An ellipse intersects the hyperbola 2 x^2-2 y^2=1 orthogonally. The...

Given, $2 x^2-2 y^2=1$ \Rightarrow \quad $\frac{x^2}{\left(\frac{1}{2}\right)}-\frac{y^2}{\left(\frac{1}{2}\right)}=$1 Eccentricity of hyperbola $=\sqrt{2}$ So, eccentricity of ellipse $=1 / \sqrt{2}$ Let equation of ellipse be $$\begin{aligned}& \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b) \Rightarrow \frac{1}{\sqrt{2}}=\sqrt{1-\frac{b^2}{a^2}}\\ & \Rightarrow \quad \frac{b^2}{a^2}=\frac{1}{2} \Rightarrowa^2=2 b² \\ & \therefore \quad x²+2 y²=2 b² \\ & \end{aligned}$$Let ellipse and hyperbola intersect at $A\left(\frac{1}{\sqrt{2}} \sec \theta, \frac{1}{\sqrt{2}} \tan \theta\right)$On differentiating Eq. (i), 4 x-4 y $\frac{d y}{d x}=$0 \Rightarrow $\frac{d y}{d x}=\frac{x}{y} \therefore \quad\left(\frac{d y}{d x}\right)_{\text{at } A}=\frac{\sec \theta}{\tan \theta}=\operatorname{cosec} \theta$On differentiating Eq. (ii), 2 x+4 y $\frac{d y}{d x}=$0 \therefore \quad$\left(\frac{d y}{d x}\right)_{\text{at } A}=-\frac{x}{2 y}=-\frac{1}{2} \operatorname{cosec} \theta$Since, ellipse and hyperbola are orthogonal. $$\begin{aligned}& \therefore \quad-\frac{1}{2} \operatorname{cosec}^2 \theta=-1 \\ & \Rightarrow \quad \operatorname{cosec}^2 \theta=2 \Rightarrow \theta=\pm \frac{\pi}{4}\\ & \therefore A\left(1, \frac{1}{\sqrt{2}}\right) \text{ or }\left(1,-\frac{1}{\sqrt{2}}\right) \end{aligned}$$From Eq. (i), $1+2\left(\frac{1}{\sqrt{2}}\right)^2=2 b^2\Rightarrow \quad b^2=1$ Equation of ellipse is $x^2+2 y^2=2$ Coordinate of foci (\pm a e, 0)=\left(\pm \sqrt{2} \cdot \frac{1}{\sqrt{2}},0\right)=($\pm$1,0) Hence, options (a) and (b) are correct. If major axis is along $Y$-axis, then $$\begin{aligned}& \frac{1}{\sqrt{2}}=\sqrt{1-\frac{a^2}{b^2}} \Rightarrowb^2=2 a² \\ & \therefore 2 x²+y²=2 a² \Rightarrow y^{\prime}=-\frac{2 x}{y}\\ & \Rightarrow \quad y^{\prime}\left(\frac{1}{\sqrt{2}} \sec \theta \frac{1}{\sqrt{2}} \tan \theta\right)=\frac{-2}{\sin \theta}\\ & \end{aligned}$$As ellipse and hyperbola are orthogonal. $$\begin{array}{ll} \therefore& -\frac{2}{\sin \theta} \cdot \operatorname{cosec} \theta=-1 \\ \Rightarrow & \operatorname{cosec}^2 \theta=1 \Rightarrow \theta=\pm \frac{\pi}{4}\\ \therefore & 2 x²+y²=2 a² \Rightarrow 2+\frac{1}{2}=2 a² \\ \Rightarrow & a²=\frac{5}{4} \Rightarrow2 x²+y²=\frac{5}{2} \end{array}$$Corresponding foci are $(0, \pm 1)$.