Chord of curvature through pole of cardioid

The chord of curvature through the pole is 2ρ sin φ, where φ is the angle between the radius vector and the tangent.

Cardioid: r = a(1 + cos θ) = 2a cos²(θ/2).

Angle φ: dr/dθ = −a sin θ, so

tan φ = r dθ/dr = a(1 + cos θ)/(−a sin θ) = −cot(θ/2) = tan(π/2 + θ/2)

so φ = π/2 + θ/2 and sin φ = cos(θ/2).

Radius of curvature: ρ = (r² + r₁²)^(3/2) / (r² + 2r₁² − r r₂), with r₁ = −a sin θ, r₂ = −a cos θ:

r² + r₁²          = a²(2 + 2cos θ) = 2a r
r² + 2r₁² − r r₂  = a²[(1+cosθ)² + 2sin²θ + cosθ(1+cosθ)]
                  = 3a²(1 + cos θ) = 3a r
ρ = (2ar)^(3/2) / (3ar) = (2/3)√(2ar)

Since √(2ar) = 2a cos(θ/2), ρ = (4a/3) cos(θ/2).

Chord of curvature through the pole:

2ρ sin φ = 2 · (4a/3) cos(θ/2) · cos(θ/2) = (8a/3) cos²(θ/2)

Since r = 2a cos²(θ/2), cos²(θ/2) = r/2a, so the chord = (8a/3)(r/2a).

Answer: chord of curvature through the pole = 4r/3 = (4a/3)(1 + cos θ).