Solution : Let m be the slope of the tangent, since the tangent is perpendicular to the line x – y = 0 \(\therefore\) 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 \(\therefore\); \(a^2\) […]
Solution : Equation of the conjugate hyperbola to the hyperbola \(x^2-3y^2\) = 1 is \(-x^2-3y^2\) = 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 Similar Questions Angle between asymptotes of hyperbola xy=8 is Find the
The eccentricity of the conjugate hyperbola to the hyperbola \(x^2-3y^2\) = 1 is Read More »
Solution : For ellipse e = \(4\over 5\), so foci = (\(\pm\)4, 0) 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 Similar Questions Find the equation of the ellipse
Solution : f(x) = \(e^{x-[x]+|cos\pi x|+|cos2\pi x|+ ….. + |cosn\pi x|}\) Period of x – [x] = 1 Period of \(|cos\pi x|\) = 1 Period of \(|cos2\pi x|\) = \(1\over 2\) ………………………………. Period of \(|cosn\pi x|\) = \(1\over n\) So period of f(x) will be L.C.M of all period = 1. Similar Questions If y
Solution : Given f(x) = \(log_a(x + \sqrt{(x^2+1)})\) f'(x) = \(log_ae\over {\sqrt{1+x^2}}\) > 0 which is strictly increasing functions. Thus, f(x) is injective, given that f(x) is onto. Hence the given function f(x) is invertible. Interchanging x & y \(\implies\) \(log_a(y + \sqrt{(y^2+1)})\) = x \(\implies\) \(y + \sqrt{(y^2+1)}\) = \(a^x\) ……..(1) and \(\sqrt{(y^2+1)}\) –
Solution : Now 1 \(\le\) \(16sin^2x\) + 1) \(\le\) 17 0 \(\le\) \(log_2(16sin^2x+1)\) \(\le\) \(log_217\) 2 – \(log_217\) \(\le\) 2 – \(log_2(16sin^2x+1)\) \(\le\) 2 Now consider 0 < 2 – \(log_2(16sin^2x+1)\) \(\le\) 2 -\(\infty\) < \(log_{\sqrt{2}}(2-log_2(16sin^2x+1))\) \(\le\) \(log_{\sqrt{2}}2\) = 2 the range is (-\(\infty\), 2] Similar Questions If y = 2[x] + 3 & y
Find the range of the function \(log_{\sqrt{2}}(2-log_2(16sin^2x+1))\) Read More »
Solution : Let m be the slope of the tangent, since the tangent is perpendicular to the line y + 2x = 4 \(\therefore\) mx – 2 = -1 \(\implies\) m = \(1\over 2\) Since \(3x^2+4y^2\) = 12 or \(x^2\over 4\) + \(y^2\over 3\) = 1 Comparing this with \(x^2\over a^2\) + \(y^2\over b^2\) =
Solution : \(\because\) Equation of ellipse is \(9x^2 + 16y^2\) = 144 or \(x^2\over 16\) + \({(y-3)}^2\over 9\) = 1 comparing this with \(x^2\over a^2\) + \(y^2\over b^2\) = 1 then we get \(a^2\) = 16 and \(b^2\) = 9 and comparing the line y = x + k with y = mx + c
For what value of k does the line y = x + k touches the ellipse \(9x^2 + 16y^2\) = 144. Read More »
Solution : Here S = (2, 3) & S’ is (-2, 3) and b = \(\sqrt{5}\) \(\implies\) SS’ = 4 = 2ae \(\implies\) ae = 2 but \(b^2\) = \(a^2(1-e^2)\) \(\implies\) 5 = \(a^2\) – 4 \(\implies\) a = 3 Hence the equation to major axis is y = 3. Centre of ellipse is midpoint
Solution : Family of circles is \(x^2 + y^2 – 2x – 4y + 1\) + \(\lambda\)(\(x^2 + y^2 – 1\)) = 0 (1 + \(\lambda\))\(x^2\) + (1 + \(\lambda\))\(y^2\) – 2x – 4y + (1 – \(\lambda\))) = 0 \(x^2 + y^2 – {2\over {1 + \lambda}} x – {4\over {1 + \lambda}}y +
