{"id":28989,"date":"2021-05-05T13:00:11","date_gmt":"2021-05-05T13:00:11","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=28989"},"modified":"2024-09-22T01:53:25","modified_gmt":"2024-09-22T01:53:25","slug":"equation-of-hyperbolas-and-its-deduction","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/en\/equation-of-hyperbolas-and-its-deduction\/","title":{"rendered":"Equation of Hyperbolas and Its Deduction"},"content":{"rendered":"<p><center><\/p>\n<h1>Equation of Hyperbolas and Its Deduction<\/h1>\n<p><em><strong>Summary:<\/strong><br \/>\nIn this lesson, we will explore the geometric definition of the hyperbola, contrast it with the ellipse, and deduce its general and canonical equations.<br \/>\n   <\/em><\/p>\n<p>   <strong>Learning Objectives:<\/strong><br \/>\n   At the end of this lesson, the student will be able to:<\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Define<\/strong> geometrically what a hyperbola is.<\/li>\n<li><strong>Deduce<\/strong> the general and canonical equations of hyperbolas based on their geometric definition.<\/li>\n<li><strong>Identify<\/strong> the differences between ellipses and hyperbolas in terms of focal distances.<\/li>\n<\/ol>\n<p>   <strong>CONTENT INDEX<\/strong><br \/>\n<a href=\"#1\">Geometric Definition of Hyperbola<\/a><br \/>\n<a href=\"#2\">Deduction of the Equation of Hyperbolas<\/a><br \/>\n<a href=\"#3\">General Equation of Hyperbolas<\/a><br \/>\n<a href=\"#4\">Canonical Equation of Hyperbolas<\/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>Geometric Definition of Hyperbola<\/h2>\n<p style=\"text-align: justify;\">Previously, we reviewed the equation of ellipses and circles and discovered that they take the form <span class=\"katex-eq\" data-katex-display=\"false\">ax^2 + bx + cy^2 + dy + e = 0<\/span>, where <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">b<\/span> are two nonzero quantities with the same sign. We mentioned that if <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">b<\/span> have opposite signs, then instead of an ellipse, we would have a hyperbola. We said nothing more about these curves, and now we will fill that gap. We will complete our study by defining what a hyperbola is geometrically, and from this, we will obtain the general and canonical equations of hyperbolas.<\/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>On the one hand, we have that an ellipse is defined<\/strong><\/a> as the set of all points such that the sum of their distances from two other points, called foci, is always the same value. Similarly, and in contrast, a hyperbola is defined as the collection of all points such that the absolute value of the difference between their distances from the focal points is always the same value.<\/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=\"Geometric definition of Hyperbola\" 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=\"Geometric definition of Hyperbola\" class=\" aligncenter lazyload\" width=\"493\" height=\"340\" \/><\/noscript><\/p>\n<p style=\"text-align: justify;\">In other words, the relationship<\/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;\">is satisfied, where <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> is any fixed real number.<\/p>\n<p style=\"text-align: justify;\">This actually produces two equations, namely: <span class=\"katex-eq\" data-katex-display=\"false\">d(f_1,P) - d(f_2,P) = 2a<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">d(f_2,P) - d(f_1,P) = 2a<\/span>, one for each branch of the hyperbola.<\/p>\n<p><a name=\"2\"><\/a>  <\/p>\n<h2>Deduction of the Equation of Hyperbolas<\/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>From the geometric definition, it is possible to obtain<\/strong><\/a> the algebraic representation of hyperbolas. To do this, we will start from the simplest case, and from there, we will extend the generalizations. Our reasoning will be carried out for one branch of the hyperbola; the reasoning for the other branch is completely analogous.<\/p>\n<h3>Deduction of the Simplified Form<\/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>Consider two focal points<\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">f_1 = (-c,0)<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">f_2 = (c,0).<\/span> The point <span class=\"katex-eq\" data-katex-display=\"false\">p = (x,y)<\/span> will be on the hyperbola if<\/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=\"Hyperbola centered at origin\" 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=\"Hyperbola centered at origin\" 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;\">From here, the following reasoning follows:<\/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>; equation of the hyperbolas<\/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>; expanding the squares<\/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>; redistributing terms<\/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>; squaring both sides<\/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>; canceling like terms<\/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>; redistributing like terms<\/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>; simplifying like terms<\/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>; simplifying like terms<\/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>; squaring both sides<\/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>; operating parentheses<\/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>; canceling like terms<\/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>; regrouping terms<\/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>; regrouping terms<\/td>\n<\/tr>\n<tr><\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\">For this final expression, as with ellipses, we take <span class=\"katex-eq\" data-katex-display=\"false\">b^2=c^2-a^2<\/span> and arrive at the equation for ellipses:<\/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>General Equation of Hyperbolas<\/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>To obtain the general equation<\/strong><\/a> of hyperbolas, we simply take the one we just obtained and apply the position transformations:<\/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;\">and with this, we automatically obtain the general equation of ellipses with center <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>Canonical Equation of Hyperbolas<\/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>And if we now take the general equation<\/strong><\/a> of ellipses and expand it, we arrive at the canonical expression:<\/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>; General equation of hyperbolas<\/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>; solving squares and multiplying everything by <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>; solving parentheses<\/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>; Grouping like terms<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\">This last expression is of the form <span class=\"katex-eq\" data-katex-display=\"false\">Ax^2+Bx + Cy^2 + Dy + E = 0,<\/span> where <span class=\"katex-eq\" data-katex-display=\"false\">A<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">C<\/span> are always nonzero and have opposite signs, as we anticipated when studying ellipses.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Equation of Hyperbolas and Its Deduction Summary: In this lesson, we will explore the geometric definition of the hyperbola, contrast it with the ellipse, and deduce its general and canonical equations. Learning Objectives: At the end of this lesson, the student will be able to: Define geometrically what a hyperbola is. Deduce the general and [&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":9,"footnotes":""},"categories":[583,567],"tags":[],"class_list":["post-28989","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-algebra-and-geometry","category-mathematics"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Equation of Hyperbolas and Its Deduction - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Learn everything about hyperbolas: geometric definition, deduction of their general and canonical equations, differences with ellipses, and how to solve equations step by step. 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