{"id":3216,"date":"2022-06-06T16:27:00","date_gmt":"2022-06-06T14:27:00","guid":{"rendered":"https:\/\/studijni-svet.cz\/kurzy\/courses\/matematika-k-maturite\/lekce\/hyperbola-reseny-priklad\/"},"modified":"2022-06-07T22:03:20","modified_gmt":"2022-06-07T20:03:20","slug":"hyperbola-reseny-priklad","status":"publish","type":"lesson","link":"https:\/\/studijni-svet.cz\/kurzy\/courses\/matematika-k-maturite\/lekce\/hyperbola-reseny-priklad\/","title":{"rendered":"Hyperbola \u0159e\u0161en\u00fd p\u0159\u00edklad"},"content":{"rendered":"<p><strong>Zad\u00e1n\u00ed:<\/strong> Ur\u010dete st\u0159edovou rovnici a asymptoty hyperboly se st\u0159edem S[2; -1], ohniskem E[7; -1] a vrcholem A[5; -1].<\/p>\n<p>&nbsp;<\/p>\n<p><strong>\u0158e\u0161en\u00ed:<\/strong><\/p>\n<p style=\"text-align: center\">Na\u0161\u00edm c\u00edlem je zjistit, jak hyperbola vypad\u00e1 a n\u00e1sledn\u011b dosadit do rovnice. K tomu <strong>pot\u0159ebujeme zn\u00e1t sou\u0159adnice st\u0159edu, d\u00e9lku hlavn\u00ed osy a vedlej\u0161\u00ed osy hyperboly<\/strong>. Proto\u017ee sou\u0159adnice st\u0159edu zn\u00e1me, m\u016f\u017eeme se rovnou vrhnout na v\u00fdpo\u010det d\u00e9lky osy <em><strong>a<\/strong><\/em> a poloosy <strong><em>b<\/em><\/strong>.<\/p>\n<p style=\"text-align: center\">Nejprve za\u010dneme t\u00edm, \u017ee si dopo\u010d\u00edt\u00e1me poloosu <strong><em>a<\/em><\/strong>, proto\u017ee v\u00edme, \u017ee<strong><em> a=|SA|<\/em><\/strong>. Dosad\u00edme do vzorce pro v\u00fdpo\u010det vzd\u00e1lenosti:<\/p>\n<p>$$|AB| = \\sqrt{(b_{1} \u2013 a_{1})^{2} + (b_{2} \u2013 a_{2})^{2}}$$<\/p>\n<p style=\"text-align: center\">dostaneme:<\/p>\n<p>$$a=|SA| = \\sqrt{(a_{1} \u2013 m)^{2} + (a_{2} \u2013 n)^{2}}= \\sqrt{(5\u2013 2)^{2} + (-1+1)^{2}}= \\sqrt{9}=3$$<\/p>\n<p style=\"text-align: center\">Stejn\u00fdm zp\u016fsobem m\u016f\u017eeme dopo\u010d\u00edtat excentricitu, proto\u017ee ta je rovn\u00e1 vzd\u00e1lenosti ohniska od st\u0159edu, tedy <em><strong>e=|SE|<\/strong><\/em><\/p>\n<p>$$e=|SE| = \\sqrt{(e_{1} \u2013 m)^{2} + (e_{2} \u2013 n)^{2}}= \\sqrt{(\u2013 2)^{2} + (-1+1)^{2}}= \\sqrt{25}=5$$<\/p>\n<p style=\"text-align: center\">Nyn\u00ed m\u016f\u017eeme dopo\u010d\u00edtat poloosu<strong><em> b<\/em><\/strong><em>,\u00a0<\/em>kterou vyj\u00e1d\u0159\u00edme ze vztahu<\/p>\n<p style=\"text-align: center\">$$e=\\sqrt{a^2+b^2}$$<\/p>\n<p style=\"text-align: center\">tak\u017ee dostaneme:<\/p>\n<p style=\"text-align: center\">$$b=\\sqrt{e^2-a^2}=\\sqrt{5^2-3^2}=\\sqrt{16}=4$$<\/p>\n<p style=\"text-align: center\">Pokud si um\u00edst\u011bn\u00ed st\u0159edu, vrcholu a ohniska na\u010drtneme a dotvo\u0159\u00edme z n\u011bj hyperbolu,<\/p>\n<p style=\"text-align: center\">zjist\u00edme, \u017ee je <strong>rovnob\u011b\u017en\u00e1 s osou x<\/strong>, tud\u00ed\u017e budeme dosazovat do rovnice:<\/p>\n<p style=\"text-align: center\">$$\\frac{(x-m)^2}{a^2}-\\frac{(y-n)^2}{b^2}=1$$<\/p>\n<p style=\"text-align: center\">Na\u0161e rovnice hyperboly tedy vypad\u00e1 n\u00e1sledovn\u011b:<\/p>\n<p style=\"text-align: center\">$$\\frac{(x-2)^2}{9}-\\frac{(y+1)^2}{16}=1$$<\/p>\n","protected":false},"comment_status":"closed","ping_status":"closed","template":"","class_list":["post-3216","lesson","type-lesson","status-publish","hentry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.5 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Hyperbola \u0159e\u0161en\u00fd p\u0159\u00edklad | Studijni-svet.cz<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/studijni-svet.cz\/kurzy\/courses\/matematika-k-maturite\/lekce\/hyperbola-reseny-priklad\/\" \/>\n<meta property=\"og:locale\" content=\"cs_CZ\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Hyperbola \u0159e\u0161en\u00fd p\u0159\u00edklad | Studijni-svet.cz\" \/>\n<meta property=\"og:description\" content=\"Zad\u00e1n\u00ed: Ur\u010dete st\u0159edovou rovnici a asymptoty hyperboly se st\u0159edem S[2; -1], ohniskem E[7; -1] a vrcholem A[5; -1]. &nbsp; \u0158e\u0161en\u00ed: Na\u0161\u00edm c\u00edlem je zjistit, jak hyperbola vypad\u00e1 a n\u00e1sledn\u011b dosadit do rovnice. 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