An ellipse intersects the hyperbola 2x^2 - 2y | vedclass

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(D) Given hyperbola: $2x^2 - 2y^2 = 1 \Rightarrow \frac{x^2}{1/2} - \frac{y^2}{1/2} = 1$.Here,$a^2 = 1/2, b^2 = 1/2$. Eccentricity $e_h = \sqrt{1 + \f

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$(A, B)$

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(D) Given hyperbola: $2x^2 - 2y^2 = 1 \Rightarrow \frac{x^2}{1/2} - \frac{y^2}{1/2} = 1$.Here,$a^2 = 1/2, b^2 = 1/2$. Eccentricity $e_h = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2}$.Eccentricity of ellipse $e = \frac{1}{e_h} = \frac{1}{\sqrt{2}}$.Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Since $e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{2}$,we have $\frac{b^2}{a^2} = \frac{1}{2} \Rightarrow a^2 = 2b^2$.Equation of ellipse: $x^2 + 2y^2 = 2b^2$.For orthogonal intersection,the product of slopes of tangents at the point of intersection $(x_0, y_0)$ is $-1$.For hyperbola: $4x - 4y y' = 0 \Rightarrow y' = \frac{x}{y}$.For ellipse: $2x + 4y y' = 0 \Rightarrow y' = -\frac{x}{2y}$.Product: $(\frac{x_0}{y_0})(-\frac{x_0}{2y_0}) = -1 \Rightarrow x_0^2 = 2y_0^2$.Substitute $x_0^2 = 2y_0^2$ into the hyperbola equation: $2(2y_0^2) - 2y_0^2 = 1$ $\Rightarrow 2y_0^2 = 1$ $\Rightarrow y_0^2 = 1/2$ and $x_0^2 = 1$.Substitute $(x_0^2, y_0^2) = (1, 1/2)$ into the ellipse equation: $1 + 2(1/2) = 2b^2$ $\Rightarrow 2b^2 = 2$ $\Rightarrow b^2 = 1$.Thus,$a^2 = 2(1) = 2$. The equation is $x^2 + 2y^2 = 2$.Foci are $(\pm ae, 0) = (\pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}, 0) = (\pm 1, 0)$.Therefore,options $(A)$ and $(B)$ are correct.

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