An ellipse 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, then

Options

A. Equation of the ellipse is `x^(2)+2y^(2)=2` with foci at `(+-1,0)`
B. Equation of the ellipse is `x^(2)+2y^(2)=2` with foci at `(+-sqrt(2),0)`
C. Equation of the ellipse is `x^(2)+2y^(2)=4` with foci at `(+-1,0)`
D. Equation of the ellipse is `x^(2)+2y^(2)=4` with foci at `(+-sqrt(2),0)`

Explanation

To solve the problem step by step, we need to find the equation of the ellipse that intersects the given hyperbola orthogonally and has an eccentricity that is the reciprocal of the hyperbola's eccentricity.

### Step 1: Identify the hyperbola and its properties
The equation of the hyperbola is given as:
\[ 2x^2 - 2y^2 = 1 \]
We can rewrite it in standard form:
\[ \frac{x^2}{\frac{1}{2}} - \frac{y^2}{\frac{1}{2}} = 1 \]
From this, we can identify \( a^2 = \frac{1}{2} \) and \( b^2 = \frac{1}{2} \).
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