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:

![Desmos.com](useruploads.socratic.org)