A) 192
B) 64
C) 16
D) 32
E) 128
Correct Answer: A
Solution :
Eccentricity of ellipse\[=\sqrt{1-\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1-\frac{3}{4}}\] \[=\frac{1}{2}\] \[\therefore \]Eccentricity of hyperbola\[=2\] \[\therefore \] \[2=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1+\frac{{{b}^{2}}}{64}}\] \[\Rightarrow \] \[4=1+\frac{{{b}^{2}}}{64}\Rightarrow 3\times 64={{b}^{2}}\] \[\Rightarrow \] \[192={{b}^{2}}\]You need to login to perform this action.
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