Question:

<p><strong>Solution:</strong></p><p><strong>Step 1:</strong> For the hyperbola \(2x^2 - 2y^2 = 1\) or \(\frac{x^2}{1/2} - \frac{y^2}{1/2} = 1\), we have \(a^2 = b^2 = \frac{1}{2}\), so \(e_h = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{2}\).</p><p><strong>Step 2:</strong> The eccentricity of the ellipse is \(e_e = \frac{1}{\sqrt{2}}\).</p><p><strong>Step 3:</strong> For orthogonal intersection, the tangents at intersection point are perpendicular. This gives the constraint \(a_e^2 + b_e^2 = a_h^2 + b_h^2\).</p><p><strong>Step 4:</strong> With \(e_e = \frac{1}{\sqrt{2}}\), we have \(b_e^2 = a_e^2(1 - e_e^2) = \frac{a_e^2}{2}\). Combined with orthogonality and solving, we get \(a_e^2 = 2\), \(b_e^2 = 1\).</p><p><strong>Step 5:</strong> The ellipse equation is \(\frac{x^2}{2} + y^2 = 1\) or \(x^2 + 2y^2 = 2\). The distance between foci: \(c = \sqrt{a_e^2 - b_e^2} = \sqrt{2 - 1} = 1\), so foci are \((\pm 1, 0)\).</p><p>∴ Answers are (a) and (b).</p>