 + 96x - 7h^2 + 9(y - 0)^2 = 144 - 7h^2#

Complete the square for the x term:

#-7(x^2 - 96/7x + h^2) + 9(y - 0)^2 = 144 - 7h^2#

#-2hx = -96/7x#
#h = 48/7#

#-7(x - 48/7)^2 + 9(y - 0)^2 = -1296/7#

#(x - 48/7)^2/(36/7)^2 - (y - 0)^2/(12sqrt(7)/7)^2 = 1#

Center#(48/7, 0)#

#x_1 = 48/7 - 36/7 = 12/7#

#x_2 = 48/7 + 36/7 = 12#

Vertices: #(12/7, 0) and (12,0)#

#b/a = (12sqrt(7)/7)/(36/7) = sqrt(7)/3#

asymptotes:

#y = -(sqrt(7)/3)(x - 48/7)#

#y = (sqrt(7)/3)(x - 48/7)#

Here is the graph:

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