A) \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{12}=1\]
B) \[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{12}=1\]
C) \[\frac{{{x}^{2}}}{12}+\frac{{{y}^{2}}}{4}=1\]
D) \[\frac{{{x}^{2}}}{12}-\frac{{{y}^{2}}}{4}=1\]
Correct Answer: B
Solution :
Here for given ellipse \[a=5,\ b=3,\ {{b}^{2}}={{a}^{2}}(1-{{e}^{2}})\]Þ \[e=\frac{4}{5}\] Therefore, focus is (? 4, 0), (4, 0). Given eccentricity of hyperbola = 2 \[a=\frac{ae}{e}=\frac{4}{2}=2\] and \[b=2\sqrt{(4-1)}=2\sqrt{3}\] Hence hyperbola is \[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{12}=1\].
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