{"id":8065,"date":"2021-11-13T19:40:44","date_gmt":"2021-11-13T14:10:44","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8065"},"modified":"2021-11-13T21:48:27","modified_gmt":"2021-11-13T16:18:27","slug":"check-the-orthogonality-of-the-curves-y2-x-and-x2-y","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/check-the-orthogonality-of-the-curves-y2-x-and-x2-y\/","title":{"rendered":"Check the orthogonality of the curves \\(y^2\\) = x and \\(x^2\\) = y."},"content":{"rendered":"<h2>Solution :<\/h2>\n<p>Solving the curves simultaneously we get points of intersection as (1, 1) and (0, 0).<\/p>\n<p>At (1, 1) for first curve \\(2y({dy\\over dx})_1\\) = 1\u00a0 \\(\\implies\\)\u00a0 \\(m_1\\) = \\(1\\over 2\\)<\/p>\n<p>&amp; for second curve 2x = \\(({dy\\over dx})_2\\) \\(\\implies\\)\u00a0 \\(m_2\\) = 2<\/p>\n<p>\\(m_1m_2\\) = -1 at (1, 1).<\/p>\n<p>But at (0, 0) clearly x-axis &amp; y-axis are their respective tangents hence they are orthogonal at (0, 0) but not at (1, 1). Hence these curves are not said to be orthogonal.<\/p>\n<hr \/>\n<h3 style=\"text-align: center;\">Similar Questions<\/h3>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/find-the-equations-of-the-tangent-and-the-normal-at-the-point-t-on-the-curve-x-a-sin3-t-y-b-cos3-t\/\" aria-label=\"\u201cFind the equations of the tangent and the normal at the point \u2018t\u2019 on the curve x = \\(a sin^3 t\\), y = \\(b cos^3 t\\).\u201d (Edit)\">Find the equations of the tangent and the normal at the point \u2018t\u2019 on the curve x = \\(a sin^3 t\\), y = \\(b cos^3 t\\).<\/a><\/p>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/find-the-equation-of-the-normal-to-the-curve-y-2x2-3-sin-x-at-x-0\/\" aria-label=\"\u201cFind the equation of the normal to the curve y = \\(2x^2 + 3 sin x\\) at x = 0.\u201d (Edit)\">Find the equation of the normal to the curve y = \\(2x^2 + 3 sin x\\) at x = 0.<\/a><\/p>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/find-the-angle-between-the-curves-xy-6-and-x2-y-12\/\" aria-label=\"\u201cFind the angle between the curves xy = 6 and \\(x^2 y\\) =12.\u201d (Edit)\">Find the angle between the curves xy = 6 and \\(x^2 y\\) =12.<\/a><\/p>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/find-the-equation-of-the-tangent-to-curve-y-5x2-6x-7-at-the-point-1-2-35-4\/\" aria-label=\"\u201cFind the equation of the tangent to curve y = \\(-5x^2 + 6x + 7\\)\u00a0 at the point (1\/2, 35\/4).\u201d (Edit)\">Find the equation of the tangent to curve y = \\(-5x^2 + 6x + 7\\)\u00a0 at the point (1\/2, 35\/4).<\/a><\/p>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/the-angle-of-intersection-between-the-curve-x2-32y-and-y2-4x-at-point-16-8-is\/\" aria-label=\"\u201cThe angle of intersection between the curve \\(x^2\\) = 32y and \\(y^2\\) = 4x at point (16, 8) is\u201d (Edit)\">The angle of intersection between the curve \\(x^2\\) = 32y and \\(y^2\\) = 4x at point (16, 8) is<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Solution : Solving the curves simultaneously we get points of intersection as (1, 1) and (0, 0). At (1, 1) for first curve \\(2y({dy\\over dx})_1\\) = 1\u00a0 \\(\\implies\\)\u00a0 \\(m_1\\) = \\(1\\over 2\\) &amp; for second curve 2x = \\(({dy\\over dx})_2\\) \\(\\implies\\)\u00a0 \\(m_2\\) = 2 \\(m_1m_2\\) = -1 at (1, 1). But at (0, 0) clearly &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathemerize.com\/check-the-orthogonality-of-the-curves-y2-x-and-x2-y\/\"> <span class=\"screen-reader-text\">Check the orthogonality of the curves \\(y^2\\) = x and \\(x^2\\) = y.<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[66,43],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Check the orthogonality of the curves \\(y^2\\) = x and \\(x^2\\) = y.<\/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:\/\/mathemerize.com\/check-the-orthogonality-of-the-curves-y2-x-and-x2-y\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Check the orthogonality of the curves \\(y^2\\) = x and \\(x^2\\) = y.\" \/>\n<meta property=\"og:description\" content=\"Solution : Solving the curves simultaneously we get points of intersection as (1, 1) and (0, 0). At (1, 1) for first curve (2y({dyover dx})_1) = 1\u00a0 (implies)\u00a0 (m_1) = (1over 2) &amp; for second curve 2x = (({dyover dx})_2) (implies)\u00a0 (m_2) = 2 (m_1m_2) = -1 at (1, 1). But at (0, 0) clearly &hellip; Check the orthogonality of the curves (y^2) = x and (x^2) = y. 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At (1, 1) for first curve (2y({dyover dx})_1) = 1\u00a0 (implies)\u00a0 (m_1) = (1over 2) &amp; for second curve 2x = (({dyover dx})_2) (implies)\u00a0 (m_2) = 2 (m_1m_2) = -1 at (1, 1). But at (0, 0) clearly &hellip; Check the orthogonality of the curves (y^2) = x and (x^2) = y. 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