![]() Gerstner Center for Cancer Diagnostics The Gerstner Center is developing next-generation diagnostic technology for cancer detection and tracking disease progression.Carlos Slim Center for Health Research The Slim Center aims to bring the benefits of genomics-driven medicine to Latin America, gleaning new insights into diseases with relevance to the region.Collaborations and consortia We join with institutions and scientists the world over to address foundational challenges in science and health.Resources, services, and tools Key scientific datasets and computational tools developed by our scientists and their collaborators.Publications A catalog of scientific papers published by our members and staff scientists.Partnering and licensing We work closely with pharmaceutical, biotech, and technology partners to accelerate the translation of our discoveries.COVID-19 Our community is deeply engaged in the local, national, and global effort to respond to COVID-19.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. ![]() A hyperbola is the locus of points that has a constant ratio of distance between a focus (point) and a directrix (line), where that constant ratio is greater than 1.Here, the eccentricity is $\frac C A$, which, by this definition, must be a constant less than 1 for every point on the ellipse. An ellipse is the locus of points that has a constant ratio of distance between a focus (point) and a directrix (line), where that constant ratio is between 0 and 1.We have two definitions of an ellipse and a hyperbola. Note that has a proof for this question with Dandelin spheres.) Also, I've made a YouTube video explaining this whole proof. (NOTE: This is my attempt at answering this question and this question, but I rewrote it in order to make it easier for me to solve. ![]()
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