The eccentricity of the hyperbola x^2 - y^2 = | vedclass

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(A) The given equation of the hyperbola is $x^2 - y^2 = 25$.Dividing both sides by $25$,we get $\frac{x^2}{25} - \frac{y^2}{25} = 1$.This is in the st

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$\sqrt{2}$

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(A) The given equation of the hyperbola is $x^2 - y^2 = 25$.Dividing both sides by $25$,we get $\frac{x^2}{25} - \frac{y^2}{25} = 1$.This is in the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = 25$ and $b^2 = 25$.Thus,$a = 5$ and $b = 5$.The eccentricity $e$ of a hyperbola is given by the formula $e = \sqrt{1 + \frac{b^2}{a^2}}$.Substituting the values,$e = \sqrt{1 + \frac{25}{25}} = \sqrt{1 + 1} = \sqrt{2}$.

The eccentricity of the hyperbola $x^2 - y^2 = 25$ is

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