Sketch conic r = 10cosecθ/(2cosecθ+3)

Multiply the numerator and denominator by sinθ (cosecθ · sinθ = 1):

r = 10 cosecθ/(2 cosecθ + 3) = 10/(2 + 3 sinθ) = 5/(1 + (3/2) sinθ)

Compare with r = l/(1 + e sinθ): e = 3/2 > 1, so the conic is a hyperbola; l = 5.

Key facts

• ed = 5, so d = 10/3: directrix y = 10/3.
• The transverse axis is along the y-axis.
• θ = π/2: r = 10/5 = 2, vertex (0, 2).
• θ = 3π/2: r = 10/(2 − 3) = −10, giving the vertex (0, 10) (a negative r reverses the direction).
• Centre = (0, 6); a = 4; ae = 6, so the foci are (0, 0) and (0, 12).
• b² = a²(e² − 1) = 16 · 5/4 = 20, so b = 2√5.
• Asymptotes: 1 + (3/2)sinθ = 0, sinθ = −2/3, θ ≈ 221.8° and 318.2° (r → ∞). Through the centre their slopes are ±a/b = ±2/√5.
• θ = 0 or π: r = 5, so the latus rectum ends are (±5, 0).

                       y
         \             |             /
          \            * F2 (0,12)  /
           '.          |          .'
             '-.______ V' ______.-'      upper branch, vertex V'(0,10)
                       |
                       + C (0,6)         centre
                       |
   - - - - - - - - - - | - - - - - - - - directrix y = 10/3
              .-'''''' V ''''''-.        lower branch, vertex V(0,2)
           .'          |          '.
   -------/------------F------------\------- x   F = pole (0,0)
       (-5,0)          |          (5,0)
         /             |             \

Sketch: a hyperbola opening up and down, with vertices V(0, 2) and V'(0, 10) and centre C(0, 6). The lower branch (vertex V) curves around the focus at the pole and crosses the x-axis at (±5, 0); the upper branch (vertex V') curves around the focus F2(0, 12). The branches are not mirror images about the x-axis; they are symmetric about the centre (0, 6). The directrix y = 10/3 lies between the vertex V and the centre C.