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

An ellipse intersects the hyperbola 2x² - 2y² = 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 $\text{Hyperbola: } 2x^{2} - 2y^{2} = 1 \Rightarrow \dfrac{x^{2}}{1/2} - \dfrac{y^{2}}{1/2} = 1$ $\text{Ellipse: axes along coordinates; } e_{\text{ellipse}} = \dfrac{1}{e_{\text{hyperbola}}}$ $\text{Ellipse and hyperbola intersect orthogonally}$ $\text{Check which of (1) – (4) is/are correct}$ $2x^{2} - 2y^{2} = 1 \Rightarrow \dfrac{x^{2}}{1/2} - \dfrac{y^{2}}{1/2} = 1$ $a^{2} = b^{2} = \dfrac{1}{2},\quad e_{h} = \sqrt{1 + \dfrac{b^{2}}{a^{2}}} = \sqrt{2}$ $e_{e} = \dfrac{1}{e_{h}} = \dfrac{1}{\sqrt{2}}$ $\dfrac{x^{2}}{A^{2}} + \dfrac{y^{2}}{B^{2}} = 1,\quad e_{e} = \sqrt{1 - \dfrac{B^{2}}{A^{2}}} = \dfrac{1}{\sqrt{2}}$ $\Rightarrow \dfrac{B^{2}}{A^{2}} = \dfrac{1}{2} \Rightarrow A^{2} = 2B^{2}$ $\text{Orthogonal intersection: tangents } \perp$ $\Rightarrow \text{normals } \perp \Rightarrow \nabla F \cdot \nabla G = 0$ $F = 2x^{2} - 2y^{2} - 1,\quad \nabla F = (4x,\ -4y)$ $G = x^{2} + 2y^{2} - 2B^{2},\quad \nabla G = (2x,\ 4y)$ $(4x)(2x) + (-4y)(4y) = 0$ $8x^{2} - 16y^{2} = 0 \Rightarrow x^{2} = 2y^{2}$ $2x^{2} - 2y^{2} = 1,\quad x^{2} = 2y^{2}$ $2(2y^{2}) - 2y^{2} = 1 \Rightarrow 2y^{2} = 1 \Rightarrow y^{2} = \dfrac{1}{2}$ $x^{2} = 1$ $x^{2} + 2y^{2} = 2B^{2} \Rightarrow 1 + 2\cdot\dfrac{1}{2} = 2B^{2}$ $2 = 2B^{2} \Rightarrow B^{2} = 1,\quad A^{2} = 2$ $\text{Ellipse: } x^{2} + 2y^{2} = 2$ $(1)\ x^{2} + 2y^{2} = 2 \quad \checkmark$ $(3)\ x^{2} + 2y^{2} = 4 \quad ✗$ $c^{2} = A^{2} - B^{2} = 2 - 1 = 1 \Rightarrow c = 1$ $\text{Foci: } (\pm 1, 0)$ $(2)\ (\pm 1, 0) \quad \checkmark$ $(4)\ (\pm\sqrt{2}, 0) \quad ✗$ $\boxed{\text{Correct: (1) and (2)}}$