{"id":3282,"date":"2021-07-21T17:02:02","date_gmt":"2021-07-21T17:02:02","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3282"},"modified":"2021-11-24T23:27:04","modified_gmt":"2021-11-24T17:57:04","slug":"equation-of-normal-to-hyperbola-in-all-forms","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-normal-to-hyperbola-in-all-forms\/","title":{"rendered":"Equation of Normal to Hyperbola in all Forms"},"content":{"rendered":"
The equation of normal to the given hyperbola at the point P(\\(x_1, y_1\\))<\/strong> is<\/p>\n \\(a^2x\\over x_1\\) + \\(b^2y\\over y_1\\) = \\(a^2+b^2\\) = \\(a^2e^2\\)<\/p>\n<\/blockquote>\n\n\n Example : <\/span> Find the equation of normal to the hyperbola \\(x^2\\over 25\\) – \\(y^2\\over 16\\) = 1 at (5, 1).<\/p>\n Solution : <\/span>We have, \\(x^2\\over 25\\) – \\(y^2\\over 16\\) = 1 The normal to the given hyperbola whose slope is ‘m’, is<\/p>\n y = mx \\(\\mp\\) \\({m(a^2+b^2)}\\over \\sqrt{a^2 – m^2b^2}\\)<\/p>\n Foot of normal are (\\(\\mp\\)\\({a^2}\\over \\sqrt{a^2 – m^2b^2}\\), \\(\\mp\\)\\({mb^2}\\over \\sqrt{a^2 – m^2b^2}\\)).<\/p>\n<\/blockquote>\n\n\n Example : <\/span> Find the normal to the hyperbola \\(x^2\\over 16\\) – \\(y^2\\over 9\\) = 1 whose slope is 1.<\/p>\n Solution : <\/span>We have, \\(x^2\\over 16\\) – \\(y^2\\over 9\\) = 1 The equation of normal to the given hyperbola at its point (asec\\(\\theta\\), btan\\(\\theta\\))<\/strong>, is<\/p>\n \\(ax\\over sec\\theta \\) + \\(by\\over tan\\theta \\) = \\(a^2+b^2\\) = \\(a^2e^2\\)<\/p>\n<\/blockquote>\n\n\n Example : <\/span> Line \\(xcos\\alpha\\) + \\(ysin\\alpha\\) = p is a normal to the hyperbola \\(x^2\\over a^2\\) – \\(y^2\\over b^2\\) = 1, if<\/p>\n Solution : <\/span>We have, \\(x^2\\over a^2\\) – \\(y^2\\over b^2\\) = 1\n
\nCompare given equation with \\(x^2\\over a^2\\) – \\(y^2\\over b^2\\) = 1
\na = 5 and b = 4
\nHence, required equation of normal is \\(25x\\over 5\\) + \\(16y\\over 1\\) = 41
\n= 5x + 16y = 41
<\/p>\n\n\n(b) Slope form :<\/h3>\n
\n
\nCompare given equation with \\(x^2\\over a^2\\) – \\(y^2\\over b^2\\) = 1
\na = 4 and b = 3
\nHence, required equation of normal is y = x \\(\\mp\\) \\({25}\\over \\sqrt{7}\\)
\n<\/p>\n\n\n(c) Parametric form :<\/h3>\n
\n
\nThe normal to hyperbola is \\(ax\\over sec\\theta \\) + \\(by\\over tan\\theta \\) = \\(a^2+b^2\\)
\ncomparing it with the given line equation
\n\\(acos\\theta\\over cos\\alpha\\) = \\(bcot\\theta\\over sin\\alpha\\) = \\(a^2 + b^2\\over p\\)
\n\\(\\implies\\) \\(sec\\theta\\) = \\(ap\\over cos\\alpha(a^2+b^2)\\), \\(tan\\theta\\) = \\(bp\\over sin\\alpha(a^2+b^2)\\)
\nEliminating \\(\\theta\\), we get
\n\\(a^2sec^2\\alpha\\) – \\(b^2cosec^2\\alpha\\) = \\((a^2 + b^2)^2)\\over p^2\\)
<\/p>\n\n\n\n