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Radius of curvature of a² = r²cos2θ

Short answer

By the pedal method: p = a²/r, dp/dr = –a²/r², so ρ = r dr/dp = r³/a² in magnitude.

Find the radius of curvature at any point (r,θ) for the curve a² = r²cos2θ.

Note: the curve a² = r² cos 2θ is r²(cos²θ − sin²θ) = a², i.e. x² − y² = a², a rectangular hyperbola (not a lemniscate). We find ρ by the pedal method ρ = r dr/dp.

Take logs: ln a² = 2 ln r + ln cos 2θ. Differentiate with respect to θ:

0 = (2/r)(dr/dθ) − 2 tan 2θ, so r dθ/dr = cot 2θ = tan(π/2 − 2θ)

So φ = π/2 − 2θ, and

p = r sin φ = r cos 2θ = r · a²/r² = a²/r (since cos 2θ = a²/r²).

Differentiate: dp/dr = −a²/r². Therefore

ρ = r dr/dp = r · (−r²/a²) = −r³/a².

Magnitude: ρ = r³/a² (the negative sign only shows the direction of the centre of curvature).

✓ Answer checked by Claude Opus 5.5 (high effort)

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