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For the hyperbola $H : x^{2} - y^{2} = 1$ and the ellipse
$E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b > 0, l e t t h e (1) e c c e n t r i c i t y o f E b e r e c i p r o c a l o f t h e e c c e n t r i c i t y o f H, a n d (2) t h e l i n e y = \sqrt{\frac{5}{2}} x + K b e a c o m m o n \tan g e n t o f E a n d H . T h e n 4 (a^{2} + b^{2}) i s e q u a l t o____________.$

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answer is 3.

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Detailed Solution

$e_{E} = \sqrt{1 - \frac{b^{2}}{a^{2}},} e_{H} = \sqrt{2} I f \Rightarrow e_{E} = \frac{1}{e_{H}} \Rightarrow \frac{a^{2} - b^{2}}{a^{2}} = \frac{1}{2} \Rightarrow 2 a^{2} - 2 b^{2} = a^{2} ∴ a^{2} = 2 b^{2} a n d y = \sqrt{\frac{5}{2}} x + k i s \tan g e n t t o e l l i p s e a n d h y p e r b o l a t h e n K^{2} = a^{2} \times \frac{5}{2} + b^{2} = \frac{3}{2} \Rightarrow 6 b^{2} = \frac{3}{2} \Rightarrow b^{2} = \frac{1}{4} a n d a^{2} = \frac{1}{2} ∴ 4 (a^{2} + b^{2}) = 3$

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For the hyperbola H : x2-y2=1 and the ellipseE : x2a2+y2b2=1, a>b>0, let the (1) eccentricity of E be reciprocal of the eccentricity of H, and (2) the line y=52x+K be a common tangent of E and H. Then 4a2+b2 is equal to ____________.