If e_{1} and e_{2} are the eccentricities of | vedclass

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(B) The given equation of the hyperbola is $3x^{2} - 3y^{2} = 25$,which can be written as $x^{2} - y^{2} = \frac{25}{3}$.Comparing this with the stand

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$e_{1}^{2} + e_{2}^{2} = 4$

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(B) The given equation of the hyperbola is $3x^{2} - 3y^{2} = 25$,which can be written as $x^{2} - y^{2} = \frac{25}{3}$.Comparing this with the standard form $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$,we get $a^{2} = \frac{25}{3}$ and $b^{2} = \frac{25}{3}$.The eccentricity $e_{1}$ is given by $e_{1} = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{25/3}{25/3}} = \sqrt{1 + 1} = \sqrt{2}$.The equation of the conjugate hyperbola is $\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$,which is $-x^{2} + y^{2} = \frac{25}{3}$.The eccentricity $e_{2}$ of the conjugate hyperbola is given by $e_{2} = \sqrt{1 + \frac{a^{2}}{b^{2}}} = \sqrt{1 + \frac{25/3}{25/3}} = \sqrt{1 + 1} = \sqrt{2}$.Therefore,$e_{1}^{2} + e_{2}^{2} = (\sqrt{2})^{2} + (\sqrt{2})^{2} = 2 + 2 = 4$.

If $e_{1}$ and $e_{2}$ are the eccentricities of a hyperbola $3x^{2} - 3y^{2} = 25$ and its conjugate,then

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