Sketch polar conic r = 12/(...) (2071 Shrawan)

Note: the denominator is cut off in the scan, so the exact curve cannot be fixed. The standard form of such papers is r = 12/(p + q cosθ) (or with sinθ); the method below gives the result for any constants p, q > 0.

Step 1: standard form

r = 12/(p + q cosθ) = (12/p) / (1 + (q/p) cosθ)

Compare with r = l/(1 + e cosθ):

• e = q/p, semi-latus rectum l = 12/p
• ed = l, so d = 12/q: directrix x = 12/q (to the right of the pole)
• the pole is a focus; the axis of the conic is the x-axis

Step 2: identify the conic

(If the sign is p − q cosθ, the curve is the mirror image in the y-axis, and the directrix is x = −12/q.)

Step 3: vertices and size

• θ = 0: r = 12/(p + q), vertex (12/(p + q), 0).
• θ = π: r = 12/(p − q), vertex (−12/(p − q), 0).
• θ = π/2: r = 12/p, the ends of the latus rectum (0, ±12/p).
• For an ellipse (q < p): a = 12p/(p² − q²), centre (−12q/(p² − q²), 0), b² = a²(1 − e²).
• For a hyperbola (q > p): a = 12p/(q² − p²), b² = a²(e² − 1).

Step 4: sketch

Mark the focus at the origin, the vertices, the latus rectum ends and the directrix; draw a closed curve for e < 1, a parabola opening to the left for e = 1 (focus at the origin, vertex (6/p, 0)), or two branches for e > 1.

Worked example (p = 3, q = 2): r = 12/(3 + 2cosθ), e = 2/3, an ellipse with vertices (12/5, 0) and (−12, 0), a = 36/5, centre (−24/5, 0), b² = 144/5, directrix x = 6.

            y
            |         x = 6
         .-'+'-.      |
  ---(---F2--C--F1)-----------x
  -12 -9.6 -4.8 0 2.4
         '-.+.-'