Mathematics - Hyperbola Question with Solution | TestHub

Here, equation of ellipse $$\begin{array}{rl}& \frac{x^2}{4}+\frac{y^2}{1}=1\\ & \Rightarrow e^2=1-\frac{b^2}{a^2}=1-\frac{1}{4}=\frac{3}{4}\\ & \therefore e=\frac{\sqrt{3}}{2}\text{ and focus }(\pm ae,0)\\ & =(\pm \sqrt{3},0)\\ & \end{array}$$ For hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, $e_1^2=1+\frac{b^2}{a^2}$ where, $e_1^2=\frac{1}{e^2}=\frac{4}{3}$ $$\Rightarrow 1+\frac{b^2}{a^2}=\frac{4}{3}\Rightarrow \frac{b^2}{a^2}=\frac{1}{3}$$ and hyperbola passes through $(\pm \sqrt{3},0)$. Now, $\frac{3}{a^2}=1\Rightarrow a^2=3$ 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 $(\pm ae,0)$. Now, $\left(\pm \sqrt{3}\cdot \frac{2}{\sqrt{3}},0\right)\Rightarrow (\pm 2,0)$ Hence, both options (b) and (d) are correct.