If e_1 is the eccentricity of the hyperbola x | vedclass

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(C) The given equations are in parametric form $x = a \sec 	heta, y = b 	an 	heta$,which represents the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2

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$\frac{1}{2}$

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(C) The given equations are in parametric form $x = a \sec 	heta, y = b 	an 	heta$,which represents the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2} = \frac{a^2 + b^2}{a^2}$. For the first hyperbola,$a = 1$ and $b = \sqrt{2}$,so $e_1^2 = 1 + \frac{(\sqrt{2})^2}{1^2} = 1 + 2 = 3$. For the second hyperbola,$a = \sqrt{2}$ and $b = 1$,so $e_2^2 = 1 + \frac{1^2}{(\sqrt{2})^2} = 1 + \frac{1}{2} = \frac{3}{2}$. Therefore,$\frac{e_2^2}{e_1^2} = \frac{3/2}{3} = \frac{1}{2}$.

If $e_1$ is the eccentricity of the hyperbola $x = \sec \theta, y = \sqrt{2} \tan \theta$ and $e_2$ is the eccentricity of the hyperbola $x = \sqrt{2} \sec \theta, y = \tan \theta$,then $\frac{e_2^2}{e_1^2} = $

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