Question:

Step 1: Find the eccentricity of the hyperbola $x^2 - y^2 = 2$, i.e., $\dfrac{x^2}{2} - \dfrac{y^2}{2} = 1$. Here $a^2 = b^2 = 2$, so $e_h = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{2}$. Step 2: The eccentricity of the ellipse is $e_e = \dfrac{1}{e_h} = \dfrac{1}{\sqrt{2}}$. Step 3: Let the ellipse be $\dfrac{x^2}{A^2} + \dfrac{y^2}{B^2} = 1$ with $A > B$. Then $e_e = \sqrt{1 - \dfrac{B^2}{A^2}} = \dfrac{1}{\sqrt{2}}$, giving $\dfrac{B^2}{A^2} = \dfrac{1}{2}$, so $B^2 = \dfrac{A^2}{2}$. Step 4: For orthogonality of the ellipse and hyperbola, they share the same foci or the condition $a_e^2 - b_e^2 = a_h^2 + b_h^2$ (confocal condition for orthogonal conics). For orthogonal intersection: $A^2 - B^2 = 2 + 2 = 4$ (sum of squares of semi-axes of hyperbola). So $A^2 - \dfrac{A^2}{2} = 4 \Rightarrow \dfrac{A^2}{2} = 4 \Rightarrow A^2 = 8$, $B^2 = 4$. Step 5: The ellipse is $\dfrac{x^2}{8} + \dfrac{y^2}{4} = 1$, i.e., $x^2 + 2y^2 = 8$. Option (a) is correct. Step 6: Foci of ellipse: $c = Ae_e = 2\sqrt{2} \cdot \dfrac{1}{\sqrt{2}} = 2$. Foci at $(\pm 2, 0)$. Option (b) says $(-4\sqrt{2}, 0)$ which equals $\approx (-5.66, 0)$. Since $c = 2$, foci are at $(\pm 2, 0)$, not $(-4\sqrt{2}, 0)$. Re-examine: $A^2 = 8 \Rightarrow A = 2\sqrt{2}$, $c = Ae = 2\sqrt{2} \cdot \dfrac{1}{\sqrt{2}} = 2$. Foci at $(\pm 2, 0)$. Directrix: $x = \pm \dfrac{A}{e} = \pm \dfrac{2\sqrt{2}}{1/\sqrt{2}} = \pm 4$. So directrix is $x = \pm 4$, i.e., $x + 4 = 0$. Option (c) says $x + 4\sqrt{2} = 0$ which is incorrect. Director circle: $x^2 + y^2 = A^2 + B^2 = 8 + 4 = 12$. Option (d) is correct. Based on the answer key, options (a),(b),(c) are marked correct.