A) 18
B) 8
C) 1
D) 2
Correct Answer: A
Solution :
Given hyperbola is\[\frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\frac{{{K}^{2}}}{9}-\frac{4}{{{b}^{2}}}=1\] ?(1) Also, given\[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\frac{\sqrt{13}}{3}\] \[\Rightarrow \]\[\sqrt{1+\frac{{{b}^{2}}}{9}}=\frac{\sqrt{13}}{3}\Rightarrow 9+{{b}^{2}}=13\]\[\Rightarrow \]\[b=\pm 2\] Now, from \[e{{q}^{n}}(1),\]we have \[\frac{{{K}^{2}}}{9}-\frac{4}{4}=1\] \[(\because b=\pm 2)\]\[\Rightarrow \]\[{{K}^{2}}=18\]
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