Let the eccentricity of the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 be...

Here, equation of ellipse $$\begin{aligned}& \frac{x^2}{4}+\frac{y^2}{1}=1 \\ & \Rightarrow \quad e²=1-\frac{b^2}{a^2}=1-\frac{1}{4}=\frac{3}{4}\\ & \therefore \quad e=\frac{\sqrt{3}}{2} \text{\text{ and } focus }(\pma e, 0) \\ & =(\pm \sqrt{3},0) \\ & \end{aligned}$$ For hyperbola\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,$e_1^2=1+\frac{b^2}{a^2}$where,$e_1^2=\frac{1}{e^2}=\frac{4}{3}$\Rightarrow \quad 1+$\frac{b^2}{a^2}=\frac{4}{3} \Rightarrow \frac{b^2}{a^2}=\frac{1}{3}$and hyperbola passes through$(\pm \sqrt{3}, 0)$. Now,$\quad \frac{3}{a^2}=1 \Rightarrow a^2=3$From Eqs. (i) and (ii), we get$b^2=1\therefore$Equation of hyperbola is$\frac{x^2}{3}-\frac{y^2}{1}=$1 Focus is$(\pm a e, 0)$. Now,$\quad\left(\pm \sqrt{3} \cdot \frac{2}{\sqrt{3}}, 0\right) \Rightarrow(\pm 2,0) Hence, both options (b) and (d) are correct.