An elllipse intersects the hyperbola `2x^(2) - 2y^(2) =1` orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes and the equation of the ellipse is `x^(2) + ky^(2) =k` then the value of k is ______ .

Answer

To solve the problem, we need to find the value of \( k \) for the ellipse given that it intersects the hyperbola orthogonally and the eccentricity of the ellipse is the reciprocal of that of the hyperbola.

### Step-by-Step Solution:

1. **Identify the equations**:
The hyperbola is given by:
\[ 2x^2 - 2y^2 = 1 \]
We can rewrite this as:
\[ \frac{x^2}{\frac{1}{2}} - \frac{y^2}{\frac{1}{2}} = 1 \]
This shows that \( a^2 = \frac{1}{2} \) and \( b^2 = \frac{1}{2} \) for the hyperbola.

2. **Find the eccentricity of the hyperbola**:
The eccentricity \( e \) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{\frac{1}{2}}{\frac{1}{2}}} = \sqrt{2} \]

3. **Set up the ellipse**:
The equation of the ellipse is given as:
\[ x^2 + ky^2 = k \]
We can rewrite this in standard form:
\[ \frac{x^2}{k} + \frac{y^2}{\frac{k}{k}} = 1 \]
This indicates that \( a^2 = k \) and \( b^2 = \frac{k}{k} = 1 \).

4. **Find the eccentricity of the ellipse**:
The eccentricity \( e' \) of the ellipse is given by:
\[ e' = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{k}} \]

5. **Use the relationship between eccentricities**:
According to the problem, the eccentricity of the ellipse is the reciprocal of that of the hyperbola:
\[ e' = \frac{1}{e} = \frac{1}{\sqrt{2}} \]
Therefore, we have:
\[ \sqrt{1 - \frac{1}{k}} = \frac{1}{\sqrt{2}} \]

6. **Square both sides**:
Squaring both sides gives:
\[ 1 - \frac{1}{k} = \frac{1}{2} \]

7. **Solve for \( k \)**:
Rearranging the equation:
\[ \frac{1}{k} = 1 - \frac{1}{2} = \frac{1}{2} \]
Thus, taking the reciprocal:
\[ k = 2 \]

### Final Answer:
The value of \( k \) is \( \boxed{2} \).