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Pedal equation of x^(2/3)+y^(2/3)=a^(2/3)

Short answer

Parametrize x = a cos³t, y = a sin³t; get p = a sin t cos t, r² = a² – 3p², so the pedal equation is r² = a² – 3p².

Find the pedal equation of the curve x^(2/3) + y^(2/3) = a^(2/3).

Astroid: x^(2/3) + y^(2/3) = a^(2/3). Use the parametric form x = a cos³t, y = a sin³t.

dx/dt = −3a cos²t sin t, dy/dt = 3a sin²t cos t, so dy/dx = −tan t.

Tangent: y − a sin³t = −tan t (x − a cos³t). Multiply by cos t:

x sin t + y cos t = a sin t cos t (sin²t + cos²t) = a sin t cos t

Perpendicular from the origin:

p = a sin t cos t / √(sin²t + cos²t) = a sin t cos t …(1)

Radius vector:

r² = x² + y² = a²(cos⁶t + sin⁶t) = a²(1 − 3 sin²t cos²t)

since cos⁶t + sin⁶t = (cos²t + sin²t)³ − 3 sin²t cos²t(cos²t + sin²t). Using (1):

r² = a² − 3p²

Pedal equation: r² = a² − 3p² (or 3p² = a² − r²).

✓ Answer checked by Claude Opus 5.5 (high effort)

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