{"id":2544,"date":"2021-07-12T13:00:18","date_gmt":"2021-07-12T13:00:18","guid":{"rendered":"https:\/\/mathemerize.com\/?p=2544"},"modified":"2021-11-24T23:31:39","modified_gmt":"2021-11-24T18:01:39","slug":"equation-of-conjugate-hyperbola","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-conjugate-hyperbola\/","title":{"rendered":"Equation of Conjugate Hyperbola"},"content":{"rendered":"
Here you will learn what is conjugate hyperbola, equation of conjugate hyperbola and basic definitions of like eccentricity and latus rectum.<\/p>\n
Let’s begin –<\/p>\n
The hyperbola whose transverse & conjugate axes are respectively the conjugate and transverse axes of given hyperbola is called the conjugate hyperbola of given hyperbola.\u00a0<\/p>\n
The equation of the conjugate hyperbola<\/strong> of the hyperbola \\(x^2\\over a^2\\) – \\(y^2\\over b^2\\) = 1 is<\/p>\n -\\(x^2\\over a^2\\) + \\(y^2\\over b^2\\) = 1.<\/p>\n<\/blockquote>\n The graph of the conjugate hyperbola is shown in figure.<\/p>\n The eccentricity of the conjugate hyperbola<\/strong> is given by \\(a^2\\) = \\(b^2(e^2-1)\\) and the length of latus rectum<\/strong> is \\(2a^2\\over b\\).<\/p>\n The equation of the conjugate hyperbola is -\\(x^2\\over a^2\\) + \\(y^2\\over b^2\\) = 1.<\/strong><\/p>\n (a) Centre (0,0).<\/p>\n (b) Vertices (0,b) & (0,-b)<\/p>\n (c) foci \\((0, \\pm ae)\\)<\/p>\n (d) Length of\u00a0 transverse axis is 2b<\/p>\n (e) Length of\u00a0 conjugate axis is 2a<\/p>\n (f) Equation of\u00a0 directrices is x = \\(\\pm {b\\over e}\\)<\/p>\n (g) Eccentricity is given by \\(a^2\\) = \\(b^2(e^2-1)\\)<\/p>\n (h) Length of latus rectum is \\(2a^2\\over b\\).<\/p>\n (i) Equation of the transverse axis is x = 0.<\/p>\n (j) Equation of the conjugate axis is y = 0.<\/p>\n\n\n Example : <\/span> Find the eccentricity of the conjugate hyperbola to the hyperbola \\(x^2 – 3y^2\\) = 1.\n<\/p> Solution : <\/span>Equation of the conjugate hyperbola to the hyperbola \\(x^2 – 3y^2\\) = 1 is \\(-x^2 + 3y^2\\) = 1\n\\(\\implies\\) -\\(x^2\\over 1\\) + \\(y^2\\over 1\/3\\) = 1. Hope you learnt conjugate hyperbola, learn more concepts of hyperbola and practice more questions to get ahead in the competition. Good luck!<\/p>\n\n\n\n
Conjugate Hyperbola & Basic Definitions :<\/h2>\n
\nHere \\(a^2\\) = 1, \\(b^2\\) = 1\/3
\nTherefore eccentricity is e = \\(\\sqrt{1 + a^2\/b^2}\\) = 2.<\/p>\n\n\n