The eccentricity of the hyperbola frac{sqrt{1 | vedclass

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(B) The given equation is $\frac{\sqrt{1999}}{3}(x^2 - y^2) = 1$.This can be rewritten as $x^2 - y^2 = \frac{3}{\sqrt{1999}}$.Dividing both sides by $

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$\sqrt{2}$

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(B) The given equation is $\frac{\sqrt{1999}}{3}(x^2 - y^2) = 1$.This can be rewritten as $x^2 - y^2 = \frac{3}{\sqrt{1999}}$.Dividing both sides by $\frac{3}{\sqrt{1999}}$,we get $\frac{x^2}{\frac{3}{\sqrt{1999}}} - \frac{y^2}{\frac{3}{\sqrt{1999}}} = 1$.This is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = b^2 = \frac{3}{\sqrt{1999}}$.Since $a = b$,the hyperbola is a rectangular hyperbola.The eccentricity $e$ of a hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.For a rectangular hyperbola,$a = b$,so $e = \sqrt{1 + 1} = \sqrt{2}$.

The eccentricity of the hyperbola $\frac{\sqrt{1999}}{3}(x^2 - y^2) = 1$ is

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