Let the eccentricity of the hyperbola frac{x^ | vedclass

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(C) For the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$,we have $a_e^2 = 4$ and $b_e^2 = 1$. The eccentricity $e_e$ is given by $b_e^2 = a_e^2(1 - e_e

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$(B, D)$

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(C) For the ellipse $\frac{x^2}{4} + \frac{y^2}{1} = 1$,we have $a_e^2 = 4$ and $b_e^2 = 1$. The eccentricity $e_e$ is given by $b_e^2 = a_e^2(1 - e_e^2)$,so $1 = 4(1 - e_e^2)$,which gives $e_e = \frac{\sqrt{3}}{2}$.Given the eccentricity of the hyperbola $e_h = \frac{1}{e_e} = \frac{2}{\sqrt{3}}$.For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $e_h^2 = 1 + \frac{b^2}{a^2}$,so $\frac{4}{3} = 1 + \frac{b^2}{a^2}$,which implies $\frac{b^2}{a^2} = \frac{1}{3}$,or $a^2 = 3b^2$.The foci of the ellipse are $(\pm ae_e, 0) = (\pm 2 	imes \frac{\sqrt{3}}{2}, 0) = (\pm \sqrt{3}, 0)$.Since the hyperbola passes through $(\sqrt{3}, 0)$,we have $\frac{(\sqrt{3})^2}{a^2} - 0 = 1$,so $a^2 = 3$.Then $b^2 = \frac{a^2}{3} = 1$.The equation of the hyperbola is $\frac{x^2}{3} - \frac{y^2}{1} = 1$,which is $x^2 - 3y^2 = 3$.Thus,options $A$ and $D$ are incorrect (as $A$ has $b^2=2$ and $D$ is $x^2-3y^2=3$ which is correct,but $A$ is wrong). Wait,$x^2-3y^2=3$ is $D$. Let's recheck: $a^2=3, b^2=1$. Equation is $\frac{x^2}{3} - y^2 = 1 \implies x^2 - 3y^2 = 3$. So $D$ is correct.The focus of the hyperbola is $(ae_h, 0) = (\sqrt{3} 	imes \frac{2}{\sqrt{3}}, 0) = (2, 0)$. So $B$ is correct.The eccentricity $e_h = \frac{2}{\sqrt{3}} 
eq \sqrt{\frac{5}{3}}$. So $C$ is incorrect.Therefore,the correct options are $(B, D)$.

Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be the reciprocal of the eccentricity of the ellipse $x^2+4y^2=4$. If the hyperbola passes through a focus of the ellipse,then:

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