A) \[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{12}=1\]
B) \[\frac{{{x}^{2}}}{12}-\frac{{{y}^{2}}}{4}=1\]
C) \[\frac{{{y}^{2}}}{4}-\frac{{{x}^{2}}}{12}=1\]
D) \[\frac{{{y}^{2}}}{12}-\frac{{{x}^{2}}}{4}=1\]
Correct Answer: C
Solution :
Foci \[(0,\pm 4)\]\[\equiv (0,\pm \,be)\]Þ \[be=4\] Vertices \[(0,\pm 2)\equiv (0,\pm b)\Rightarrow b=2\Rightarrow a=2\sqrt{3}\] Hence equation is \[\frac{-{{x}^{2}}}{{{(2\sqrt{3})}^{2}}}+\frac{{{y}^{2}}}{{{(2)}^{2}}}=1\]or\[\frac{{{y}^{2}}}{4}-\frac{{{x}^{2}}}{12}=1\].
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