Hyperbola with Eccentricity Reciprocal to Ellipse

To find the correct statements, we analyze the given information step by step:

1. Eccentricity of the Ellipse:
   The equation of the given ellipse is:
   $x^2 + 4y^2 = 4 \implies \frac{x^2}{4} + \frac{y^2}{1} = 1$
   Here, the semi-major axis is $a_E = 2$ and the semi-minor axis is $b_E = 1$.
   The eccentricity $e_E$ of the ellipse is given by:
   $e_E = \sqrt{1 - \frac{b_E^2}{a_E^2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$

2. Eccentricity of the Hyperbola:
   The eccentricity $e_H$ of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is the reciprocal of $e_E$:
   $e_H = \frac{1}{e_E} = \frac{2}{\sqrt{3}}$
   Therefore, option (C) is incorrect.

3. Foci of the Ellipse:
   The foci of the ellipse lie on the x-axis at:
   $(\pm a_E e_E, 0) = \left(\pm 2 \cdot \frac{\sqrt{3}}{2}, 0\right) = (\pm \sqrt{3}, 0)$

4. Equation of the Hyperbola:
   The hyperbola passes through a focus of the ellipse, so the point $(\sqrt{3}, 0)$ lies on the hyperbola:
   $\frac{(\sqrt{3})^2}{a^2} - \frac{0^2}{b^2} = 1 \implies \frac{3}{a^2} = 1 \implies a^2 = 3$

   Using the eccentricity relation for the hyperbola:
   $e_H^2 = 1 + \frac{b^2}{a^2}$
   Substituting $e_H = \frac{2}{\sqrt{3}}$ and $a^2 = 3$:
   $\left(\frac{2}{\sqrt{3}}\right)^2 = 1 + \frac{b^2}{3} \implies \frac{4}{3} = 1 + \frac{b^2}{3} \implies b^2 = 1$

   Thus, the equation of the hyperbola is:
   $\frac{x^2}{3} - \frac{y^2}{1} = 1 \implies x^2 - 3y^2 = 3$
   Therefore, option (A) is incorrect and option (D) is correct.

5. Foci of the Hyperbola:
   The foci of the hyperbola are located at $(\pm a e_H, 0)$:
   $a e_H = \sqrt{3} \times \frac{2}{\sqrt{3}} = 2$
   Hence, the foci of the hyperbola are $(\pm 2, 0)$, which means $(2, 0)$ is a focus of the hyperbola.
   Therefore, option (B) is correct.

Correct Options: B, D