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By Guest M., Miyaoka R., Ohnita Y. (eds.)

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11, right). We can rephrase this property by claiming that the tangent tP bisects the exterior angle of the triangle F1 P F2 at P . Similar arguments hold for tangents tP of hyperbolas. 12, left). On the other hand, the standard definition ∣r1 − r2 ∣ = const. implies that r˙1 = r˙2 . 12, right). Consequently, the tangent tP at P to the hyperbola c bisects the interior angle of the triangle F1 P F2 at P . 12. The tangent line tP at any point P to the hyperbola c bisects the interior angle of the triangle F1 P F2 .

I, with f˙ = dτ The new parameter τ needs not be the time anymore. Nevertheless, we want to retain the notations ‘velocity’ and ‘acceleration’ since they provide the derivatives with an intuitive meaning. The derivatives of the new parametrization c(τ ) ∶= c (f (τ )), τ ∈ I, are dc dc dτ = ⋅ = f˙ c˙ ≠ 0, c˙ = dτ dτ dτ 2 ¨ = d c = f¨c˙ + f˙2 ¨c . c dτ 2 We note that the velocity vectors v = c˙ and v = c˙ are linearly dependent. ¨ is a linear The spanned tangent line remains the same. The vector a ∶= c ¨.

Proof: We differentiate the equation d2 (τ ) = (c(τ ) − m)2 twice and obtain ˙ )⟩, dd˙ = ⟨c(τ ) − m, c(τ ˙ 2 + ⟨c(τ ) − m, c ¨(τ )⟩. 1 Classical definitions We denote the Frenet frame at c(τ0 ) by (e1 , e2 ), extend it by e3 to an orthonormal frame in E3 , and set 0 = (τ0 ) and v0 = v(τ0 ). 20), we get at τ = τ0 dd˙ ∣τ =τ0 = v0 ⟨c(τ0 ) − m, e1 ⟩, v2 d˙ 2 + dd¨∣τ =τ0 = v02 + ⟨c(τ0 ) − m, v(τ ˙ 0 ) e1 + 0 e2 ⟩ . , m = Hence, d(τ c(τ0 ) + 0 e2 + μ e3 = c∗ (τ0 ) + μ, e3 for all μ ∈ R3 . , in the osculating plane.

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