what is the rectangular equation for r=12/(3-6cosX).
in terms of x,y, and r,
r^2 = x^2 + y^2, and cosX = x/r
so r = 12/(3 - 6(x/r)) cross-multiply
3r - 6x = 12
r = 4 + 2x, square both sides
r^2 = 16 + 16x + 4x^2 , replace r^2
x^2 + y^2 = 16 + 16x + 4x^2
3x^2 - y^2 + 16x = -16
according to the pattern of the equation it sure looks like a hyperbola.