{"id":3280,"date":"2021-07-21T17:02:12","date_gmt":"2021-07-21T17:02:12","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3280"},"modified":"2021-11-24T23:25:54","modified_gmt":"2021-11-24T17:55:54","slug":"equation-of-tangent-to-hyperbola-in-all-forms","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-tangent-to-hyperbola-in-all-forms\/","title":{"rendered":"Equation of Tangent to Hyperbola in all Forms"},"content":{"rendered":"
The equation of tangent to the given hyperbola at its point (\\(x_1, y_1\\))<\/strong> is<\/p>\n \\(x{x_1}\\over a^2\\) – \\(y{y_1}\\over b^2\\) = 1<\/p>\n<\/blockquote>\n\n\n Example : <\/span> Find the equation of tangent to the hyperbola \\(16x^2\\) – \\(9y^2\\) = 144 at (5, 16\/3).<\/p>\n Solution : <\/span>We have, \\(16x^2\\) – \\(9y^2\\) = 144 The equation of tangent to the given hyperbola whose slope is ‘m’, is<\/p>\n y = mx \\(\\pm\\) \\(\\sqrt{a^2m^2 – b^2}\\)<\/p>\n The Point of contact are (\\({\\mp} a^2m\\over \\sqrt{a^2m^2 – b^2}\\), \\({\\mp} b^2\\over \\sqrt{a^2m^2 – b^2}\\))<\/p>\n Note that there are two parallel tangents having the same slope m.<\/p>\n<\/blockquote>\n\n\n Example : <\/span> Find the tangent to the hyperbola \\(x^2\\over 25\\) – \\(y^2\\over 16\\) = 1 whose slope is 1.<\/p>\n Solution : <\/span>We have, \\(x^2\\over 25\\) – \\(y^2\\over 16\\) = 1 The equation of tangent to the given hyperbola at the point (asec\\(\\theta\\), btan\\(\\theta\\))<\/strong>, is<\/p>\n \\(xsec\\theta\\over a\\) – \\(ytan\\theta\\over b\\) = 1<\/p>\n<\/blockquote>\n\n\n Example : <\/span> Find the tangent to the hyperbola \\(x^2 – 4y^2\\) = 36 which is perpendicular to the line x – y + 4 = 0<\/p>\n Solution : <\/span>Let m be the slope of the tangent, since the tangent is perpendicular to the line x – y = 0\n
\n\\(\\implies\\) \\(x^2\\over 9\\) – \\(y^2\\over 16\\) = 1
\nCompare given equation with \\(x^2\\over a^2\\) – \\(y^2\\over b^2\\) = 1
\na = 3 and b = 16
\nHence, required equation of tangent is \\(5x\\over 9\\) – \\(16\/3\\over 16\\) = 1
\n= \\(5x\\over 9\\) – \\(y\\over 3\\) = 1
<\/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 = 5 and b = 4
\nHence, required equation of normal is y = x \\(\\pm\\) \\(\\sqrt{9}\\)
\n\\(\\implies\\) y = x \\(\\pm\\) 3
<\/p>\n\n\n(c) Parametric form :<\/h3>\n
\n
\\(\\therefore\\) m\\(\\times\\)1 = -1 \\(\\implies\\) m = -1
Since \\(x^2-4y^2\\) = 36 or \\(x^2\\over 36\\) – \\(y^2\\over 9\\) = 1
Comparing this with \\(x^2\\over a^2\\) – \\(y^2\\over b^2\\) = 1
\\(\\therefore\\) \\(a^2\\) = 36 and \\(b^2\\) = 9
So the equation of the tangent are y = -1x \\(\\pm\\) \\(\\sqrt{36\\times {-1}^2 – 9}\\)
\\(\\implies\\) y = x \\(\\pm\\) \\(\\sqrt{27}\\) \\(\\implies\\) x + y \\(\\pm\\) 3\\(\\sqrt{3}\\) = 0<\/p>\n\n\n\n