Let frac{x^2}{a^2} + frac{y^2}{b^2} = 1, a | vedclass

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(C) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the eccentricity $e$ is given by $e^2 = 1 - \frac{b^2}{a^2}$.Given $e = \frac{1}{\sqrt{2}}

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$3/2$

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(C) For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,the eccentricity $e$ is given by $e^2 = 1 - \frac{b^2}{a^2}$.Given $e = \frac{1}{\sqrt{2}}$,so $e^2 = \frac{1}{2}$.Thus,$1 - \frac{b^2}{a^2} = \frac{1}{2} \Rightarrow \frac{b^2}{a^2} = \frac{1}{2}$.The length of the latus rectum is $\frac{2b^2}{a} = \sqrt{14}$.From $\frac{b^2}{a^2} = \frac{1}{2}$,we have $b^2 = \frac{a^2}{2}$.Substituting this into the latus rectum formula: $\frac{2(a^2/2)}{a} = \sqrt{14} \Rightarrow a = \sqrt{14}$.Then $b^2 = \frac{14}{2} = 7$.For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,the eccentricity $e_H$ is given by $e_H^2 = 1 + \frac{b^2}{a^2}$.Substituting $\frac{b^2}{a^2} = \frac{1}{2}$,we get $e_H^2 = 1 + \frac{1}{2} = \frac{3}{2}$.

Let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ be an ellipse,whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is:

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