Sketch polar conic r = 12/(...) (2068 Chaitra)

Note: the denominator is cut off in the scan, so the exact curve cannot be fixed. The method below works for r = 12/(p + q cosθ) (or sinθ) with unknown constants p, q > 0; the answer is decided by the ratio q/p.

Step 1: standard form

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

This has the form r = l/(1 + e cosθ), so

• eccentricity e = q/p
• semi-latus rectum l = 12/p
• directrix at distance d = l/e = 12/q from the pole: x = 12/q
• the pole is at a focus and the axis is the x-axis

Step 2: classify

• q < p: ellipse (0 < e < 1)
• q = p: parabola (e = 1)
• q > p: hyperbola (e > 1)

Step 3: key points

• Put θ = 0: r₁ = 12/(p + q), the vertex (12/(p + q), 0).
• Put θ = π: r₂ = 12/(p − q), the vertex (−12/(p − q), 0).
• Put θ = π/2: r = 12/p, so the latus rectum has length 24/p, with ends (0, ±12/p).
• For an ellipse, 2a = r₁ + r₂ = 24p/(p² − q²); the centre is the mid-point of the vertices and b² = a²(1 − e²).

Step 4: draw

Draw the x- and y-axes, put the focus at the pole, plot the two vertices and the latus rectum ends, and join them by a smooth closed curve (ellipse), one open curve (parabola) or two branches (hyperbola). If the denominator contains sinθ instead of cosθ, the same working applies with the axis along the y-axis and the directrix y = 12/q.

Example (p = 2, q = 2): r = 12/(2 + 2cosθ) = 6/(1 + cosθ), a parabola with focus at the origin, vertex (3, 0), directrix x = 6, opening to the left.