A) \[(\pm \sqrt{11},0)\]
B) \[(0,\pm \sqrt{12})\]
C) \[(\pm \sqrt{12},0)\]
D) \[(\pm \sqrt{13},0)\]
Correct Answer: D
Solution :
Given, hyperbola is\[4{{x}^{2}}-9{{y}^{2}}-36=0\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{4}=1\] Here, \[{{a}^{2}}=9,\text{ }{{\text{b}}^{2}}=4\] \[\Rightarrow \] \[a=3,\text{ }b=2\] \[\therefore \] \[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}\] \[=\sqrt{1+\frac{4}{9}}=\frac{\sqrt{13}}{3}\] Hence, focus of hyperbola\[=(\pm \text{ }ae,0)\] \[=\left( \pm 3.\frac{\sqrt{13}}{3},0 \right)\] \[=(\pm \sqrt{13},0)\]
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