{"id":7184,"date":"2021-10-22T23:58:21","date_gmt":"2021-10-22T18:28:21","guid":{"rendered":"https:\/\/mathemerize.com\/?p=7184"},"modified":"2021-10-25T00:01:01","modified_gmt":"2021-10-24T18:31:01","slug":"find-the-normal-to-the-hyperbola-x2over-16-y2over-9-1-whose-slope-is-1","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-normal-to-the-hyperbola-x2over-16-y2over-9-1-whose-slope-is-1\/","title":{"rendered":"Find the normal to the hyperbola \\(x^2\\over 16\\) – \\(y^2\\over 9\\) = 1 whose slope is 1."},"content":{"rendered":"

Solution :<\/h2>\n

We have, \\(x^2\\over 16\\) – \\(y^2\\over 9\\) = 1<\/p>\n

Compare given equation with \\(x^2\\over a^2\\) – \\(y^2\\over b^2\\) = 1<\/p>\n

a = 4 and b = 3<\/p>\n

Since the normal to the given hyperbola whose slope is ‘m’, is\u00a0 y = mx \\(\\mp\\) \\({m(a^2+b^2)}\\over \\sqrt{a^2 – m^2b^2}\\)<\/p>\n

Hence, required equation of normal is y = x \\(\\mp\\) \\({25}\\over \\sqrt{7}\\).<\/p>\n

\n

Similar Questions<\/h3>\n

Angle between asymptotes of hyperbola xy=8 is<\/a><\/p>\n

Find the asymptotes of the hyperbola \\(2x^2 + 5xy + 2y^2 + 4x + 5y\\) = 0. Find also the general equation of all the hyperbolas having the same set of asymptotes.<\/a><\/p>\n

Find the equation of the tangent to the hyperbola \\(x^2 \u2013 4y^2\\) = 36 which is perpendicular to the line x \u2013 y + 4 = 0<\/a><\/p>\n

The eccentricity of the conjugate hyperbola to the hyperbola \\(x^2-3y^2\\) = 1 is<\/a><\/p>\n

If the foci of a hyperbola are foci of the ellipse \\(x^2\\over 25\\) + \\(y^2\\over 9\\) = 1. If the eccentricity of the hyperbola be 2, then its equation is :<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : We have, \\(x^2\\over 16\\) – \\(y^2\\over 9\\) = 1 Compare given equation with \\(x^2\\over a^2\\) – \\(y^2\\over b^2\\) = 1 a = 4 and b = 3 Since the normal to the given hyperbola whose slope is ‘m’, is\u00a0 y = mx \\(\\mp\\) \\({m(a^2+b^2)}\\over \\sqrt{a^2 – m^2b^2}\\) Hence, required equation of normal is …<\/p>\n

Find the normal to the hyperbola \\(x^2\\over 16\\) – \\(y^2\\over 9\\) = 1 whose slope is 1.<\/span> Read More »<\/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":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[51,43],"tags":[],"yoast_head":"\nFind the normal to the hyperbola \\(x^2\\over 16\\) - \\(y^2\\over 9\\) = 1 whose slope is 1.<\/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\/find-the-normal-to-the-hyperbola-x2over-16-y2over-9-1-whose-slope-is-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Find the normal to the hyperbola \\(x^2\\over 16\\) - \\(y^2\\over 9\\) = 1 whose slope is 1.\" \/>\n<meta property=\"og:description\" content=\"Solution : We have, (x^2over 16) – (y^2over 9) = 1 Compare given equation with (x^2over a^2) – (y^2over b^2) = 1 a = 4 and b = 3 Since the normal to the given hyperbola whose slope is ‘m’, is\u00a0 y = mx (mp) ({m(a^2+b^2)}over sqrt{a^2 – m^2b^2}) Hence, required equation of normal is … Find the normal to the hyperbola (x^2over 16) – (y^2over 9) = 1 whose slope is 1. 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