For the hyperbola H : x^{2} - y^{2} = 1 and t | vedclass

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(D) For hyperbola $H: x^{2} - y^{2} = 1$,$e_{H} = \sqrt{1 + 1} = \sqrt{2}$.Given $e_{E} = \frac{1}{e_{H}} = \frac{1}{\sqrt{2}}$.Since $e_{E}^{2} = 1 -

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$3$

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(D) For hyperbola $H: x^{2} - y^{2} = 1$,$e_{H} = \sqrt{1 + 1} = \sqrt{2}$.Given $e_{E} = \frac{1}{e_{H}} = \frac{1}{\sqrt{2}}$.Since $e_{E}^{2} = 1 - \frac{b^{2}}{a^{2}}$,we have $\frac{1}{2} = 1 - \frac{b^{2}}{a^{2}}$ $\Rightarrow \frac{b^{2}}{a^{2}} = \frac{1}{2}$ $\Rightarrow a^{2} = 2b^{2}$.The line $y = mx + K$ is a tangent to $H: x^{2} - y^{2} = 1$ if $K^{2} = a_{H}^{2}m^{2} - b_{H}^{2} = 1(\frac{5}{2}) - 1 = \frac{3}{2}$.The line $y = mx + K$ is a tangent to $E: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ if $K^{2} = a^{2}m^{2} + b^{2}$.Equating $K^{2} = \frac{3}{2} = a^{2}(\frac{5}{2}) + b^{2}$.Substituting $a^{2} = 2b^{2}$,we get $\frac{3}{2} = (2b^{2})(\frac{5}{2}) + b^{2} = 5b^{2} + b^{2} = 6b^{2}$.Thus,$b^{2} = \frac{3}{12} = \frac{1}{4}$ and $a^{2} = 2(\frac{1}{4}) = \frac{1}{2}$.Therefore,$4(a^{2} + b^{2}) = 4(\frac{1}{2} + \frac{1}{4}) = 4(\frac{3}{4}) = 3$.

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