Example 1 : <\/span> 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 : <\/p>\n Solution : <\/span>For ellipse e = \\(4\\over 5\\), so foci = (\\(\\pm\\)4, 0) Example 2 : <\/span> The eccentricity of the conjugate hyperbola to the hyperbola \\(x^2-3y^2\\) = 1 is-<\/p>\n Solution : <\/span>Equation of the conjugate hyperbola to the hyperbola \\(x^2-3y^2\\) = 1 is Example 3 : <\/span> Find the equation of 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 Example 4 : <\/span> 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.<\/p>\n Solution : <\/span>Let \\(2x^2 + 5xy + 2y^2 + 4x + 5y + k\\) = 0 be asymptotes. This will represent two straight line Practice these given hyperbola examples to test your knowledge on concepts of hyperbola.<\/p>\n \n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":" Here you will learn some hyperbola examples for better understanding of hyperbola concepts. Example 1 : 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 : Solution : For ellipse e = \\(4\\over 5\\), …<\/p>\n
for hyperbola e = 2, so a = \\(ae\\over e\\) = \\(4\\over 2\\) = 2, b = \\(2\\sqrt{4-1}\\) = 2\\(\\sqrt{3}\\)
Hence the equation of the hyperbola is \\(x^2\\over 4\\) – \\(y^2\\over 12\\) = 1<\/p>\n
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\\(-x^2-3y^2\\) = 1 \\(\\implies\\) \\(-x^2\\over 1\\) + \\(y^2\\over {1\/3}\\) = 1
Here \\(a^2\\) = 1, \\(b^2\\) = \\(1\\over 3\\)
\\(\\therefore\\) eccentricity e = \\(\\sqrt{1 + a^2\/b^2}\\) = \\(\\sqrt{1+3}\\) = 2\n <\/p>
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\\(\\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
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so \\(abc + 2fgh – af^2 – bg^2 – ch^2\\) = 0 \\(\\implies\\) 4k + 25 – \\(25\\over 2\\) – 8 – \\(25\\over 4\\)k = 0
\\(\\implies\\) k = 2
\\(\\implies\\) \\(2x^2 + 5xy + 2y^2 + 4x + 5y + 2\\) = 0 are asymptotes
\\(\\implies\\) (2x+y+2) = 0 and (x+2y+1) = 0 are asymptotes
and \\(2x^2 + 5xy + 2y^2 + 4x + 5y + c\\) = 0 is general equation of hyperbola.<\/p>\n
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