Sketch polar conic r = 10cosecθ/(...)

Note: the denominator is cut off in the scan, so the exact curve cannot be fixed. The paper's form is r = 10 cosecθ/(m cosecθ + n); the method below works for any constants m, n > 0, and the result is read off from n/m.

Step 1: convert to standard polar form

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

r = 10/(m + n sinθ) = (10/m) / (1 + (n/m) sinθ)

Compare with r = l/(1 + e sinθ):

• eccentricity e = n/m
• semi-latus rectum l = 10/m
• directrix: ed = l, so d = 10/n, and the directrix is y = 10/n (above the pole)
• the pole is a focus and the axis of the conic is the y-axis

Step 2: identify the conic

Step 3: vertices

• θ = π/2: r = 10/(m + n), vertex (0, 10/(m + n)).
• θ = 3π/2: r = 10/(m − n), vertex (0, −10/(m − n)).
• θ = 0 and π: r = 10/m, the ends of the latus rectum (±10/m, 0).

Step 4: sketch

• For n < m: an ellipse with major axis along the y-axis, a = 10m/(m² − n²), centre (0, −10n/(m² − n²)), b² = a²(1 − e²), one focus at the origin.
• For n = m: a parabola opening downward, with focus at the origin and directrix y = 10/n.
• For n > m: a hyperbola with transverse axis on the y-axis.

Worked example (m = 3, n = 2): r = 10/(3 + 2 sinθ), e = 2/3, ellipse; vertices (0, 2) and (0, −10), so a = 6, centre (0, −4), b² = 20; directrix y = 5.

          y
          |   y = 5
  - - - - | - - - - - -
          | V(0,2)
       .--+--.
      /   F   \
  ---(----+----)---- x
      \   C   /
       '--.--'
          V'(0,-10)