![]() ![]() This balance between duration and effort carries with it a good part of the challenges of the moment, and of the possible scenarios in the relatively short term.Īll the protagonists know that no negotiations are possible today because both parties, Russians and Ukrainians, are preparing for looming offensives and counter-offensives. But above all, if we do not wish it, we must collectively be credible in our ability to endure this effort." But he immediately added, "By saying this, I do not wish it. In his speech, just a few days before the one-year anniversary of the Russian invasion, Macron said he was prepared for a prolonged conflict in Ukraine. Timing is a topic that weighs on the destiny of the war, on the outlines of a possible peace, on the rest of the world that suffers the consequences. Still, time management is crucial in any military conflict, and the French President mentioned it in his speech at the Munich Security Conference this past weekend, and in the interview he gave to France Inter public radio afterwards. ![]() $$f_1:d_1=f_2:d_2=c/a$$ that proves the statement.Stay up-to-date with the latest on the Russia-Ukraine war, with our exclusive international coverage. So, also the triangles $H_2PF_2$ and $F_2PN$ are similar. The triangles $H_1PF_1$ and $F_1PN$ are similar (they have the same angle $\alpha$ and two alternate interior angles). Let’s call $N$ the intersection of this line with the line joining $F_1$ and $F_2$, and we’ll call $NF_1=m_1$ $NF_2=m_2$ ![]() The perpendicular bisector of the segment $F_1F_2$ meets the circle in $V$ (and the quadrilateral $VF_1F_2P$ is a cyclic quadrilateral). Given the locus of points for which is constant the sum of the distances from two fixed points $F_1$ and $F_2$ then for any point $P$ belonging to the locus it is $PF_1=e\cdot PH_1$, where $PH_1$ is the distance form an appropriate fixed line (directrix) perpendicular to the line joining $F_1$ and $F_2$. I'm not much satisfied with it (especially in the second part that is just sketched) and I'm positive it could be significantly improved. ![]() If it can help here is a pure geometric demonstration (rather complex but it uses plane geometry and not the Dandelin spheres) that I devised some years ago for the ellipse (reference: ). Thus, we are done and both definitions are equivalent with equivalent eccentricities for both ellipses and hyperbolas. However, this is the exact same eccentricity we derived for the second equation, so the eccentricities are equal.
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