A) \[\left( \sqrt{5},2\sqrt{2} \right)\]
B) \[\left( 5,2\sqrt{3} \right)\]
C) \[\left( 0,\,\,2 \right)\]
D) \[\left( \sqrt{10},\,\,2\sqrt{3} \right)\]
Correct Answer: C
Solution :
Ellipse\[\frac{{{x}^{2}}}{12}+\frac{{{y}^{2}}}{16}=1\] foci \[(0,\pm be)\] \[{{e}_{e}}=\sqrt{1-\frac{12}{16}}=\frac{1}{2}\]= for hyperbola \[{{h}_{h}}=2\] \[{{e}_{H}}=\frac{3}{2}\] equation \[\Rightarrow \,\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=-1\] \[{{e}_{H}}\frac{3}{2}=\sqrt{1+\frac{{{a}^{2}}}{{{b}^{2}}}}\] \[\Rightarrow \] \[\frac{9}{4}-1=\frac{{{a}^{2}}}{{{b}^{2}}}\] \[\frac{{{a}^{2}}}{{{b}^{2}}}=\frac{5}{4}\] \[\Rightarrow \] \[{{a}^{2}}=5\] \[\frac{{{x}^{2}}}{5}-\frac{{{y}^{2}}}{4}=-1\]You need to login to perform this action.
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