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Ellipse curvature at end of major axis

Short answer

At (a,0) (t=0), parametric form gives ρ = b²/a. The latus rectum is 2b²/a, so half is b²/a, proving ρ equals half the latus rectum.

Show that for the ellipse x²/a² + y²/b² = 1, the radius of curvature at the extremity of the major axis is equal to half of the latus rectum.

Ellipse: x²/a² + y²/b² = 1 with a > b, so the major axis lies along the x-axis, with extremity (a, 0). The latus rectum is 2b²/a, so half of it is b²/a.

Radius of curvature at (a, 0). Use the parametric form x = a cos t, y = b sin t. The point (a, 0) corresponds to t = 0.

x′ = −a sin t,  y′ = b cos t
x″ = −a cos t,  y″ = −b sin t
ρ = (x′² + y′²)^(3/2) / (x′y″ − y′x″)
  = (a² sin²t + b² cos²t)^(3/2) / (ab sin²t + ab cos²t)
  = (a² sin²t + b² cos²t)^(3/2) / (ab)

At t = 0: ρ = (b²)^(3/2)/(ab) = b³/(ab) = b²/a.

Since the latus rectum is l = 2b²/a, we get

ρ = b²/a = l/2 = half of the latus rectum. Shown.

✓ Answer checked by Claude Opus 5.5 (high effort)

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