If the eccentricity of the hyperbola x^2 - y^ | vedclass

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(B) For the hyperbola $\frac{x^2}{5} - \frac{y^2}{5 \cos^2 \alpha} = 1$,the eccentricity $e_1$ is given by $e_1^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{5

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$\pi / 4$

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(B) For the hyperbola $\frac{x^2}{5} - \frac{y^2}{5 \cos^2 \alpha} = 1$,the eccentricity $e_1$ is given by $e_1^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{5 \cos^2 \alpha}{5} = 1 + \cos^2 \alpha$.For the ellipse $\frac{x^2}{25 \cos^2 \alpha} + \frac{y^2}{25} = 1$,the eccentricity $e_2$ is given by $e_2^2 = 1 - \frac{a^2}{b^2} = 1 - \frac{25 \cos^2 \alpha}{25} = 1 - \cos^2 \alpha = \sin^2 \alpha$.Given $e_1 = \sqrt{3} e_2$,we have $e_1^2 = 3 e_2^2$.Substituting the values: $1 + \cos^2 \alpha = 3 \sin^2 \alpha$.Using $\cos^2 \alpha = 1 - \sin^2 \alpha$,we get $1 + (1 - \sin^2 \alpha) = 3 \sin^2 \alpha$.$2 = 4 \sin^2 \alpha \Rightarrow \sin^2 \alpha = \frac{1}{2}$.Thus,$\sin \alpha = \frac{1}{\sqrt{2}}$,which implies $\alpha = \frac{\pi}{4}$.

If the eccentricity of the hyperbola $x^2 - y^2 \sec^2 \alpha = 5$ is $\sqrt{3}$ times the eccentricity of the ellipse $x^2 \sec^2 \alpha + y^2 = 25$,then a value of $\alpha$ is:

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