翻訳と辞書 Words near each other ・ Deris ・ Deris Benavides ・ Deris Rural District ・ Deris Umanzor ・ Derita (Charlotte neighborhood) ・ Deritend ・ Deritend ware ・ Deriugins School ・ Derius ・ Derivae ・ Derivart ・ Derivate (disambiguation) ・ Derivation ・ Derivation (differential algebra) ・ Derivation of self inductance ・ Derivation of the Cartesian form for an ellipse ・ Derivation of the conjugate gradient method ・ Derivation of the Navier–Stokes equations ・ Derivation of the Routh array ・ Derivations of the Lorentz transformations ・ Derivative ・ Derivative (chemistry) ・ Derivative (disambiguation) ・ Derivative (film) ・ Derivative (finance) ・ Derivative algebra ・ Derivative algebra (abstract algebra) ・ Derivative and Commodity Exchange Nepal Ltd. ・ Derivative chromosome ・ Derivative Dribble Dictionary Lists mini英和辞書 mini和英辞書 Webster 1913 Latin-English FOLDOC Wikipedia English ウィキペディア

翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Derivation of the Cartesian form for an ellipse ： ウィキペディア英語版 Derivation of the Cartesian form for an ellipse The derivation of the Cartesian form for an ellipse is simple and instructive. One simple definition of the ellipse is the "locus of all points of the plane whose distances to two fixed points (called the foci) add to the same constant". See ellipse for other definitions. Let the foci be points (-c,0) and (c,0). Then the ellipse center is (0, 0).If (x,y) is any point on the ellipse and if $d\_1$ is the distance between (x,y) and (-c,0) and $d\_2$ is the distance between (x,y) and (c,0), i.e. then we can define $a$ :$d\_1\; +\; d\_2\; =\; 2a\backslash ,$ (a here is the semi-major axis, although this is irrelevant for the sake of the proof). From this simple definition we can derive the cartesian equation. Substituting: :$\backslash sqrt\; +\; \backslash sqrt\; =\; 2a$ To simplify we isolate the radical and square both sides. :$\backslash sqrt\; =\; 2a\; -\; \backslash sqrt$ :$(x+c)^2\; +\; y^2\; =\; \backslash left\; (\; 2a\; -\; \backslash sqrt\; \backslash right\; )^2$ :$(x+c)^2\; +\; y^2\; =\; 4a^2\; -\; 4a\backslash sqrt\; +\; (x-c)^2\; +y^2$ Solving for the root and simplifying: :$(x+c)^2\; +\; y^2\; -\; (x-c)^2\; -\; y^2\; -\; 4a^2\; =\; -\; 4a\backslash sqrt$ :$-((x+c)^2\; -\; 4a^2\; -\; (x-c)^2)\; =\; \backslash sqrt$ Swap sides to return to original format and continue: :$\backslash sqrt\; =\; -\; ((x+c)^2-4a^2-(x-c)^2)$ :$\backslash sqrt\; =\; -\; (x^2\; +\; 2xc\; +\; c^2\; -4a^2\; -x^2\; +2xc\; -c^2)$ :$\backslash sqrt\; =\; -\; (4xc\; -\; 4a^2)$ :$\backslash sqrt\; =\; a\; -\; x$ A final squaring :$(x-c)^2+y^2\; =\; a^2\; -\; 2xc\; +\; x^2$ :$x^2\; -\; 2xc\; +\; c^2\; +\; y^2\; =\; a^2\; -2xc\; +\; x^2$ :$x^2\; +\; c^2\; +\; y^2\; =\; a^2\; +\; x^2$ Grouping the x-terms and dividing by $a^2-c^2\backslash ,$ :$x^2\; -\; x^2\; +\; y^2\; =\; a^2\; -\; c^2$ :$x^2\; \backslash left(\; 1\; -\; \backslash right)\; +\; y^2\; =\; a^2\; -\; c^2$ Where: $1\; =$ :$x^2\; \backslash left(\; \backslash right)\; +\; y^2\; =\; a^2\; -\; c^2$ :$+\; =\; 1$ If x = 0 then :$d\_1\; =\; d\_2\; =\; a\; =\; \backslash sqrt$ (where b is the semi-minor axis) Therefore we can substitute :$b^2\; =\; a^2\; -\; c^2\backslash ,$ And we have our desired equation: :$+\; =\; 1$ 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Derivation of the Cartesian form for an ellipse」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.