A) \[(\sqrt{13},0)\]
B) \[\left( \sqrt{\frac{13}{2}},\sqrt{6} \right)\]
C) \[\left( \frac{1}{2}\sqrt{13},\frac{\sqrt{3}}{2} \right)\]
D) \[\left( \frac{\sqrt{39}}{2},\sqrt{3} \right)\]
Correct Answer: C
Solution :
hyperbola is \[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{9}=1\] \[foci\left( \pm \sqrt{13},0 \right)\] \[e=\frac{\sqrt{13}}{2}\] \[{{e}_{1}}\times \frac{\sqrt{13}}{2}=\frac{1}{2}\] \[{{e}_{1}}=\frac{1}{\sqrt{13}}\] equation of ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\left( \sqrt{13},0 \right)\] \[{{a}^{2}}=13\] foci of ellipse \[\sqrt{{{a}^{2}}-{{b}^{2}}}=a.\frac{1}{\sqrt{13}}\] \[13-{{b}^{2}}=1\] \[{{b}^{2}}=12\] equation of ellipse\[\frac{{{x}^{2}}}{13}+\frac{{{y}^{2}}}{12}=1\]
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