Sketch conic r = 10/(2−3sinθ)

Divide by 2:

r = 10/(2 − 3 sinθ) = 5/(1 − (3/2) sinθ)

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

Key facts

• ed = 5, so d = 10/3. With (1 − e sinθ) the directrix is below the pole: y = −10/3.
• The transverse axis is along the y-axis.
• θ = π/2: r = 10/(2 − 3) = −10, giving the vertex (0, −10).
• θ = 3π/2: r = 10/5 = 2, giving the vertex (0, −2).
• Centre = (0, −6); a = 4; ae = 6, so the foci are (0, 0) and (0, −12).
• b² = a²(e² − 1) = 20, b = 2√5.
• Asymptotes: sinθ = 2/3, θ ≈ 41.8° and 138.2°.
• θ = 0 or π: r = 5, so the latus rectum ends are (±5, 0).

          y
          |
     ___  | F(0,0)  ___
        \_V(0,-2)_/    upper branch
          C(0,-6)
        _/V'(0,-10)\_  lower branch
     ___/ F2(0,-12) \___

Sketch: a hyperbola opening up and down (mirror image of r = 10/(2 + 3 sinθ) in the x-axis). Its vertices are (0, −2) and (0, −10), the centre is (0, −6), the foci are the origin and (0, −12), and the asymptotes make angles of about 41.8° and 138.2° with the x-axis through the centre.