Mathematics - Hyperbola Question with Solution | TestHub

Given hyperbola is $$\frac{x^2}{9}-\frac{y^2}{b^2}=1$$ Since this passes through $(K,2)$, therefore $$\frac{K^2}{9}-\frac{4}{b^2}=1$$ Also, given $e=\sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{13}}{3}$ $$\begin{array}{rl}& \Rightarrow \sqrt{1+\frac{b^2}{9}}=\frac{\sqrt{13}}{3}\Rightarrow 9+b^2=13\\ & \Rightarrow b=\pm 2\end{array}$$ Now, from ${\mathrm{eq}}^{\mathrm{n}}$ (1), we have $$\begin{array}{rl}& \frac{K^2}{9}-\frac{4}{4}=1(\because b=\pm 2)\\ & \Rightarrow K^2=18\end{array}$$