# ℵ 0 write a short outline of the important

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Write a short outline of the important achievements and events in geometry, from ancient times to today.
4 proved the area of a circle, surface area and volume of a sphere formulas involving π. He was doing an early version of an integral, but as equation s and functions did not exist at that time, there were no calculus before the 17 th century. In around 250 BC, Apollonius wrote a book about conics to illustrate the different cuts of a plane on a cone (i.e. circle, ellipse, parabola and hyperbola). However, for the hyperbola, he was actually looking at a double cone. For the next 1000 years, everyone did mathematics using lengths and geometry was considered, at that time, the basis of mathematics. In 1100, Umar al-Khayyami found solutions to the cubic equation using geometry, still in lengths. Between the 1200s and the 1500s were the Dark Ages of Mathematics. Nothing new was invented… In 1640, Descartes invented graphs for curves, as curves can be expressed as equations, something that unified geometry and algebra. Calculus was invented shortly thereafter. Cavalieri and Wallace used Archimedes’ work to find the area “under” a general curve, something that later become an integral. In 1830, Lagrange (and others) discovered that Euclid’s methods were not up to the current standards. They looked back at the Parallel postulate. Later, Lobachevsky, Bolyai and Gauss decided to prove the parallel postulate by reductio ad absurdum and discovered non-Euclidean geometries, where triangles no longer add up to 180° and the Pythagorean Theorem no longer true. Consequently, there may be many or no parallels in such geometries, so the parallel postulate becomes different in each of those geometries. Find the foci, eccentricity and directrix of the curve given by ? 2 16 ? 2 9 = 1. ? 2 16 ? 2 9 = 1 is a hyperbola with a = 4 and b = 3. Foci are located at (- c , 0) and (0, c ) where a 2 + b 2 = c 2 . In this problem, c 2 = 4 2 + 3 2 = 25 → c = 5. Therefore, the foci are at (-5, 0) and (5, 0). For each c , the directrix is at x = ? 2 ? . Here, for c = -5, the directrix is at 4 2 −5 = −16 5 . For c = 5, the directrix is at 4 2 5 = 16 5 . The eccentricity is given by √? 2 +? 2 ? = √4 2 +3 2 4 = √25 4 = 5 4 .
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