Solution : The locus of the intersection of tangents which are at right angles is known as director circle of the hyperbola. The equation to the director circle is : \(x^2+y^2\) = \(a^2-b^2\) If \(b^2\) < \(a^2\), this circle is real ; If \(b^2\) = \(a^2\) the radius of the circle is zero & it โฆ
What is the Equation of Director Circle of Hyperbola ? Read More ยป
Solution : We have, \(x^2\over 16\) โ \(y^2\over 9\) = 1 Compare given equation with \(x^2\over a^2\) โ \(y^2\over b^2\) = 1 a = 4 and b = 3 Since the normal to the given hyperbola whose slope is โmโ, isย y = mx \(\mp\) \({m(a^2+b^2)}\over \sqrt{a^2 โ m^2b^2}\) Hence, required equation of normal is โฆ
Solution : Since given hyperbola xy = 8 is rectangular hyperbola. And eccentricity of rectangular hyperbola is \(\sqrt{2}\) Angle between asymptotes of hyperbola is \(2sec^{-1}(e)\) \(\implies\) \(\theta\) = \(2sec^{-1}(\sqrt{2})\) \(\implies\) \(\theta\) = \(2sec^{-1}(sec 45)\) \(\implies\) \(\theta\) = 2(45) = 90 Similar Questions Find the normal to the hyperbola \(x^2\over 16\) โ \(y^2\over 9\) = 1 โฆ
Solution : Let \(2x^2 + 5xy + 2y^2 + 4x + 5y + k\) = 0 be asymptotes. This will represent two straight line so \(abc + 2fgh โ af^2 โ bg^2 โ ch^2\) = 0 \(\implies\) 4k + 25 โ \(25\over 2\) โ 8 โ \(25\over 4\)k = 0 \(\implies\) k = 2 \(\implies\) โฆ
Solution : Let m be the slope of the tangent, since the tangent is perpendicular to the line x โ y = 0 \(\therefore\)ย m\(\times\)1 = -1 \(\implies\) m = -1 Since \(x^2-4y^2\) = 36 or \(x^2\over 36\) โ \(y^2\over 9\) = 1 Comparing this with \(x^2\over a^2\) โ \(y^2\over b^2\) = 1 \(\therefore\); \(a^2\) โฆ
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, \(b^2\) = \(1\over 3\) \(\therefore\)ย eccentricity e = \(\sqrt{1 + a^2/b^2}\) = \(\sqrt{1+3}\) = 2 Similar Questions Angle between asymptotes of hyperbola xy=8 is Find the โฆ
The eccentricity of the conjugate hyperbola to the hyperbola \(x^2-3y^2\) = 1 is Read More ยป
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 \(x^2\over 4\) โ \(y^2\over 12\) = 1 Similar Questions Find the equation of the ellipse โฆ
