"Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \mathrm{a}\mathrm{b}$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latusrectum 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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Given the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a > b$, the eccentricity $ e $ is given by the formula:

$ e = \sqrt{1 - \left(\frac{b}{a}\right)^2} $

It is provided that the eccentricity $ e $ is $ \frac{1}{\sqrt{2}} $ (given), so we can equate the two expressions for eccentricity:

$ \frac{1}{\sqrt{2}} = \sqrt{1 - \left(\frac{b}{a}\right)^2} $

Squaring both sides to eliminate the square root gives:

$ \frac{1}{2} = 1 - \left(\frac{b}{a}\right)^2 $

$ \left(\frac{b}{a}\right)^2 = 1 - \frac{1}{2} $

$ \left(\frac{b}{a}\right)^2 = \frac{1}{2} $

Taking the square root on both sides:

$ \frac{b}{a} = \frac{1}{\sqrt{2}} $

$ a = b\sqrt{2} $

Now, for the ellipse, the length of the latus rectum is given by the formula:

$ \text{Length of Latus Rectum (L)} = \frac{2b^2}{a} $

It's provided that the length of the latus rectum $ L $ is $ \sqrt{14} $, so substitute the known values to find $ b $:

$ \sqrt{14} = \frac{2b^2}{b\sqrt{2}} = \frac{2b}{\sqrt{2}} $

$ b\sqrt{2} = \sqrt{14} $

$ b^2 = \frac{14}{2} $

$ b^2 = 7 $

And since $ a = b\sqrt{2} $, we can find $ a^2 $:

$ a^2 = (b\sqrt{2})^2 $

$ a^2 = 7 \cdot 2 $

$ a^2 = 14 $

Now we have an ellipse with $ a^2 = 14 $ and $ b^2 = 7 $. The equation of a hyperbola similar to the given ellipse but with the terms subtracted is:

$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $

For the hyperbola, the square of the eccentricity $ e' $ is given by:

$ (e')^2 = 1 + \frac{b^2}{a^2} $

Substitute the values we've found for $ a^2 $ and $ b^2 $ into the formula for the square of the hyperbola's eccentricity:

$ (e')^2 = 1 + \frac{b^2}{a^2} $

$ (e')^2 = 1 + \frac{7}{14} $

$ (e')^2 = 1 + \frac{1}{2} $

$ (e')^2 = \frac{3}{2} $

Therefore, the square of the eccentricity of the hyperbola is $ \frac{3}{2} $, which corresponds to option C.

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Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \mathrm{a}>\mathrm{b}$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latusrectum 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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Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \mathrm{a}>\mathrm{b}$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latusrectum 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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