Let `frac{x^2}{a^2}+ frac{y^2}{b^2}=1`, ` a gt 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

Options

A. 3
B. `7/2`
C. `3/2`
D. `5/2`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Understand the given information
We have an ellipse defined by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) where \(a > b\). The eccentricity \(e\) is given as \(\frac{1}{\sqrt{2}}\) and the length of the latus rectum (LR) is given as \(\sqrt{14}\).

### Step 2: Use the formula for the length of the latus rectum
The formula for the length of the latus rectum of an ellipse is given by:
\[ LR = \frac{2b^2}{a} \]
Setting this equal to \(\sqrt{14}\):
\[ \frac{2b^2}{a} = \sqrt{14} \]
From this, we can express \(b^2\) in terms of \(a\):
\[ b^2 = \frac{a \sqrt{14}}{2} \]

### Step 3: Use the formula for eccentricity
The formula for the eccentricity of an ellipse is given by:
\[ e^2 = 1 - \frac{b^2}{a^2} \]
Substituting the value of \(e^2 = \frac{1}{2}\) into the equation:
\[ \frac{1}{2} = 1 - \frac{b^2}{a^2} \]
Rearranging gives:
\[ \frac{b^2}{a^2} = 1 - \frac{1}{2} = \frac{1}{2} \]
Thus, we have:
\[ b^2 = \frac{a^2}{2} \]

### Step 4: Equate the two expressions for \(b^2\)
We have two expressions for \(b^2\):
1. \(b^2 = \frac{a \sqrt{14}}{2}\)
2. \(b^2 = \frac{a^2}{2}\)

Setting these equal to each other:
\[ \frac{a \sqrt{14}}{2} = \frac{a^2}{2} \]
Multiplying both sides by 2:
\[ a \sqrt{14} = a^2 \]
Assuming \(a \neq 0\), we can divide by \(a\):
\[ \sqrt{14} = a \]

### Step 5: Calculate \(b^2\)
Substituting \(a = \sqrt{14}\) back into the equation for \(b^2\):
\[ b^2 = \frac{(\sqrt{14})^2}{2} = \frac{14}{2} = 7 \]

### Step 6: Write the equation of the hyperbola
Now we can write the equation of the hyperbola:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \implies \frac{x^2}{14} - \frac{y^2}{7} = 1 \]

### Step 7: Find the square of the eccentricity of the hyperbola
The eccentricity \(E\) of a hyperbola is given by:
\[ E^2 = 1 + \frac{b^2}{a^2} \]
Substituting \(b^2 = 7\) and \(a^2 = 14\):
\[ E^2 = 1 + \frac{7}{14} = 1 + \frac{1}{2} = \frac{3}{2} \]

### Final Answer
Thus, the square of the eccentricity of the hyperbola is:
\[ \boxed{\frac{3}{2}} \]