{"id":28999,"date":"2021-05-05T13:00:09","date_gmt":"2021-05-05T13:00:09","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=28999"},"modified":"2024-09-22T01:55:28","modified_gmt":"2024-09-22T01:55:28","slug":"equation-des-hyperboles-et-sa-deduction","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/equation-des-hyperboles-et-sa-deduction\/","title":{"rendered":"\u00c9quation des hyperboles et sa d\u00e9duction"},"content":{"rendered":"<p><center><\/p>\n<h1>\u00c9quation des hyperboles et sa d\u00e9duction<\/h1>\n<p><em><strong>R\u00e9sum\u00e9 :<\/strong><br \/>\nDans ce cours, nous explorerons la d\u00e9finition g\u00e9om\u00e9trique de l&#8217;hyperbole, la comparerons \u00e0 l&#8217;ellipse et d\u00e9duirons son \u00e9quation g\u00e9n\u00e9rale et canonique.<br \/>\n   <\/em><\/p>\n<p>   <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>D\u00e9finir<\/strong> g\u00e9om\u00e9triquement ce qu&#8217;est une hyperbole.<\/li>\n<li><strong>D\u00e9montrer<\/strong> l&#8217;\u00e9quation g\u00e9n\u00e9rale et canonique des hyperboles \u00e0 partir de leur d\u00e9finition g\u00e9om\u00e9trique.<\/li>\n<li><strong>Identifier<\/strong> les diff\u00e9rences entre les ellipses et les hyperboles en termes de distances focales.<\/li>\n<\/ol>\n<p>   <strong>TABLE DES MATI\u00c8RES<\/strong><br \/>\n<a href=\"#1\">D\u00e9finition g\u00e9om\u00e9trique de l&#8217;hyperbole<\/a><br \/>\n<a href=\"#2\">D\u00e9duction de l&#8217;\u00e9quation des hyperboles<\/a><br \/>\n<a href=\"#3\">\u00c9quation g\u00e9n\u00e9rale des hyperboles<\/a><br \/>\n<a href=\"#4\">\u00c9quation canonique des hyperboles<\/a>\n   <\/p>\n<p>   <\/center><\/p>\n<p>   <center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/1Aearz-E3bk\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p><a name=\"1\"><\/a>   <\/p>\n<h2>D\u00e9finition g\u00e9om\u00e9trique de l&#8217;hyperbole<\/h2>\n<p style=\"text-align: justify;\">Auparavant, nous avons examin\u00e9 l&#8217;\u00e9quation des ellipses et des cercles et d\u00e9couvert qu&#8217;elles prennent la forme <span class=\"katex-eq\" data-katex-display=\"false\">ax^2 + bx + cy^2 + dy + e = 0<\/span>, o\u00f9 <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">b<\/span> sont deux valeurs non nulles du m\u00eame signe. Nous avons mentionn\u00e9 que si <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">b<\/span> ont des signes oppos\u00e9s, alors au lieu d&#8217;une ellipse, nous obtenons une hyperbole. Nous n&#8217;avons rien dit de plus sur ces courbes, et nous allons maintenant combler cette lacune. Nous compl\u00e9terons notre \u00e9tude en d\u00e9finissant g\u00e9om\u00e9triquement ce qu&#8217;est une hyperbole et, \u00e0 partir de l\u00e0, nous obtiendrons l&#8217;\u00e9quation g\u00e9n\u00e9rale et canonique des hyperboles.<\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=1Aearz-E3bk&amp;t=176s\" target=\"_blank\" rel=\"noopener\"><strong>D&#8217;une part, une ellipse est d\u00e9finie<\/strong><\/a> comme l&#8217;ensemble de tous les points dont la somme des distances \u00e0 deux autres points, appel\u00e9s foyers, est toujours la m\u00eame. De m\u00eame, et en opposition, une hyperbole est d\u00e9finie comme l&#8217;ensemble de tous les points dont la valeur absolue de la diff\u00e9rence entre les distances aux foyers est toujours constante.<\/p>\n<p>   <img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-BNQiwaq_OJs\/YJGda6VHOmI\/AAAAAAAAFDo\/edTQHWwLGGQwyGszR7c-7H74a09ASsK2gCLcBGAsYHQ\/s0\/hiperbola%2Bdefinici%25C3%25B3n%2Bgr%25C3%25A1fica.PNG\" alt=\"D\u00e9finition g\u00e9om\u00e9trique de l'hyperbole\" class=\" aligncenter lazyload\" width=\"493\" height=\"340\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-BNQiwaq_OJs\/YJGda6VHOmI\/AAAAAAAAFDo\/edTQHWwLGGQwyGszR7c-7H74a09ASsK2gCLcBGAsYHQ\/s0\/hiperbola%2Bdefinici%25C3%25B3n%2Bgr%25C3%25A1fica.PNG\" alt=\"D\u00e9finition g\u00e9om\u00e9trique de l'hyperbole\" class=\" aligncenter lazyload\" width=\"493\" height=\"340\" \/><\/noscript><\/p>\n<p style=\"text-align: justify;\">Autrement dit, la relation suivante est satisfaite :<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">|d(f_1,P) - d(f_2,P)| = 2a<\/span>\n<p style=\"text-align: justify;\">O\u00f9 <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> est un nombre r\u00e9el fix\u00e9 quelconque.<\/p>\n<p style=\"text-align: justify;\">Cela donne en r\u00e9alit\u00e9 deux \u00e9quations : <span class=\"katex-eq\" data-katex-display=\"false\">d(f_1,P) - d(f_2,P) = 2a<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">d(f_2,P) - d(f_1,P) = 2a<\/span>, une pour chaque branche de l&#8217;hyperbole.<\/p>\n<p><a name=\"2\"><\/a>  <\/p>\n<h2>D\u00e9duction de l&#8217;\u00e9quation des hyperboles<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=1Aearz-E3bk&amp;t=331s\" target=\"_blank\" rel=\"noopener\"><strong>\u00c0 partir de la d\u00e9finition g\u00e9om\u00e9trique, il est possible de d\u00e9river<\/strong><\/a> la repr\u00e9sentation alg\u00e9brique des hyperboles. Pour cela, nous commencerons par le cas le plus simple, et \u00e0 partir de l\u00e0, nous g\u00e9n\u00e9raliserons. Notre raisonnement portera sur une seule branche de l&#8217;hyperbole, et le raisonnement pour l&#8217;autre branche est tout \u00e0 fait analogue.<\/p>\n<h3>D\u00e9duction de la forme simplifi\u00e9e<\/h3>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=1Aearz-E3bk&amp;t=356s\" target=\"_blank\" rel=\"noopener\"><strong>Consid\u00e9rons deux points focaux<\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">f_1 = (-c,0)<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">f_2 = (c,0).<\/span> Le point <span class=\"katex-eq\" data-katex-display=\"false\">p = (x,y)<\/span> se trouve sur l&#8217;hyperbole si<\/p>\n<p>   <img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-SMOUgyC1lM4\/YJGg_MIkJTI\/AAAAAAAAFDw\/6JzXOcfZi70lpvTZtbC6y26AvTQzcnWNgCLcBGAsYHQ\/s0\/hiperbola%2Bcentrada%2Ben%2Bel%2Borigen.PNG\" alt=\"Hyperbole centr\u00e9e \u00e0 l'origine\" class=\"aligncenter  lazyload\" width=\"342\" height=\"288\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-SMOUgyC1lM4\/YJGg_MIkJTI\/AAAAAAAAFDw\/6JzXOcfZi70lpvTZtbC6y26AvTQzcnWNgCLcBGAsYHQ\/s0\/hiperbola%2Bcentrada%2Ben%2Bel%2Borigen.PNG\" alt=\"Hyperbole centr\u00e9e \u00e0 l'origine\" class=\"aligncenter  lazyload\" width=\"342\" height=\"288\" \/><\/noscript><\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sqrt{(x+c)^2+y^2} - \\sqrt{(x-c)^2+y^2} = 2a<\/span>\n<p>   &nbsp;<\/p>\n<p style=\"text-align: justify;\">Et de l\u00e0, nous obtenons le raisonnement suivant :<\/p>\n<table style=\"text-align: justify;\">\n<tbody>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\sqrt{(x+c)^2+y^2} - \\sqrt{(x-c)^2+y^2} = 2a<\/span><\/td>\n<td>; \u00e9quation des hyperboles<\/td>\n<\/tr>\n<tr><\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\sqrt{x^2 + 2xc + c^2 + y^2} - \\sqrt{x^2 - 2xc + c^2 + y^2} = 2a<\/span><\/td>\n<td>; d\u00e9veloppement des carr\u00e9s<\/td>\n<\/tr>\n<tr><\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\sqrt{x^2 + 2xc + c^2 + y^2} = 2a + \\sqrt{x^2 - 2xc + c^2 + y^2}<\/span><\/td>\n<td>; redistribution des termes<\/td>\n<\/tr>\n<tr><\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> \\color{red}{x^2} + 2xc + \\color{purple}{c^2} + \\color{violet}{y^2} = 4a^2 + 4a\\sqrt{x^2 - 2xc + c^2 + y^2} + \\color{red}{x^2} - 2xc + \\color{purple}{c^2} + \\color{violet}{y^2}<\/span><\/td>\n<td>; en \u00e9levant les deux c\u00f4t\u00e9s au carr\u00e9<\/td>\n<\/tr>\n<tr><\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> 2xc = 4a^2 + 4a\\sqrt{x^2 - 2xc + c^2 + y^2} - 2xc <\/span><\/td>\n<td>; \u00e9limination des termes identiques<\/td>\n<\/tr>\n<tr><\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> 4xc = 4a^2 + 4a\\sqrt{x^2 - 2xc + c^2 + y^2} <\/span><\/td>\n<td>; redistribution des termes identiques<\/td>\n<\/tr>\n<tr><\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> xc = a^2 + a\\sqrt{x^2 - 2xc + c^2 + y^2} <\/span><\/td>\n<td>; simplification des termes identiques<\/td>\n<\/tr>\n<tr><\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> xc - a^2 = a\\sqrt{x^2 - 2xc + c^2 + y^2} <\/span><\/td>\n<td>; simplification des termes identiques<\/td>\n<\/tr>\n<tr><\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> x^2c^2 -2xca^2 + a^4 = a^2(x^2 - 2xc + c^2 + y^2) <\/span><\/td>\n<td>; en \u00e9levant les deux c\u00f4t\u00e9s au carr\u00e9<\/td>\n<\/tr>\n<tr><\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> x^2c^2 \\color{red}{-2xca^2} + a^4 = a^2x^2 \\color{red}{- 2xca^2} + a^2c^2 + a^2y^2 <\/span><\/td>\n<td>; op\u00e9ration des parenth\u00e8ses<\/td>\n<\/tr>\n<tr><\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> x^2c^2 + a^4 = a^2x^2 + a^2c^2 + a^2y^2 <\/span><\/td>\n<td>; \u00e9limination des termes identiques<\/td>\n<\/tr>\n<tr><\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> x^2(c^2 - a^2) - a^2y^2 = a^2c^2 - a^4 = a^2(c^2 - a^2) <\/span><\/td>\n<td>; r\u00e9arrangement des termes<\/td>\n<\/tr>\n<tr><\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{x^2}{a^2} - \\frac{y^2}{c^2 - a^2} = 1 <\/span><\/td>\n<td>; r\u00e9arrangement des termes<\/td>\n<\/tr>\n<tr><\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\">Pour cette derni\u00e8re expression, comme pour les ellipses, on prend <span class=\"katex-eq\" data-katex-display=\"false\">b^2=c^2-a^2<\/span> et on obtient l&#8217;\u00e9quation des hyperboles :<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{ \\left(\\frac{x}{a}\\right)^2 - \\left(\\frac{y}{b}\\right)^2 = 1 }<\/span>\n<p><a name=\"3\"><\/a>     <\/p>\n<h2>\u00c9quation g\u00e9n\u00e9rale des hyperboles<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=1Aearz-E3bk&amp;t=801s\" target=\"_blank\" rel=\"noopener\"><strong>Pour obtenir l&#8217;\u00e9quation g\u00e9n\u00e9rale<\/strong><\/a> des hyperboles, il suffit de prendre celle que nous venons d&#8217;obtenir et d&#8217;appliquer les transformations de position :<\/p>\n<table style=\"text-align: justify;\">\n<tbody>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">x\\longmapsto x-h<\/span><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y\\longmapsto y-k<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\">et avec cela, nous obtenons automatiquement l&#8217;\u00e9quation g\u00e9n\u00e9rale des ellipses avec un centre en <span class=\"katex-eq\" data-katex-display=\"false\">(h,k)<\/span>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\color{blue}{ \\left(\\frac{x-h}{a}\\right)^2 - \\left(\\frac{y-k}{b}\\right)^2 = 1 }<\/span>\n<p><a name=\"4\"><\/a>     <\/p>\n<h2>\u00c9quation canonique des hyperboles<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=1Aearz-E3bk&amp;t=974s\" target=\"_blank\" rel=\"noopener\"><strong>Et si maintenant nous prenons l&#8217;\u00e9quation g\u00e9n\u00e9rale<\/strong><\/a> des hyperboles et la d\u00e9veloppons, nous arriverons \u00e0 l&#8217;expression canonique :<\/p>\n<table style=\"text-align: justify;\">\n<tbody>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\left(\\frac{x-h}{a}\\right)^2 - \\left(\\frac{y-k}{b}\\right)^2 = 1<\/span><\/td>\n<td>; \u00c9quation g\u00e9n\u00e9rale des hyperboles<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">b^2 (x^2 - 2xh + h^2) - a^2(y^2-2ky + y^2) = a^2b^2<\/span><\/td>\n<td>; d\u00e9veloppement des carr\u00e9s et multiplication de tout par <span class=\"katex-eq\" data-katex-display=\"false\">a^2b^2<\/span><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> b^2 x^2 - 2hb^2x + h^2b^2 - a^2 y^2+ 2k a^2 y - a^2 k^2 = a^2b^2<\/span><\/td>\n<td>; op\u00e9ration des parenth\u00e8ses<\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> b^2 x^2 - (2hb^2) x - a^2 y^2+ (2k a^2) y - (a^2b^2 + a^2 k^2 - h^2b^2) = 0 <\/span><\/td>\n<td>; regroupement des termes identiques<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\">Cette derni\u00e8re expression est de la forme <span class=\"katex-eq\" data-katex-display=\"false\">Ax^2+Bx + Cy^2 + Dy + E = 0,<\/span> o\u00f9 <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> et <span class=\"katex-eq\" data-katex-display=\"false\">C<\/span> sont toujours non nuls et de signes oppos\u00e9s, comme nous l&#8217;avons pr\u00e9dit en \u00e9tudiant les ellipses.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00c9quation des hyperboles et sa d\u00e9duction R\u00e9sum\u00e9 : Dans ce cours, nous explorerons la d\u00e9finition g\u00e9om\u00e9trique de l&#8217;hyperbole, la comparerons \u00e0 l&#8217;ellipse et d\u00e9duirons son \u00e9quation g\u00e9n\u00e9rale et canonique. Objectifs d&#8217;apprentissage : \u00c0 la fin de ce cours, l&#8217;\u00e9tudiant sera capable de : D\u00e9finir g\u00e9om\u00e9triquement ce qu&#8217;est une hyperbole. D\u00e9montrer l&#8217;\u00e9quation g\u00e9n\u00e9rale et canonique des [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":28988,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":20,"footnotes":""},"categories":[585,569],"tags":[],"class_list":["post-28999","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 v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>\u00c9quation des hyperboles et sa d\u00e9duction - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Apprenez tout sur les hyperboles : d\u00e9finition g\u00e9om\u00e9trique, d\u00e9duction de leur \u00e9quation g\u00e9n\u00e9rale et canonique, diff\u00e9rences avec les ellipses, et comment r\u00e9soudre les \u00e9quations \u00e9tape par \u00e9tape. 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