Sketch conic r = 12/(2−6cosθ)

Divide by 2 to get the form r = l/(1 − e cosθ):

r = 12/(2 − 6cosθ) = 6/(1 − 3cosθ)

So e = 3 > 1: the conic is a hyperbola, with the pole at a focus and l = 6.

Key facts

• ed = 6, so d = 2. The sign (1 − e cosθ) puts the directrix to the left of the pole: x = −2.
• θ = 0: r = 6/(1 − 3) = −3, giving the vertex (−3, 0) (a negative r means the point lies in the opposite direction).
• θ = π: r = 6/4 = 3/2, giving the vertex (−3/2, 0).
• θ = π/2: r = 6, so the latus rectum ends are (0, ±6).
• Centre = (−9/4, 0); a = 3/4; ae = 9/4, so the foci are (0, 0) and (−9/2, 0).
• b² = a²(e² − 1) = (9/16)(8) = 9/2.
• Asymptote directions: 1 − 3cosθ = 0, cosθ = 1/3, θ = ±70.5° (r → ∞).

                                                y
            \                       /           |
             \                     /            |
              \                   /             |
               )        :        (              |
 ----------F2--V-----C--:--------V'-------------F1------ x
               )        :        (              |
              /         :         \             |
             /          :          \            |
            /           :           \           |
                     x = -2
 F2(-9/2, 0)   V(-3, 0)   C(-9/4, 0)   V'(-3/2, 0)   F1 = pole (0, 0)

Sketch: a hyperbola opening left and right, with vertices V(−3, 0) and V'(−3/2, 0) and centre C(−9/4, 0). The right branch (vertex V') curves around the focus F1 at the pole and passes through (0, ±6); the left branch (vertex V) curves around F2(−9/2, 0). The asymptotes pass through C at about ±70.5° to the x-axis, and the directrix x = −2 lies between C and V'.