{"id":28942,"date":"2021-04-27T13:00:18","date_gmt":"2021-04-27T13:00:18","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=28942"},"modified":"2024-09-22T02:16:43","modified_gmt":"2024-09-22T02:16:43","slug":"caracterisation-des-paraboles-et-leurs-graphiques","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/caracterisation-des-paraboles-et-leurs-graphiques\/","title":{"rendered":"Caract\u00e9risation des Paraboles et leurs Graphiques"},"content":{"rendered":"<p><center><\/p>\n<h1>Caract\u00e9risation des Paraboles et leurs Graphiques<\/h1>\n<p><em><strong>R\u00e9sum\u00e9 :<\/strong><br \/>\n   Dans ce cours, nous examinerons la caract\u00e9risation des paraboles \u00e0 partir de leur \u00e9quation g\u00e9n\u00e9rale et forme canonique, en expliquant comment identifier des \u00e9l\u00e9ments cl\u00e9s tels que le sommet, le foyer, la directrice, l&#8217;axe de sym\u00e9trie et les \u00e9ventuelles intersections avec l&#8217;axe des abscisses.<br \/>\n   <\/em><br \/>\n   <strong>Objectifs d&#8217;Apprentissage :<\/strong><br \/>\n   \u00c0 la fin de ce cours, l&#8217;\u00e9tudiant sera capable de :<\/p>\n<ol style=\"text-align: left;\">\n<li><strong>Calculer<\/strong> la position du sommet, du foyer et de la directrice de la parabole \u00e0 partir de sa forme g\u00e9n\u00e9rale et canonique.<\/li>\n<li><strong>Transformer<\/strong> l&#8217;\u00e9quation canonique en forme g\u00e9n\u00e9rale pour extraire des informations g\u00e9om\u00e9triques.<\/li>\n<li><strong>Esquisser<\/strong> le graphique de la parabole avec les informations obtenues.<\/li>\n<\/ol>\n<p>   <strong>TABLE DES MATI\u00c8RES<\/strong><br \/>\n   <a href=\"#1\">Forme g\u00e9n\u00e9rale et canonique des paraboles<\/a><br \/>\n   <a href=\"#2\">Caract\u00e9risation des Paraboles \u00e0 partir de l&#8217;\u00c9quation G\u00e9n\u00e9rale<\/a><br \/>\n   <a href=\"#3\">Caract\u00e9risation des Paraboles \u00e0 partir de l&#8217;\u00c9quation Canonique<\/a><br \/>\n   <a href=\"#4\">Caract\u00e9risation automatique avec Excel<\/a>\n   <\/p>\n<p><\/center><br \/>\n<center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/C6DbrJDiZTM\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><br \/>\n<a name=\"1\"><\/a><\/p>\n<h2>Forme g\u00e9n\u00e9rale et canonique des paraboles<\/h2>\n<p style=\"text-align: justify;\">Dans le cours pr\u00e9c\u00e9dent, nous avons vu que les paraboles peuvent \u00eatre exprim\u00e9es alg\u00e9briquement \u00e0 travers l&#8217;\u00e9quation g\u00e9n\u00e9rale des paraboles comme suit.<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">(x-x_0)^2 =4f(y-y_0)<\/span>\n<p style=\"text-align: justify;\">O\u00f9 le couple <span class=\"katex-eq\" data-katex-display=\"false\">(x_0,y_0)<\/span> repr\u00e9sente la position du sommet et <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> est la distance focale. Si <span class=\"katex-eq\" data-katex-display=\"false\">f \\gt 0<\/span>, alors le foyer est \u00e0 une distance <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> au-dessus du sommet, et si <span class=\"katex-eq\" data-katex-display=\"false\">f\\lt 0,<\/span> alors le foyer est \u00e0 une distance <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> en dessous du sommet.<\/p>\n<p style=\"text-align: justify;\">Nous avons \u00e9galement vu que l&#8217;\u00e9quation des paraboles, lorsqu&#8217;elle est ramen\u00e9e \u00e0 sa forme canonique, est \u00e9quivalente \u00e0 un polyn\u00f4me de degr\u00e9 2.<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">y(x) = ax^2 + bx + c,<\/span> avec <span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0<\/span>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=C6DbrJDiZTM&amp;t=228s\" target=\"_blank\" rel=\"noopener\"><strong>Caract\u00e9riser une parabole consiste<\/strong><\/a> \u00e0 r\u00e9v\u00e9ler les informations suivantes :<\/p>\n<ul style=\"text-align: justify;\">\n<li>Les coordonn\u00e9es du sommet<\/li>\n<li>Les coordonn\u00e9es du foyer<\/li>\n<li>L&#8217;\u00e9quation de la directrice<\/li>\n<li>L&#8217;\u00e9quation de l&#8217;axe de sym\u00e9trie<\/li>\n<li>Les intersections avec l&#8217;axe des abscisses (s&#8217;il en existe)<\/li>\n<li>Enfin, construire une esquisse du graphique avec les informations collect\u00e9es.<\/li>\n<\/ul>\n<p><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-xAUdvfTRbjw\/YIbmIXDdT-I\/AAAAAAAAFAI\/8NH0t_EWbH0KIuFDnsRu2IyHdyN4WU54wCLcBGAsYHQ\/s0\/caracterizaci%25C3%25B3n%2Bde%2Bpar%25C3%25A1bolas.PNG\" alt=\"Caract\u00e9risation des Paraboles\" class=\" aligncenter lazyload\" width=\"537\" height=\"414\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-xAUdvfTRbjw\/YIbmIXDdT-I\/AAAAAAAAFAI\/8NH0t_EWbH0KIuFDnsRu2IyHdyN4WU54wCLcBGAsYHQ\/s0\/caracterizaci%25C3%25B3n%2Bde%2Bpar%25C3%25A1bolas.PNG\" alt=\"Caract\u00e9risation des Paraboles\" class=\" aligncenter lazyload\" width=\"537\" height=\"414\" \/><\/noscript><br \/>\n<a name=\"2\"><\/a><\/p>\n<h2>Caract\u00e9risation des Paraboles \u00e0 partir de l&#8217;\u00c9quation G\u00e9n\u00e9rale<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=C6DbrJDiZTM&amp;t=316s\" target=\"_blank\" rel=\"noopener\"><strong>Si vous avez la parabole d\u00e9crite par <\/strong><\/a>l&#8217;\u00e9quation g\u00e9n\u00e9rale, alors vous avez d\u00e9j\u00e0 presque toutes les informations n\u00e9cessaires pour compl\u00e9ter la caract\u00e9risation, seules les intersections avec l&#8217;axe des abscisses n\u00e9cessiteront une analyse suppl\u00e9mentaire.<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">(x-x_0)^2 =4f(y-y_0)<\/span>\n<p style=\"text-align: justify;\">De l\u00e0, vous obtenez :<\/p>\n<ul style=\"text-align: justify;\">\n<li><strong>Sommet :<\/strong> Le point de coordonn\u00e9es <span class=\"katex-eq\" data-katex-display=\"false\">(x_0,y_0)<\/span><\/li>\n<li><strong>Position focale : <\/strong> \u00e0 <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> unit\u00e9s au-dessus du sommet<\/li>\n<li><strong>Foyer :<\/strong> le point de coordonn\u00e9es <span class=\"katex-eq\" data-katex-display=\"false\">(x_0,y_0 + f)<\/span><\/li>\n<li><strong>Directrice :<\/strong> la droite d&#8217;\u00e9quation <span class=\"katex-eq\" data-katex-display=\"false\">y= y_0 - f<\/span><\/li>\n<li><strong>L&#8217;axe de sym\u00e9trie :<\/strong> la droite d&#8217;\u00e9quation <span class=\"katex-eq\" data-katex-display=\"false\">x= x_0<\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify;\">Pour trouver les intersections avec l&#8217;axe des abscisses, vous devrez transformer l&#8217;\u00e9quation g\u00e9n\u00e9rale en forme canonique, \u00e9galer le polyn\u00f4me de second degr\u00e9 r\u00e9sultant \u00e0 z\u00e9ro. Si des solutions existent, elles seront les intersections avec l&#8217;axe des abscisses.<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>Caract\u00e9risation des Paraboles \u00e0 partir de l&#8217;\u00c9quation Canonique<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=C6DbrJDiZTM&amp;t=393s\" target=\"_blank\" rel=\"noopener\"><strong>Lorsque l&#8217;\u00e9quation des paraboles<\/strong><\/a> est pr\u00e9sent\u00e9e sous forme canonique, vous avez deux possibilit\u00e9s : 1) Caract\u00e9riser en transformant en \u00e9quation g\u00e9n\u00e9rale ou 2) En utilisant la sym\u00e9trie et les intersections avec l&#8217;axe des abscisses. Les deux m\u00e9thodes ont leurs vertus. La seconde est g\u00e9n\u00e9ralement plus rapide, mais les paraboles ne coupent pas toujours l&#8217;axe des abscisses, la premi\u00e8re est plus laborieuse mais, comme nous le verrons plus tard, est facile \u00e0 automatiser. Nous examinerons les deux alternatives afin que vous puissiez choisir, selon vos pr\u00e9f\u00e9rences et vos besoins, quelle voie suivre.<\/p>\n<h3>Transformer en \u00e9quation g\u00e9n\u00e9rale<\/h3>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=C6DbrJDiZTM&amp;t=475s\" target=\"_blank\" rel=\"noopener\"><strong>La transformation en forme g\u00e9n\u00e9rale<\/strong><\/a> se fait selon le raisonnement suivant, o\u00f9 <span class=\"katex-eq\" data-katex-display=\"false\">a,b,c\\in\\mathbb{R}<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0.<\/span>\n<table style=\"text-align: justify;\">\n<tbody>\n<tr>\n<td width=\"50\">(1)<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y=ax^2 + bx + c<\/span><\/td>\n<td>; \u00c9quation canonique des paraboles<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y=a\\left[x^2 + \\dfrac{b}{a}x + \\dfrac{c}{a}\\right]<\/span><\/td>\n<td>; Factorisation par <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y=a\\left[ \\left(x + \\dfrac{b}{2a}\\right)^2 - \\dfrac{b^2}{4a^2} + \\dfrac{c}{a}\\right]<\/span><\/td>\n<td>; Parce que <span class=\"katex-eq\" data-katex-display=\"false\">\\left(x + \\dfrac{b}{2a}\\right)^2 = x^2 + \\dfrac{b}{a}x + \\dfrac{b^2}{4a^2}<\/span><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y=a\\left[ \\left(x + \\dfrac{b}{2a}\\right)^2 + \\dfrac{4ac - b^2}{4a^2} \\right]<\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y=a \\left(x + \\dfrac{b}{2a}\\right)^2 + \\dfrac{4ac - b^2}{4a}<\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y=a \\left(x + \\dfrac{b}{2a}\\right)^2 + \\left(c - \\dfrac{b^2}{4a}\\right)<\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> \\left[x - \\left(- \\dfrac{b}{2a}\\right)\\right]^2 = \\dfrac{1}{a} \\left[y - \\left(c - \\dfrac{b^2}{4a}\\right)\\right]<\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> \\left[x - \\left( -\\dfrac{b}{2a}\\right)\\right]^2 = 4\\left(\\dfrac{1}{4a}\\right) \\left[y - \\left(c - \\dfrac{b^2}{4a}\\right)\\right]<\/span><\/td>\n<td>; \u00c9quation des paraboles en forme g\u00e9n\u00e9rale<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\">\u00c0 partir de cela, nous pouvons extraire toutes les informations que nous avions de l&#8217;\u00e9quation g\u00e9n\u00e9rale en reliant ses param\u00e8tres \u00e0 ceux de l&#8217;\u00e9quation canonique, nous avons donc :<\/p>\n<ul style=\"text-align: justify;\">\n<li><strong>Sommet :<\/strong> Le point de coordonn\u00e9es <span class=\"katex-eq\" data-katex-display=\"false\">(x_0,y_0) = \\left(-\\dfrac{b}{2a}, c -\\dfrac{b^2}{4a} \\right)<\/span><\/li>\n<li><strong>Position focale : <\/strong> \u00e0 <span class=\"katex-eq\" data-katex-display=\"false\">f = \\dfrac{1}{4a}<\/span> unit\u00e9s au-dessus du sommet<\/li>\n<li><strong>Foyer :<\/strong> le point de coordonn\u00e9es <span class=\"katex-eq\" data-katex-display=\"false\">(x_0,y_0 + f) = \\left(-\\dfrac{b}{2a}, c -\\dfrac{b^2}{4a} + \\dfrac{1}{4a}\\right) =\\left(-\\dfrac{b}{2a}, c +\\dfrac{1-b^2}{4a}\\right) <\/span><\/li>\n<li><strong>Directrice :<\/strong> la droite d&#8217;\u00e9quation <span class=\"katex-eq\" data-katex-display=\"false\">y=y_0 - f= c -\\dfrac{b^2}{4a} - \\dfrac{1}{4a} = c -\\dfrac{1 + b^2}{4a}<\/span><\/li>\n<li><strong>L&#8217;axe de sym\u00e9trie :<\/strong> la droite d&#8217;\u00e9quation <span class=\"katex-eq\" data-katex-display=\"false\">x= x_0 = -\\dfrac{b}{2a}<\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify;\">Et \u00e0 partir de l\u00e0, la caract\u00e9risation des paraboles se fait comme nous l&#8217;avons d\u00e9j\u00e0 vu en utilisant l&#8217;\u00e9quation g\u00e9n\u00e9rale.<\/p>\n<h3>Utiliser la sym\u00e9trie et les intersections avec l&#8217;axe des abscisses<\/h3>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=C6DbrJDiZTM&amp;t=769s\" target=\"_blank\" rel=\"noopener\"><strong>Lorsque nous avons l&#8217;\u00e9quation <\/strong><\/a>des paraboles \u00e9crite sous forme canonique <span class=\"katex-eq\" data-katex-display=\"false\">y=ax^2 + bx+c<\/span>, il est relativement simple de calculer ses intersections avec l&#8217;axe des abscisses, il suffit de r\u00e9soudre l&#8217;\u00e9quation suivante :<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">ax^2 + bx + c = 0<\/span>\n<p style=\"text-align: justify;\">Quand cela est possible, nous obtenons les intersections <span class=\"katex-eq\" data-katex-display=\"false\">x_1<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">x_2<\/span> donn\u00e9es par<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1 = \\dfrac{-b + \\sqrt{b^2-4ac}}{2a}<\/span>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">x_2 = \\dfrac{-b - \\sqrt{b^2-4ac}}{2a}<\/span>\n<p style=\"text-align: justify;\">Comme les paraboles sont sym\u00e9triques, l&#8217;axe de sym\u00e9trie aura l&#8217;\u00e9quation suivante :<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">x = x_0 = \\dfrac{x_1 + x_2}{2}= -\\dfrac{b}{2a}<\/span>\n<p style=\"text-align: justify;\">L&#8217;axe de sym\u00e9trie passe n\u00e9cessairement par le sommet de la parabole, dont les coordonn\u00e9es sont :<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_0, y_0) = (x_0, y(x_0)) = \\left( -\\dfrac{b}{2a}, y\\left(-\\dfrac{b}{2a}\\right) \\right)<\/span>\n<p style=\"text-align: justify;\">O\u00f9<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">y_0 = y\\left(-\\dfrac{b}{2a} \\right) = a\\left(-\\dfrac{b}{2a}\\right)^2 + b\\left(-\\dfrac{b}{2a}\\right) + c = \\dfrac{b^2}{4a} - \\dfrac{b^2}{2a} + c = c - \\dfrac{b^2}{4a}<\/span>\n<p style=\"text-align: justify;\">C&#8217;est ainsi que nous obtenons les coordonn\u00e9es du sommet que nous connaissions d\u00e9j\u00e0 par d&#8217;autres m\u00e9thodes :<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_0, y_0) = \\left( -\\dfrac{b}{2a},c - \\dfrac{b^2}{4a} \\right)<\/span>\n<p style=\"text-align: justify;\">La position focale est, comme nous l&#8217;avons d\u00e9j\u00e0 vu, <span class=\"katex-eq\" data-katex-display=\"false\">f=\\dfrac{1}{4a},<\/span> et \u00e0 partir de cela, nous pouvons calculer la position de la directrice, du foyer et toutes les informations que nous avions d\u00e9j\u00e0 \u00e0 partir de l&#8217;\u00e9quation g\u00e9n\u00e9rale.<\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Caract\u00e9risation automatique avec Excel<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=C6DbrJDiZTM&amp;t=1086s\" target=\"_blank\" rel=\"noopener\"><strong>Apr\u00e8s avoir effectu\u00e9<\/strong><\/a> tous ces raisonnements, il est maintenant tr\u00e8s simple d&#8217;automatiser la caract\u00e9risation de toute parabole via Excel. Vous pouvez trouver un exemple <a href=\"https:\/\/drive.google.com\/file\/d\/1LbNOKHHfzlPgHI3_NSzuB_6_b7KmXlTd\/view?usp=sharing\" rel=\"noopener\" target=\"_blank\">ici.<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Caract\u00e9risation des Paraboles et leurs Graphiques R\u00e9sum\u00e9 : Dans ce cours, nous examinerons la caract\u00e9risation des paraboles \u00e0 partir de leur \u00e9quation g\u00e9n\u00e9rale et forme canonique, en expliquant comment identifier des \u00e9l\u00e9ments cl\u00e9s tels que le sommet, le foyer, la directrice, l&#8217;axe de sym\u00e9trie et les \u00e9ventuelles intersections avec l&#8217;axe des abscisses. Objectifs d&#8217;Apprentissage : [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":28929,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":35,"footnotes":""},"categories":[585,569],"tags":[],"citadela-post-location":[],"class_list":["post-28942","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-algebre-et-geometrie","category-mathematiques"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v28.0 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Caract\u00e9risation des Paraboles et leurs Graphiques - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Apprends \u00e0 caract\u00e9riser des paraboles en identifiant le sommet, le foyer, la directrice, l&#039;axe de sym\u00e9trie et les intersections avec l&#039;axe des abscisses.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, 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