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

Let the eccentricity of the hyperbola x²/a² - y²/b² = 1 be reciprocal to that of the ellipse x² + 4y² = 4. If the hyperbola passes through a focus of the ellipse, then

$\text{Ellipse: } x^2 + 4y^2 = 4 \ \Rightarrow\ \dfrac{x^2}{4} + \dfrac{y^2}{1} = 1$

$\text{Hyperbola: } \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$

e_(hyp) = 1/eₑₗₗ, hyp passes through a focus of the ellipse

$\text{To find: the correct statement(s) about the hyperbola}$

$\text{(its equation, a focus, and eccentricity)}$

$\dfrac{x^2}{4} + \dfrac{y^2}{1} = 1 \ \Rightarrow\ a^2 = 4,\ b^2 = 1$

eₑₗₗ = √(1 - b²/a²) = √(1 - 1/4) = √3/2

$c = \sqrt{a^2 - b^2} = \sqrt{4 - 1} = \sqrt3$

$\text{Foci of the ellipse: } (\pm\sqrt3,\ 0)$

e_(hyp) = 1/eₑₗₗ = 2/√3 e_(hyp)² = 1 + b²/a² = 4/3

$1 + \dfrac{b^2}{a^2} = \dfrac{4}{3} \ \Rightarrow\ \dfrac{b^2}{a^2} = \dfrac13$

$\therefore a^2 = 3b^2$

$\dfrac{(\sqrt3)^2}{a^2} - 0 = 1 \ \Rightarrow\ \dfrac{3}{a^2} = 1$

$\therefore a^2 = 3$

$b^2 = 1$

$\dfrac{x^2}{3} - \dfrac{y^2}{1} = 1 \ \Leftrightarrow\ x^2 - 3y^2 = 3$

$c = \sqrt{a^2 + b^2} = \sqrt{3 + 1} = 2$

$\text{Foci of the hyperbola: } (\pm 2,\ 0)$

e_(hyp) = 2/√3 = √(4/3) ≠ √(5/3)

$(1)\ \times \qquad (2)\ \checkmark \qquad (3)\ \times \qquad (4)\ \checkmark$

$\boxed{\dfrac{x^2}{3} - y^2 = 1 \ \Leftrightarrow\ x^2 - 3y^2 = 3}$

$\boxed{\text{Correct options: } (2)\ \text{and}\ (4)}$