Answer:
Option B,D
Explanation:
Here , equation of ellipse
\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} and focus (\pm ae,0)
= ( \pm \sqrt{3},0)
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}....(i)
and hyperbola passes through (\pm \sqrt{3},0)
Now, \frac{3}{a^{2}}=1 \Rightarrow a^{2}=3 ....(ii)
from eqs. (i) and (ii) , we get
b^{2}=1 ......(iiii)
\therefore Equation of hyperbola is
\frac{x^{2}}{3}-\frac{y^{2}}{1}=1
Focus is (\pm ae,0)
Now, \left(\pm \sqrt{3}.\frac{2}{\sqrt{3}},0\right) \Rightarrow (\pm2,0)
hence , both options (b) and (d) are correct